id string | solution string | answer string | metadata dict | problem string | candidates list | relevance_scores list | relevance_scores_full null |
|---|---|---|---|---|---|---|---|
aops_363016 | $I_{k} = \int_{0}^{\frac{\pi}{4}} \tan^{k} x dx$
$ = \int_{0}^{\frac{\pi}{4}} \left(\tan^{2} x + 1 - 1\right) \tan^{k-2} x dx $
$= \boxed{\int_{0}^{\frac{\pi}{4}} \left(\tan^{2} x + 1\right) \tan^{k-2} x dx} - \int_{0}^{\frac{\pi}{4}} \tan^{k-2} x dx$
Applying integration by parts to the boxed and rearrange things a... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "For a non negative integer $n$, set t $I_n=\\int_0^{\\frac{\\pi}{4}} \\tan ^ n x\\ dx$ to answer the following questions:\n\n(1) Calculate $I_{n+2}+I_n.$\n\n(2) Evaluate the values of $I_1,\\ I_2$ and $I_3... | For a nonnegative integer \(n\), set
\[
I_n=\int_0^{\pi/4}\tan^n x\,dx.
\]
(1) Calculate \(I_{n+2}+I_n.\)
(2) Evaluate \(I_1,\ I_2,\) and \(I_3.\)
1978 Niigata University entrance exam | [
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numina_10206486 | To evaluate the integral \( \int_{-1}^{a^2} \frac{1}{x^2 + a^2} \, dx \) without using \( \tan^{-1} x \) or complex integrals, we can use a substitution method.
1. **Substitution**:
Let \( x = a \tan(\theta) \). Then, \( dx = a \sec^2(\theta) \, d\theta \).
2. **Change of Limits**:
When \( x = -1 \):
\[
-... | \frac{\pi}{2a} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "aops_forum"
} | Evaluate $ \int_{ \minus{} 1}^ {a^2} \frac {1}{x^2 \plus{} a^2}\ dx\ (a > 0).$
You may not use $ \tan ^{ \minus{} 1} x$ or Complex Integral here. | [
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aops_1158403 | As per question..
$2xy= {(\,x^2+y^3)\,}^{\frac{3}{2}}$
$\Rightarrow$ $2y+2x\frac{dy}{dx}=\frac{3}{2}{(\,x^2+y^3)\,}^{\frac{3}{2}-1}\times(\,2x+3y^2\frac{dy}{dx})\,$
$\Rightarrow$ $2y+2x\frac{dy}{dx}$$=3x{(\,x^2+y^3)\,}^{\frac{1}{2}}$ $ + \frac{9}{2}y^2{(\,x^2+y^3)\,}^{\frac{1}{2}}\frac{dx}{dy}$
$\Rightarrow$ $2y-3x{(\... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "find dy/dx given that 2xy=(x^2+y^3)^(3/2). Solve step by step\n",
"content_html": "find dy/dx given that 2xy=(x^2+y^3)^(3/2). Solve step by step",
"post_id": 5503981,
"post_number": 1,
... | Find \(\dfrac{dy}{dx}\) given that
\[
2xy = (x^2 + y^3)^{3/2}.
\]
Solve step by step. | [
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aops_2235305 | [b]Only the limit: [/b]
$\frac{{{a}_{n+1}}}{n}=\frac{1}{n}+\frac{n}{{{a}_{n}}}\Leftrightarrow \frac{n+1}{n}\frac{{{a}_{n+1}}}{n+1}=\frac{1}{n}+\frac{n}{{{a}_{n}}}\text{ and if }x=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{a}_{n}}}{n}\Rightarrow x=\frac{1}{x}\Rightarrow x=1>0$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Sequence $\\{a_n\\}$ defined by recurrence relation $a_{n+1} = 1+\\frac{n^2}{a_n}$. Given $a_1>1$, find the value of $\\lim\\limits_{n\\to\\infty} \\frac{a_n}{n}$ with proof.",
"content_html": "Seque... | Sequence \{a_n\} is defined by the recurrence
\[
a_{n+1}=1+\frac{n^2}{a_n},\qquad a_1>1.
\]
Find, with proof, the value of \(\displaystyle\lim_{n\to\infty}\frac{a_n}{n}\). | [
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aops_540033 | \begin{align*} \int \frac{t^2+1}{t^2(t^2-1)+1} \ \mathrm{d}t &=- \int \frac{\frac{-(t^2+1)}{(t^2-1)^2}}{\left(\frac{t}{t^2-1}\right)^2+1}\ \mathrm{d}t = \int\frac{1}{\left(\frac{t}{t^2-1}\right)^2+1} \mathrm{d}\left(\frac{t}{t^2-1}\right)^2 .\\ &=- \tan^{-1}\left(\frac{t}{t^2-1}\right)+C. \end{align*}
The antiderivativ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find:\n\n$\\int_{0}^{\\infty} \\frac{t^2+1}{t^4 -t^2 +1}dt$.\n\nI suppose residue teorem will kill it, but I'd like to see another approach :)",
"content_html": "Find:<br>\n<br>\n<span style=\"white-... | Find
\[
\int_{0}^{\infty} \frac{t^{2}+1}{t^{4}-t^{2}+1}\,dt.
\] | [
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aops_1935900 | $\underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1-\frac{1}{2n}-\frac{1}{3\sqrt{n}} \right)}^{n}}={{e}^{-\underset{n\to \infty }{\mathop{\lim }}\,n\left( \frac{1}{2n}+\frac{1}{3\sqrt{n}} \right)}}={{e}^{-2-\infty }}=0$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Determine whether sequence $$(1-\\frac{1}{2n}-\\frac{1}{3\\sqrt{n}})^n$$ converges(conditionally/absolutely) or diverges.",
"content_html": "Determine whether sequence <img src=\"//latex.artofproblem... | Determine whether the sequence
\[
\left(1-\frac{1}{2n}-\frac{1}{3\sqrt{n}}\right)^n
\]
converges (conditionally/absolutely) or diverges. | [
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aops_559541 | Integration by parts:
\[I_{n}=\int (\ln x)^{-n}\ dx = x \cdot (\ln x)^{-n} + n \int x \cdot (\ln x)^{-n-1}\cdot \frac{1}{x}\ dx\]
\[I_{n}=x \cdot (\ln x)^{-n} + n \cdot I_{n+1} \quad\quad\quad I_{n+1}= \frac{I_{n}-x \cdot (\ln x)^{-n}}{n}\]
Back to the question:
\[I=\int\frac{(\ln x)^{n}-n!}{(\ln x)^{n+1}}\ dx = \in... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the integral:\n$I=\\int\\frac{\\ln^n(x)-n!}{\\ln^{n+1}(x)}dx.$",
"content_html": "Find the integral:<br>\n<img src=\"//latex.artofproblemsolving.com/1/7/d/17d20b194d349fe34475cfdd3fac03fc0639bbc... | Find the integral:
\[
I=\int \frac{\ln^{n}(x)-n!}{\ln^{n+1}(x)}\,dx.
\] | [
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aops_1105143 | We have $\int \frac{dx}{a + b \cos x}=\int \frac{2dt}{a(1+t^2) + b(1-t^2)}=\int \frac{2dt}{(a-b)t^2+ a+b}$.
Take then $t=u\sqrt{\dfrac{a+b}{a-b}}$ if $a^2>b^2$ and $t=-u\sqrt{\dfrac{a+b}{a-b}}$ if $a^2<b^2$.
If $a=b$, we obtain $\int \frac{dx}{a + b \cos x}=\int \frac{dx}{2a\cos^2(x/2)}=C+\dfrac{tan(x/2)}{a}$.
If $a... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find $\\int \\frac{1}{a + b \\cos x}\\ dx$, distinguishing between the cases $a^{2} > b^{2}, a^{2} < b^{2}$ and $a^{2} = b^{2}$.",
"content_html": "Find <span style=\"white-space:nowrap;\"><img src=\... | Find
\[
\int \frac{1}{a+b\cos x}\,dx,
\]
distinguishing the cases \(a^{2}>b^{2}\), \(a^{2}<b^{2}\), and \(a^{2}=b^{2}\). | [
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aops_469860 | [quote="Moubinool"]
just find $\int_{0}^{1}\frac{\ln(1+x)}{1+x^2}$[/quote]
To evaluate this integral, use the substitution $t=\frac{1-x}{1+x}$, then:
\[\begin{aligned} \int_{0}^{1}\frac{\ln(1+x)}{1+x^2} &=\int_1 ^0\frac{\ln\left(1+\frac{1-t}{1+t}\right)}{1+\left(\frac{1-t}{1+t}\right)^2}\cdot -\frac{2}{(1+t)^2}dt\\
... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $\\left(a_n\\right)_{n\\ge 1}$ be a sequence defined by: $a_n=\\prod_{k=1}^n\\, \\left(1+\\frac kn\\right)^{\\sin\\frac n{n^2+k^2}}$ , $n\\ge 1$ . Evaluate: $\\lim_{n\\to\\infty}\\, a_n$ .",
"con... | Let \((a_n)_{n\ge 1}\) be the sequence defined by
\[
a_n=\prod_{k=1}^n\left(1+\frac{k}{n}\right)^{\sin\frac{n}{n^2+k^2}},\qquad n\ge 1.
\]
Evaluate \(\displaystyle\lim_{n\to\infty}a_n\). | [
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aops_387891 | [quote="kunny"]Evaluate
\[\int_1^{\sqrt{e}} \frac{x\{\ln (x+\sqrt{1+x^2})-1\}}{\{\ln (x+\sqrt{1+x^2})\}^2\sqrt{1+x^2}}dx\]
Own[/quote]
$\int_1^{\sqrt{e}} \frac{x\{\ln (x+\sqrt{1+x^2})-1\}}{\{\ln (x+\sqrt{1+x^2})\}^2\sqrt{1+x^2}}dx=\int_1^{\sqrt{e}}(\frac{\sqrt{x^2+1}}{ln(x+\sqrt{1+x^2}})'dx$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate \n\\[\\int_1^{\\sqrt{e}} \\frac{x\\{\\ln (x+\\sqrt{1+x^2})-1\\}}{\\{\\ln (x+\\sqrt{1+x^2})\\}^2\\sqrt{1+x^2}}dx\\]\n\nOwn",
"content_html": "Evaluate<br>\n<img src=\"//latex.artofproblemsolv... | Evaluate
\[
\int_{1}^{\sqrt{e}} \frac{x\big(\ln (x+\sqrt{1+x^2})-1\big)}{\big(\ln (x+\sqrt{1+x^2})\big)^2\sqrt{1+x^2}}\,dx.
\] | [
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0.4250... | null |
aops_163468 | [hide="Possible approach"]Start by making $ t =\tan x$, and so $ I =\int^{\infty}_{0}\frac{\ln^{n}(t)}{1+t^{2}}\,dt =\int^{\pi/2}_{0}[\ln(\tan x)]^{n}\,dx =\int^{\pi/2}_{0}[\ln(\sin x)-\ln(\cos x)]^{n}\,dx$
$ =\int^{\pi/2}_{0}\left[\sum_{k=0}^{n}(-1){}^{k}{}^{n}_{k}C(\ln\sin x)^{n-k}(\ln\cos x)^{k}\right]\,dx$
For $ ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Compute $ \\int_{0}^{\\infty}\\frac{\\ln^{n}(t)}{1\\plus{}t^{2}}\\, dt$ for all $ n\\in\\Bbb{N}$ .",
"content_html": "Compute <img src=\"//latex.artofproblemsolving.com/f/4/8/f4841d7ee5c6c1270d309111... | Compute
\[
\int_{0}^{\infty}\frac{\ln^{n}(t)}{1+t^{2}}\,dt
\]
for all \(n\in\mathbb{N}\). | [
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0.44466... | null |
aops_2078099 | [b]Solution:[/b]
$\underset{n\to \infty }{\mathop{\lim }}\,{{\left( \frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}}{\ln n} \right)}^{\ln n}}=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n}{\ln n} \right)}^{\ln n}}={{e}^{\underset{n\to \infty }{\mathop{\lim }}\... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Compute the limit \n$$\\lim _{n \\rightarrow \\infty} \\left( \\frac{1 + \\frac{1}{2} + \\frac{1}{3} + ... + \\frac{1}{n}}{\\ln n} \\right) ^{\\ln n}$$\n\nL'hopital...? Mascheroni constant...! ",
... | Compute the limit
\[
\lim_{n\to\infty}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}{\ln n}\right)^{\ln n}.
\] | [
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aops_35496 | $\displaystyle \prod_ {i=1} ^ {n} x_i ^ {a_i} \leq \prod_{i=1}^{n} a_i^{a_i} \Longleftrightarrow \prod_ {i=1} ^ {n} (\frac{x_i}{a_i}) ^ {a_i} \leq 1 \Longleftrightarrow \sum_ {i=1} ^ {n}a_i \ln (\frac{x_i}{a_i}) \leq 0$ :cool: Trivial, because of $ln(x)$ is concave :cool: | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Given $a_1 , a_2 , \\dots , a_n , x_1 , x_2 , \\dots , x_n \\in \\mathbb {R} ^ {+} $ prove that: \\[ \\displaystyle \\Pi _ {i=1} ^ {n} {x_i ^ {a_i}} \\leq \\Pi_{i=1}^{n} {a_i^{a_i}}\\]\r\nprovided $\\Sigma... | Given \(a_1,a_2,\dots,a_n,x_1,x_2,\dots,x_n\in\mathbb{R}^+\) with
\(\displaystyle\sum_{i=1}^n a_i=\sum_{i=1}^n x_i=1\),
prove that
\[
\prod_{i=1}^n x_i^{a_i}\le\prod_{i=1}^n a_i^{a_i}.
\] | [
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ours_9692 | For \(x=0\), an indeterminate form \(\frac{0}{0}\) arises. By l'Hôpital's rule, the limit in this case is given by
$$
\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{g^{\prime}(x)}
$$
Differentiating the numerator and denominator twice yields:
$$
\begin{aligned}
f^{\prime \prim... | -6 | {
"competition": "german_mo",
"dataset": "Ours",
"posts": null,
"source": "Loesungen_MaOlympiade_205_part2.md"
} | Determine the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{(1-x)^{6} \sin 3 x-\sin 3 x}{3 x^{2}}
$$ | [
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aops_16465 | yeah...just use l'Hpital to get that it equals $\frac{a_1+...+a_n}{a_1\log{a_1}+...+a_n\log{a_n}}$. since all numbers have to be positive, the numerator is not zero. if the denominator is zero, then there is no limit.
Peter | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$\\ds \\lim_{x\\rightarrow 1}\\frac{x^{a_1}+x^{a_2}+...x^{a_n}-n}{a_1^x+a_2^x+...+a_n^x-a_1-a_2-...-a_n}=?$",
"content_html": "<span class=\"aopscode-error aopscode-latex-error\">$\\ds \\lim_{x\\righ... | \[
\lim_{x\to 1}\frac{x^{a_1}+x^{a_2}+\dots+x^{a_n}-n}{a_1^x+a_2^x+\dots+a_n^x-(a_1+a_2+\dots+a_n)}.
\] | [
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aops_1576255 | [hide="oops"]
Note that by swapping $x\mapsto -x$ we get \begin{align*}2I &= \int_{-\infty}^\infty\frac{\sin^2(x+\pi/4) + \sin^2(x - \pi/4)}{e^{x^2}}\,dx \\&= \int_{-\infty}^\infty\frac{\sin^2(x+\pi/4) + \cos^2(x + \pi/4)}{e^{x^2}}\,dx = \int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt\pi,\end{align*} whence $I=\tfrac12\sqrt... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "[b]MIT Integration Bee, Qualifiers 2017/16.[/b] Compute $I = \\int_{-\\infty}^{\\infty} \\frac{\\sin^2\\left(x + \\pi/4\\right)}{e^{x^2}} \\, dx$.\n\n[hide=Solution]\nRecalling the Gaussian integral \\[\\i... | Compute
\[
I=\int_{-\infty}^{\infty}\frac{\sin^{2}\!\left(x+\frac{\pi}{4}\right)}{e^{x^{2}}}\,dx.
\] | [
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aops_1205755 | For any $-1<x<1$, $f(x)=f(x^2)$....that implies $f(x^2)=f(x^4)$....by Induction, $f(x)=f(x^k)$ for all $k=2^n$...now take limit of $n$ at infinity...as $f$ is continuous and $-1<x<1$, hence $f(x)=f(0)=1/2$...this is true also for $x=\frac{1}{4}$....so $f(1/4)=1/2$....
P.S. Edited...Thanks Chandrachur | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $f:(-1,1)\\to(-1,1)$ be continuous, $f(x)=f(x^2)$ for every $x$ and $f(0)=\\frac{1}{2}$. Then find the value of $f\\left(\\frac{1}{4}\\right)$ .",
"content_html": "Let <img src=\"//latex.artofpro... | Let \(f:(-1,1)\to(-1,1)\) be continuous, \(f(x)=f(x^2)\) for every \(x\), and \(f(0)=\tfrac{1}{2}\). Find \(f\!\left(\tfrac{1}{4}\right)\). | [
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aops_2333812 | [quote=alexheinis]Actually the integral converges for $-1<a<1$. I will use the, for me, standard integral, that $\int_0^\infty {{x^{b-1}dx}\over {x+1}}={\pi\over {\sin \pi b}}$ when $0<b<1$. If it's necessary I can post a file with a solution of that one with Complex Analysis. Now take $x=\sqrt{t}$, then we get ${\pi\... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$\\alpha$ is a constant that satisfies $0<\\alpha<1$. Evaluate $$\\int_0^\\infty \\frac{x^\\alpha}{1+x^2} dx$$",
"content_html": "<img src=\"//latex.artofproblemsolving.com/1/0/f/10f32377ac67d94f764f... | Let \(\alpha\) be a constant with \(0<\alpha<1\). Evaluate
\[
\int_0^\infty \frac{x^\alpha}{1+x^2}\,dx.
\] | [
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aops_400682 | $I = \int_0^x\sin^3{x}\;{dx} = \int_0^x \sin{x}\sin^2{x}\;{dx} = \int_0^x\sin{x}\left(1-\cos^2{x}\right)\;{dx}$. Let $t = \cos{x}$, then:
$I = -\int_1^{\cos{x}}(1-t^2)\;{dt} = -\bigg[t-\frac{1}{3}t^3\bigg]_1^{\cos{x}} = 1-\frac{1}{3}-\cos{x}+\frac{1}{3}\cos^3{x}$ and
$1-\frac{1}{3}-\cos{x}+\frac{1}{3}\cos^3{x} = \fra... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove that $\\int_{0}^x sin^{3} t dt=\\frac{2}{3} -\\frac{1}{3}(2+ sin^2 x)cos x$",
"content_html": "Prove that <img src=\"//latex.artofproblemsolving.com/0/0/d/00d08b5c875209ded0b1f2fe7ebb5162ac7461... | Prove that
\[
\int_{0}^{x} \sin^{3} t\,dt = \frac{2}{3}-\frac{1}{3}\bigl(2+\sin^{2}x\bigr)\cos x.
\] | [
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aops_459321 | Well, I think it would have something to do with
\[\displaystyle\int\left(\dfrac{\mathrm{d}x}{\sqrt{1+x^2}}\right)^3\]
which relates to the Pythagorean Theorem. Trig substitution would make this soooo much easier. No offence. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Compute \t$\\int\\frac{1}{(1+x^2)^\\frac{3}{2}}$ without using any subsitution..",
"content_html": "Compute \t<img src=\"//latex.artofproblemsolving.com/d/2/9/d29461e7557be12c71c95bf4565677551487315c... | Compute
\[
\int \frac{1}{(1+x^2)^{3/2}}\,dx.
\] | [
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aops_525890 | [quote="TocNgan"][quote="AdvitiyaBrijesh"]Use $\text{L'Hostpital}$ rule..[/quote]
Can you solve completely ?[/quote]
$I= \lim_{x\rightarrow 0}\frac{1+\sin 3x-\cos 3x}{1+\sin 4x-\cos 4x} $
Using L'Hospital Rule..
$I=\lim_{x\rightarrow 0}\frac{3\cos 3x+3\sin 3x}{4\cos 4x+4\sin 4x}$
So, $I=\frac{3}{4}$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Solve that $\\lim_{x\\rightarrow 0}\\frac{1+\\sin 3x- \\cos 3x}{1+ \\sin 4x - \\cos 4x}$",
"content_html": "Solve that <img src=\"//latex.artofproblemsolving.com/6/d/d/6dd90b86edef6f558b049c84b136e56... | Cleaned problem statement:
Solve
\[
\lim_{x\to 0}\frac{1+\sin 3x-\cos 3x}{1+\sin 4x-\cos 4x}.
\]
Solution:
As x→0, use Taylor expansions up to second order:
sin kx = kx - (k^3 x^3)/6 + o(x^3), \quad cos kx = 1 - (k^2 x^2)/2 + o(x^2).
Thus numerator = 1 + (3x) - \Big(1 - \frac{(3x)^2}{2}\Big) + o(x^2) = 3x + \frac{9... | [
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aops_170316 | For the second d.eq., $ y' \equal{} \frac{y^4 \plus{} 2xy^3 \minus{} 3x^2y^2 \minus{} 2x^3y}{2x^2y^2 \minus{} 2x^3y \minus{} 2x^4}$, let $ y \equal{} ux$, $ y' \equal{} u'x \plus{} u$, and so:
$ u'x \plus{} u \equal{} \frac{u^4x^4 \plus{} 2u^3x^4 \minus{} 3u^2x^4 \minus{} 2ux^4}{2u^2x^4 \minus{} 2ux^4 \minus{} 2x^4}... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Show that the following equation\r\n\r\n$ \\frac{dy}{dx}\\equal{}f(\\frac{y}{x})$\r\n\r\nbecomes separable after the substitution $ v\\equal{}\\frac{y}{x}$\r\n\r\n\r\n\r\nAnd use the technique from the abo... | Show that the following equation
\[
\frac{dy}{dx}=f\!\left(\frac{y}{x}\right)
\]
becomes separable after the substitution \(v=\dfrac{y}{x}\).
Use the substitution \(v=\dfrac{y}{x}\) to solve the following differential equations:
1. \(x^{2}+3xy+y^{2}-x^{2}\dfrac{dy}{dx}=0.\)
2. \(\dfrac{dy}{dx}=\dfrac{y^{4}+2xy^{3}-3... | [
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aops_2655225 | [b]A symylar solution:[/b]
We have $L=\underset{x\to 1}{\mathop{\lim }}\,\frac{\sin \pi x}{{{x}^{2}}-1}=-\underset{x\to 1}{\mathop{\lim }}\,\frac{1}{x+1}\underset{x\to 1}{\mathop{\lim }}\,\frac{\sin \pi x}{1-x}=-\frac{1}{2}\underset{x\to 1}{\mathop{\lim }}\,\frac{\sin \pi x}{1-x}$, and now let $1-x=y$, so $x=1-y\Righta... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate $\\lim_{x\\to 1}\\frac{\\sin{\\pi x}}{x^2-1}$ without using L'Hopital's rule.",
"content_html": "Evaluate <img src=\"//latex.artofproblemsolving.com/f/9/9/f99558d2fc8fc1cc9c5c2ae59d074504d86... | Evaluate
\[
\lim_{x\to 1}\frac{\sin(\pi x)}{x^2-1}
\]
without using L'Hôpital's rule. | [
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aops_24510 | We have Calculus - Analysis section for this.
You need to use the identity $2\cos ^22x=\cos 4x+1$, so your integral becomes
\[\int_0^{\pi/8}(\cos 4x+1)dx=\frac{1}{4}\sin 4x+x\big|_0^{\pi/8}=\frac{1}{4}+\frac{\pi}{8}.\] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Hi :) \r\n\r\nI'm new here so I hope I'm posting at the right place :lol: \r\n\r\n[tex]\\int_{0}^{\\frac{\\pi}{8}}(2cos^22x)dx[/tex]\r\n\r\nI'm having problem with getting the integrale.\r\n\r\nAny advice?... | Cleaned problem statement (LaTeX):
Compute the integral
\[
\int_{0}^{\frac{\pi}{8}} 2\cos^2(2x)\,dx.
\] | [
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aops_24603 | the function $f(x)=ln(sin(x))$ is defined and concave on $(0,\pi)$, so that
$\frac{ln(sin(\frac{\pi}{16}))+ln(sin(\frac{\pi}{8}))+ln(sin(\frac{9\pi}{16}))}{3}<ln(sin(\frac{1}{3}(\frac{\pi}{16}+\frac{\pi}{8}+\frac{9\pi}{16})))=ln(sin(\frac{\pi}{4}))$
The inequality required follows by exponentiation. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove the following inequality.\r\n\r\n\\[\\sin \\frac{\\pi}{4}>\\sqrt[3]{\\sin \\frac{\\pi}{16}\\sin \\frac{\\pi}{8}\\sin \\frac{9}{16}\\pi}\\]",
"content_html": "Prove the following inequality.<br>... | Prove the inequality
\[
\sin\frac{\pi}{4}>\sqrt[3]{\sin\frac{\pi}{16}\,\sin\frac{\pi}{8}\,\sin\frac{9\pi}{16}}.
\] | [
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aops_556989 | Seeing as there are no solutions posted I'll post a nudge. As JBL said, I suggest you attempt to do this problem by yourself after my nudge:
\[\left[f(g(x))\right]'=f'(g(x))g'(x)\]
In this case $f(x)=\ln x, g(x)=\text{sec}x$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "I'm probably being stupid here but\n\nFind $\\frac{dy}{dx} \\ln{\\sec{x}}$",
"content_html": "I'm probably being stupid here but<br>\n<br>\nFind <img src=\"//latex.artofproblemsolving.com/d/7/1/d71e8... | Find \(\dfrac{dy}{dx}\ln\sec x\). | [
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aops_1211659 | I've tried Abhinandan's sustutuion and got $\frac{\sqrt{\pi} \Gamma(\frac{n}{2})}{2 p^{n + \frac{1}{2}} \Gamma (\frac{n+1}{2})}$. Is this correct?
I have:$ \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{p} \sec^{2} z }{(p(1 + \tan^{2} z))^{n+1}}\ dz = \frac{1}{p^{n + \frac{1}{2}}} \int_{0}^{\frac{\pi}{2}} \cos^{n-1}z\ dz$
$= \... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate $\\int_{0}^{\\infty} \\frac{1}{(x^{2} +p)^{n+1}}\\ dx$.",
"content_html": "Evaluate <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolving.com/8/8/c/88c4818cde922cd313c51... | Evaluate
\[
\int_{0}^{\infty} \frac{1}{(x^{2}+p)^{n+1}}\,dx,
\]
where p>0 and n is a nonnegative integer. | [
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aops_318115 | put x=π - y and you will get I=-I means I=0 (I is our integral).
Ooops i see you want example. You can constract some ...if you have f(x)=-f(Π-x) then by integrating from 0 to π you can take result 0 | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "evaluate\r\n\r\n\r\nint 0 to pi (sin6x/sinx)\r\n\r\n\r\n\r\ncan i have some more examples of this type?",
"content_html": "evaluate<br>\n<br>\n<br>\nint 0 to pi (sin6x/sinx)<br>\n<br>\n<br>\n<br>\n... | Evaluate
\[
\int_{0}^{\pi}\frac{\sin 6x}{\sin x}\,dx.
\]
Can I have some more examples of this type? | [
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aops_278148 | [quote="guru_bhai"]If $ \lim_{x \to 0} \left(\frac {1 \plus{} cx}{1 \minus{} cx}\right)^{\frac {1}{x}} \equal{} 4$,then
$ \lim_{x \to 0} \left(\frac {1 \plus{} 2cx}{1 \minus{} 2cx}\right)^{\frac {1}{x}} \equal{} ?$[/quote]
use $ \lim_{y \to 0}{(1 \plus{} y)^{\frac {1}{y}}} \equal{} e$ we have
$ \lim_{x \to 0}{(1 ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $ \\lim_{x \\to 0} \\left(\\frac{1\\plus{}cx}{1\\minus{}cx}\\right)^{\\frac{1}{x}} \\equal{} 4$,then\r\n\r\n$ \\lim_{x \\to 0} \\left(\\frac{1\\plus{}2cx}{1\\minus{}2cx}\\right)^{\\frac{1}{x}} \\equal{... | If
\[
\lim_{x \to 0}\left(\frac{1+cx}{1-cx}\right)^{\frac{1}{x}}=4,
\]
then
\[
\lim_{x \to 0}\left(\frac{1+2cx}{1-2cx}\right)^{\frac{1}{x}}=?
\] | [
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aops_108660 | The function only grows at a logarithmic (slow) rate as $x\to1^{-}.$ This is a convergent integral. Don't be so quick to give up on the work you've done so far.
Hint: what is $\lim_{x\to1^{-}}(1-x)\ln(1-x)?$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the area enclosed by the curve\r\n\r\ny=$-log(1-x^{2})$ the x-axis and the line x=1",
"content_html": "Find the area enclosed by the curve<br>\n<br>\ny<span style=\"white-space:nowrap;\">=<img s... | Find the area enclosed by the curve
\[
y=-\log(1-x^{2}),
\]
the \(x\)-axis, and the line \(x=1\). | [
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aops_532015 | I'm assuming you want $\int \! \frac{x}{\log^2{x}} \, dx$.
In this case, make the sub $x = e^u$, $dx = e^u \, du$ and it becomes $\int \! \frac{e^{2u}}{u^2} \, du$.
Parts gives us $-\frac{e^{2u}}{u} + 2 \int \! \frac{e^{2u}}{u} \, du$ and this second integral is not elementary. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find : \\[\\int \\frac{x}{ \\ln^2(x)} \\ \\mathrm{d}x.\\]",
"content_html": "Find : <img src=\"//latex.artofproblemsolving.com/3/f/9/3f90817f44809b0763c15b36489bca96409237db.png\" class=\"latexcenter... | Find:
\[
\int \frac{x}{\ln^2(x)}\,dx.
\] | [
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numina_10032006 | Solution. Here $u=\cos x, y=u^{3}$. We find $y_{u}^{\prime}=3 u^{2}=3 \cos ^{2} x$, $u_{x}^{\prime}=(\cos x)^{\prime}=-\sin x$; hence $y_{x}^{\prime}=3 \cos ^{2} x(-\sin x)=-3 \cos ^{2} x \sin x$. | -3\cos^{2}x\sinx | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 327. Find $\left(\cos ^{3} x\right)^{\prime}$. | [
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aops_156650 | $ \textrm{Integral calculus}\subset \textrm{Calculus}$, and as such it belongs in the calculus forums. (You may notice the announcement at the top of this forum titled, "PLEASE do not post calculus problems in this forum!")
[hide="Solution"]$ \int\frac{x^{n}+x^{n-1}+\dots+x+1}{(x-1)^{n}}dx = \int\frac{x^{n+1}-1}{(x... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Can someone please calculate this integral?\r\n$ \\int\\frac{x^{n}+x^{n-1}+\\dots+x+1}{(x-1)^{n}}dx,\\hspace{0.2cm}x>1,$ and $ n\\in\\mathbb{N^{*}}.$\r\n\r\nMOD: Moved to the calculus forum.",
"conte... | Compute the integral
\[
\int \frac{x^{n}+x^{n-1}+\dots+x+1}{(x-1)^{n}}\,dx,\qquad x>1,\; n\in\mathbb{N}^*.
\] | [
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aops_3212198 | [quote=ddt297]We have:
\begin{align*}\left(\frac{x}{\cos x}\cdot \frac{1}{\cos x+x\sin x}\right)'&=\left(\frac{x}{\cos x}\right)'\left(\frac1{\cos x+x\sin x}\right)+\left(\frac{x}{\cos x}\right)\left(\frac1{\cos x+x\sin x}\right)'\\&=\left(\frac{\cos x+x\sin x}{\cos ^2x}\right)\left(\frac1{\cos x+x\sin x}\right)+\left(... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Calculate \n$$\\int_0^{\\frac\\pi 4}\\frac{x^2}{(\\cos x+x\\sin x)^2}$$",
"content_html": "Calculate<br>\n<img src=\"//latex.artofproblemsolving.com/b/5/5/b55aea455e8994f214f92300f8831cfe3e9df635.png... | Calculate
\[
\int_{0}^{\frac{\pi}{4}} \frac{x^{2}}{\bigl(\cos x + x\sin x\bigr)^{2}}\,dx.
\] | [
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0.7754722... | null |
aops_539084 | \[\int_1^x \ln t\, dt = \int_1^x \int_1^t \frac{ds}{s} \,dt= \int_1^x \frac{ds}{s}\int_s^x dt\]\[= \int_1^x (x-s)\frac{ds}{s}= x\int_1^x \frac{ds}{s} - \int_1^x ds = x\ln x - x +1.\] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find $\\int\\log x dx$ not by using method of by parts!! :wink:",
"content_html": "Find <img src=\"//latex.artofproblemsolving.com/f/1/5/f153a4efcc6d5e60684b85f11c62e5f08051770c.png\" class=\"latex\"... | Find \(\displaystyle \int \log x \, dx\). | [
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0.43545297... | null |
aops_499276 | First let's concern ourselves with the limit, as we can pass the limit to the inside if it exists. \[ \lim_{x \to 0} x^2 \csc\left(\frac{x}{2}\right)\cot(6x) = \lim_{x \to 0} \frac{x}{\sin\left(\frac{x}{2}\right)} \times \frac{x}{\sin(6x)} \times \cos(6x) = 2 \cdot \frac{1}{6} \cdot 1 \] Hence the limit is $ \cos\left(... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the limit $ \\lim_{x\\to 0}\\cos{(\\pi x^{2} \\csc (\\frac{x}{2}) \\cot (6x)) }$",
"content_html": "Find the limit <img src=\"//latex.artofproblemsolving.com/4/9/2/49267c68081373f106aa3e57a8dbde... | Find the limit
\[
\lim_{x\to 0}\cos\!\biggl(\pi x^{2}\csc\!\bigl(\tfrac{x}{2}\bigr)\cot(6x)\biggr).
\] | [
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0.41303083349788106,
0.5640984308585943,
0.5622799072951652,
0.6427344565556996,
0.526... | null |
aops_476416 | 2. Put $z=\frac{1}{x}$. $x\to\infty\implies z\to 0+$
$\mathop {\lim }\limits_{x \to + \infty } \left( {2x^3 \sin {1 \over x} - x^3 \sin {2 \over x}} \right)=\mathop {\lim }\limits_{z \to 0+} \left( \frac{2\sin z(1-\cos z)}{z^3} \right)\\ \\ \\=\lim_{z\to0+}\frac{\sin z}{z}\cdot\lim_{z\to0+}\frac{4\sin^2\frac{z}{2}}{z^... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "To calculate the limits:\n\n1. $\\mathop {\\lim }\\limits_{x \\to + \\infty } e^{x + \\sin x} $\n\n2. $\\mathop {\\lim }\\limits_{x \\to + \\infty } \\left( {2x^3 \\sin {1 \\over x} - x^3 \\sin {2 \\over... | To calculate the limits:
1. \(\displaystyle \lim_{x\to+\infty} e^{x+\sin x}\).
2. \(\displaystyle \lim_{x\to+\infty}\bigl(2x^3\sin\tfrac{1}{x}-x^3\sin\tfrac{2}{x}\bigr).\) | [
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aops_1743228 | Recall that
$$\frac{d}{dx}\tan(x)=1+\tan(x)^2=\frac1{\cos(x)^2}\text{ and }\int\tan(x)dx=-\log|\cos(x)|$$
Therefore apply IBP with $u=x$ and $v'=\frac{dx}{\cos(x)^2}$ so you will get
$$\int\frac{x}{\cos(x)^2}dx=x\tan(x)-\int \tan(x)dx=x\tan(x)+\log|\cos(x)|+c$$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Calculate\n\n$$\\int_{}^{} \\frac{x}{\\cos^2 x}dx$$",
"content_html": "Calculate<br>\n<br>\n<img src=\"//latex.artofproblemsolving.com/0/4/1/041d10fab5066c060b4d62af93a1ac6d862128d6.png\" class=\"lat... | Calculate
\[
\int \frac{x}{\cos^2 x}\,dx.
\] | [
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numina_10032621 | Solution. Using formulas (2.4), (2. $6^{\prime \prime \prime}$ ) and (2.8), we find
\[
\begin{gathered}
y^{\prime}=(x \cos x)^{\prime}=x^{\prime} \cos x+x(\cos x)^{\prime}=1 \cdot \cos x+x(-\sin x)= \\
=\cos x-x \sin x
\end{gathered}
\] | \cosx-x\sinx | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 3. Find the derivative of the function $y=x \cos x$. | [
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aops_1312046 | [quote=am_11235...]Prove that there exists a constant $c$ > $0$ such that for all $x$ in $[1,\infty)$ , $\sum_{n\geq x} (1/n^2)$ $\leq$ $(c/x)$ .[/quote]
[hide=Hint]
Telescope!
Or compare with an integral!
[/hide]
[hide=Solution]
We claim that $c=2$ works. To prove this, it clearly suffices to prove it for $x$ being a ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove that there exists a constant $c$ > $0$ such that for all $x$ in $[1,\\infty)$ , $\\sum_{n\\geq x} (1/n^2)$ $\\leq$ $(c/x)$ .",
"content_html": "Prove that there exists a constant <img src=\"//l... | Prove that there exists a constant \(c>0\) such that for all \(x\in[1,\infty)\),
\[
\sum_{n\ge x}\frac{1}{n^2}\le\frac{c}{x}.
\] | [
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aops_553269 | The integrand is continuous and bounded, let us name the integral $I$.
With the substitution : $x=\pi -t$, we can see that $I=-I$, wich means that $I=0$ for any integer $n$.
Two remarks : $\cos$ (in latex : [i]\cos[/i]) looks certainly better than $cos$. and also when you propose a problem write it completely, in this... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "I'm kinda stuck up on this problem. Help!\n\n$\\int_{0}^\\pi e^{cos^2x}$ $cos^3[(2n+1)x]$ $dx$",
"content_html": "I'm kinda stuck up on this problem. Help!<br>\n<br>\n<img src=\"//latex.artofproblems... | Evaluate the integral
\[
\int_{0}^{\pi} e^{\cos^{2}x}\cos^{3}\bigl((2n+1)x\bigr)\,dx.
\] | [
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aops_166978 | $ \lim_{x\to 0}x^{2}\cos(10x) \equal{} 0\cdot 1 \equal{} 0$.
$ \lim_{t\to 0}\frac{1}{t^{2}}\sin^{2}\left(\frac{t}{2}\right) \equal{}\frac{1}{4}\lim_{t\to 0}\left[\frac{\sin\left(\frac{t}{2}\right)}{\frac{t}{2}}\right]^{2}\equal{}\frac{1}{4}\cdot 1^{2}\equal{}\frac{1}{4}$.
In the second one I used the fact that $ ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Can anyone help me with these two problems?\r\n\r\nlim x^2cos10x as x->0\r\n\r\nand the second:\r\n\r\nlim (1/t^2)sin^2(t/2) as t->0",
"content_html": "Can anyone help me with these two problems?<br>... | Evaluate the following limits:
1. \(\displaystyle \lim_{x\to 0} x^2\cos(10x)\).
2. \(\displaystyle \lim_{t\to 0} \frac{\sin^2\!\bigl(\tfrac{t}{2}\bigr)}{t^2}\). | [
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aops_1268870 | Yes, there is a simple one:
$I=\int \sqrt{1 + x^{2}} \ dx = \int d(x)\sqrt{1 + x^{2}} = x \sqrt{1 + x^{2}} - \int \frac{x^2}{\sqrt{1 + x^{2}}} = x \sqrt{1 + x^{2}} - \left( \int \frac{x^2+1}{\sqrt{1 + x^{2}}} - \frac{1}{\sqrt{1+x^2}} \right) = x \sqrt{1 + x^{2}} - I +\int \frac{1}{\sqrt{1+x^2}} $. Hence $I=\frac{... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find $\\int \\sqrt{1 + x^{2}} \\ dx$.\n\nWe have the substitutions $x = \\sinh t$ and $ x = \\tan t$. Any other methods?",
"content_html": "Find <span style=\"white-space:nowrap;\"><img src=\"//latex... | Find \(\displaystyle \int \sqrt{1+x^{2}}\;dx.\)
(Consider substitutions \(x=\sinh t\) or \(x=\tan t\).) | [
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aops_1173222 | I don't see much of an easier way to do this one than tabular IBP. Fortunately, this is a product of a polynomial and an easy-to-integrate function, so this one is very easy.
\[\begin{array}{c|c} x^3 & \sin{x} \\ 3x^2 & -\cos{x} \\ 6x & -\sin{x} \\ 6 & \cos{x} \\ 0 & \sin{x} \end{array}\]
\[\int \! x^3\sin{x} \, dx = -... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find $\\int_{0}^{\\frac{\\pi}{2}} x^{3} \\sin x \\ dx$.",
"content_html": "Find <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolving.com/4/a/a/4aaa87f735cac39b911efe6728a5f6fe4b... | Find
\[
\int_{0}^{\frac{\pi}{2}} x^{3}\sin x \,dx.
\] | [
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numina_10737764 | Prove: Let $f(x)=\frac{1}{3}(x-1)-x^{\frac{1}{3}}+1$, then $f^{\prime}(x)=\frac{1}{3}-\frac{1}{3} x^{-\frac{2}{3}}=\frac{1}{3}(1-$ $\left.\frac{1}{\sqrt[3]{x^{2}}}\right)$
$\because x>1, \therefore \frac{1}{\sqrt[3]{x^{2}}}<1$, thus $\frac{1}{3}(1-\frac{1}{\sqrt[3]{x^{2}}})>0$, i.e., $f^{\prime}(x)>0$, so $f(x)$ is an ... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "inequalities"
} | Example 3: Prove that: $\frac{1}{3}(x-1)>x^{\frac{1}{3}}-1$, where $x \in(1,+\infty)$. | [
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numina_10031521 | Solution.
1. Find $\varrho^{\prime}(\varphi)$:
$$
\varrho^{\prime}(\varphi)=6 \cos \varphi
$$
2. Calculate the differential of arc length:
$$
d l=\sqrt{\varrho(\varphi)^{2}+\varrho^{\prime}(\varphi)^{2}} d \varphi=\sqrt{36 \sin ^{2} \varphi+36 \cos ^{2} \varphi} d \varphi=6 d \varphi
$$
3. Find the arc length by e... | 2\pi | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example. Calculate the length of the arc of the curve given by the equation in polar coordinates
$$
\varrho=6 \sin \varphi, \quad 0 \leq \varphi \leq \pi / 3
$$ | [
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numina_10175451 | Since
$$
\begin{aligned}
& \frac{1-\cos x}{x^{2}}=\frac{1}{2} \cdot \frac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}} ; \quad \frac{\sin 3 x}{x-\pi}=\frac{\sin 3[\pi+(x-\pi)]}{x-\pi}= \\
& =-3 \cdot \frac{\sin 3(x-\pi)}{3(x-\pi)} ; \quad \frac{\cos \frac{\pi}{2} x}{x-1}=-\frac{\pi}{2} \cdot \frac{\sin \left[\... | \frac{1}{2},-3,-\frac{\pi}{2},\pi^{2} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | The following limits need to be calculated:
a.) $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=$ ?
b.) $\lim _{x \rightarrow \pi} \frac{\sin 3 x}{x-\pi}=?$
c.) $\lim _{x \rightarrow 1} \frac{\cos \frac{\pi x}{2}}{x-1}=$ ?
d.) $\lim _{x=0} \frac{1}{x} \sin \frac{\pi}{1+\pi x}=$ ? | [
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aops_1365843 | [hide="Solution [Outline]"]
Multiply both the numerator and denominator by $2\sin(x)\cos(x)$ and rewrite in terms of sines and cosines, noting that $\cot(2x) = \frac{\cos(2x)}{\sin(2x)}$. Then our integral equals
$$\int{\frac{2\sin(x)\cos(x)}{3\cos(2x) + 2\sin^2(x) + 2\cos(x)}dx}$$
Now utilize the identity $\cos(2x) ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Integrate $$\\int \\dfrac{1}{3\\cot\\left(2x\\right)+\\tan\\left(x\\right)+\\csc\\left(x\\right)} \\, dx.$$",
"content_html": "Integrate <img src=\"//latex.artofproblemsolving.com/7/3/5/735383105e2e2... | Integrate
\[
\int \frac{1}{3\cot(2x)+\tan x+\csc x}\,dx.
\] | [
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aops_280038 | [quote="solenoid"]Cound you help me with
\[ \Int \frac {\ln (x + 1)}{x^2} dx ?
\]
Thanks! :)[/quote]
OK, now I know:
\[ \Int \frac {\ln (x + 1)}{x^2} dx = - \frac{1+x}{x} + \ln\frac{x}{x+1}.
\] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Cound you help me with\r\n\r\n\\[ \\Int \\frac{\\ln (x+1)}{x^2} dx ?\\]\r\n\r\nThanks! :)",
"content_html": "Cound you help me with<br>\n<br>\n<pre class=\"aopscode-error aopscode-latex-error\">\\[ ... | Coud you help me with
\[
\int \frac{\ln(x+1)}{x^2}\,dx ?
\] | [
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aops_464542 | [quote="RohanSingh"]Ratio test shows its convergence[/quote]
$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{(n+1)^5}{2n^5}=\frac{1}{2}$
So, $\sum\frac{n^5}{2^n}$ converges. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Does the following series converge?\n\n$\\sum\\frac{n^5}{2^n}$",
"content_html": "Does the following series converge?<br>\n<br>\n<img src=\"//latex.artofproblemsolving.com/2/5/3/253d465304d6bb8c39b46... | Determine whether the series
\[
\sum_{n=1}^{\infty}\frac{n^5}{2^n}
\]
converges. | [
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numina_10030790 | The solution is as follows. $1 \cdot \check{i} s t e p$. Introduce a new unknown function of substitution $t=x+y$, then $\frac{d t}{d x}=1+\frac{1}{t^{2}}$.
$2-\check{u}$ step. Separate the variables: $\frac{t^{2} d t}{1+t^{2}}=d x$, integrate both sides, each with respect to its own variable:
$$
\int \frac{t^{2} d t... | (x+y)-\operatorname{arctg}(x+y)=x+1-\frac{\pi}{4} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 4.3. Solve the Cauchy problem: $\frac{d y}{d x}=\frac{1}{(x+y)^{2}}, y(0)=1$. | [
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aops_2574196 | [b]Solution:[/b]
$\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{e}^{x}}}{{{(1+\frac{1}{x})}^{{{x}^{2}}}}}=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{{{e}^{x}}}{{{\left( 1+\frac{1}{x} \right)}^{x}}} \right)}^{x}}==\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{{{e}^{x}}-{{\left( 1+\frac{1}{x}... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$\\lim_{x\\rightarrow+\\infty}\\frac{e^{x}}{(1+\\frac{1}{x})^{x^2}}$=?",
"content_html": "<span style=\"white-space:pre;\"><img src=\"//latex.artofproblemsolving.com/c/9/f/c9ffa3d1bee91d39a17ad534d32... | \[
\lim_{x\to+\infty}\frac{e^{x}}{\left(1+\frac{1}{x}\right)^{x^{2}}}
\] | [
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aops_216872 | $ \frac {x}{x \plus{} 1} \plus{} \frac {y}{y \plus{} 1} \plus{} \frac {z}{z \plus{} 1} \ge \frac {x}{1 \plus{} 1} \plus{} \frac {y}{1 \plus{} 1} \plus{} \frac {z}{1 \plus{} 1}\equal{}1$. As for the right inequality, $ f(x)\equal{}\frac{x}{x\plus{}1}\equal{}1\minus{}\frac{1}{x\plus{}1}$ is concave, so we have, by Jensen... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $ x,y,z$ be real numbers with $ 0\\le x,y,z \\le 1$ and $ x \\plus{} y \\plus{} z \\equal{} 2$. Prove:\r\n\\[ 1\\le \\frac {x}{x \\plus{} 1} \\plus{} \\frac {y}{y \\plus{} 1} \\plus{} \\frac {z}{z \\p... | Let \(x,y,z\) be real numbers with \(0\le x,y,z\le 1\) and \(x+y+z=2\). Prove
\[
1\le \frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\le 2.
\] | [
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aops_488492 | Maybe you want to say "rewrite your integrand in the form $\frac{\sqrt{(x^{19})^{3}}}{x(1+(x^{19})^{2})}$ and set $x^{19} = t$" (for whatever that's worth), Dr Sonnhard. At least, let's do the algebra right :) | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Integrate:\n$ \\int \\frac{\\sqrt{ x^{55}}}{1+x^{38}} \\,dx $",
"content_html": "Integrate:<br>\n<img src=\"//latex.artofproblemsolving.com/1/5/7/157f62ec29ddd7c797e6ee1332bbb1eed2897c08.png\" class... | Integrate:
\[
\int \frac{\sqrt{x^{55}}}{1+x^{38}}\,dx
\] | [
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aops_395888 | [quote="AndrewTom"]Given that $I_{n} = \int_{0}^{1} (1+x^{2})^{n} dx$, show that
$(2n+1)I_{n}= 2^{n} + 2nI_{n-1}$.
[/quote]
[hide="show recurrence"]
Using parts, let $u=(1+x^{2})^{n}, \;\ dv=dx, \;\ du=2nx(1+x^{2})^{n-1}dx, \;\ dv=x$
$x(1+x^{2})^{n}|_{0}^{1} -2n\int_{0}^{1}x^{2}(1+x^{2})^{n-1}dx$
The left side ev... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Given that $I_{n} = \\int_{0}^{1} (1+x^{2})^{n} dx$, show that\n\n$(2n+1)I_{n}= 2^{n} + 2nI_{n-1}$.\n\n(a) Verify that the method used to obtain the reducion formula is valid for all real values of $n$.\n\... | Given \(I_{n}=\displaystyle\int_{0}^{1}(1+x^{2})^{n}\,dx\), show that
\[
(2n+1)I_{n}=2^{n}+2nI_{n-1}.
\]
(a) Verify that the method used to obtain the reduction formula is valid for all real values of \(n\).
(b) What form does the reduction formula \((2n+1)I_{n}=2^{n}+2nI_{n-1}\) take when \(n=0\)? Check that this is... | [
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aops_1167144 | Well I didn't compute it like that.... coz the relation used above is unknown to me...
My solution:
$Let \ I=\int_{\frac{1}{a}}^{a} \frac{arctanx}{x}\,dx$
Put $x=\frac{1}{t}, \therefore I=\int_{a}^{\frac{1}{a}} \frac{arctan(\frac{1}{t})}{\frac{1}{t}} \cdot \frac{-1}{t^2}\,dt=\int_{\frac{1}{a}}^{a} \frac{arccot \ t}{t}\... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "What would the definite integral of [i](arctan x)/x[/i] be between the limits [i]a[/i] and [i]1/a[/i]?",
"content_html": "What would the definite integral of <i>(arctan x)/x</i> be between the limits... | What is the value of the definite integral
\[
\int_{a}^{1/a}\frac{\arctan x}{x}\,dx ?
\] | [
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aops_613517 | The same approach can be applied to the second topic as the previous one.
The region of integration $x+1<y<2$ and $0<x<1$ is corresponding to $0<x<y-1$ and $1<y<2$, therefore
\[
\int_0^1\int_{x+1}^2 e^{\left(\frac{x}{y-1}\right)}\, dy\ dx=\int_1^2\int_0^{y-1} e^{\left(\frac{x}{y-1}\right)}\, dx\ dy
\]
Now set $u=\fra... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate $\\displaystyle I=\\int^1_0 \\int^2_{x+1} e^{\\left(\\frac{x}{y-1}\\right)} dydx$",
"content_html": "Evaluate <img src=\"//latex.artofproblemsolving.com/e/a/a/eaaae44fd0d501cffc98ee2e0535980... | Evaluate
\[
I=\int_{0}^{1}\int_{x+1}^{2} e^{\left(\frac{x}{y-1}\right)} \,dy\,dx.
\] | [
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aops_1955037 | I suppose that $x \to 0^{+}$, because there is no limit if $x \to 0$.
Note that
\[\lim_{x \to 0^{+}} (\sin x)^{\frac{1}{h \cdot \ln x}}=e^{\frac{1}{2}}\]
\[= \ln \lim_{x \to 0^{+}} (\sin x)^{\frac{1}{h \cdot \ln x}}=\ln e^{\frac{1}{2}}\]
\[= \lim_{x \to 0^{+}} \ln (\sin x)^{\frac{1}{h \cdot \ln x}}=\frac{1}{2}\]
\[= \l... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the value of $h \\in \\mathbb{R}$-{$0$} so that $\\lim_{x \\to 0} (\\sin x)^{\\frac{1}{h.In(x)}}= e^{\\frac{1}{2}}$",
"content_html": "Find the value of <span style=\"white-space:pre;\"><img sr... | Find the value of \(h\in\mathbb{R}\setminus\{0\}\) such that
\[
\lim_{x\to 0} (\sin x)^{\frac{1}{h\cdot \ln x}} = e^{\frac{1}{2}}.
\] | [
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aops_301191 | If you don't see the point of Kouichi Nakagawa's post, then fill in the correct numbers as exponents for these question marks:
$ x\sqrt{x}\equal{}x^?$
$ \frac1{\sqrt[3]{4x}}\equal{}\frac1{\sqrt[3]{4}}\cdot\frac1{\sqrt[3]{x}}\equal{}\frac1{\sqrt[3]{4}}\cdot x^?$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "I'm starting to learn calculus and I need help with some problems.\r\n\r\n1 - $ \\int x \\sqrt{x}\\,dx$\r\n\r\n2 - $ \\int (6x^2 \\minus{} x\\plus{}1)\\,dx$\r\n\r\n3- $ \\int \\frac{dx}{\\sqrt[3]{4x}}$",
... | I'm starting to learn calculus and I need help with some problems.
1. \( \displaystyle \int x \sqrt{x}\,dx \)
2. \( \displaystyle \int (6x^2 - x + 1)\,dx \)
3. \( \displaystyle \int \frac{dx}{\sqrt[3]{4x}} \) | [
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aops_204092 | Dividing both denominator and numerator by $ \cos^2 x$, we have $ \int \frac{dx}{3\plus{}\sin^2 x} \equal{} \int \frac{\sec^2 x \, dx}{3 \plus{} 4 \tan^2 x}$.
Now substitute $ t \equal{} \tan x$ to obtain $ \int \frac{dx}{3\plus{}\sin^2 x} \equal{} \int \frac{dt}{3 \plus{} 4 t^2} \equal{} \frac{1}{2\sqrt{3}} \tan^{\... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the integral $ \\int\\frac{dx}{3\\plus{}\\sin^2 x}$.",
"content_html": "Find the integral <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolving.com/0/e/f/0ef39c91d0dbb46f981... | Find the integral
\[
\int \frac{dx}{3+\sin^2 x}.
\] | [
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aops_1360576 | [quote=Nikpour]$\sum_{k=1}^n \frac1n =\infty$[/quote]
No, that's not true. The sum of $n$ copies of $\frac1n$ is just $1$.
We have $\frac{n}{n+\log 2n}\le \sum_{k=1}^{n} \frac1{n+\log(n+k)}\le 1$. Since $\frac{\log 2n}{n}\to 0$ as $n\to\infty$, that left side $\frac1{1+\frac{\log 2n}{n}}$ goes to $1$, and the limit we... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "find $\\lim_{n\\to \\infty } ( \\sum_{ k=1 }^{ n } \\frac{1}{n+ \\log (n+k)})$",
"content_html": "find <img src=\"//latex.artofproblemsolving.com/c/3/e/c3ea05fde59a0b7e9a19afc5c0e0b5e6247784f2.png\... | Find
\[
\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n+\log(n+k)}.
\] | [
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numina_10738321 | 3. Use the monotonicity of $y=-x^{2}+x$ on $\left(0, \frac{1}{4}\right)$ to complete the transition from $k$ to $k+1$. | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "inequalities"
} | 3 Let $\left\{a_{n}\right\}$ be a sequence of positive terms. If $a_{n+1} \leqslant a_{n}-a_{n}^{2}$, prove that for all $n \geqslant 2$,
$$a_{n} \leqslant \frac{1}{n+2}$$ | [
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aops_529067 | [hide="Straightforward Solution with Calculus"]
We need $\dfrac{\log(k + 1)}{k + 1} < \dfrac{\log(k)}{k}$. Thus we just need that $f(x) = \dfrac{\log x}{x}$ is decreasing for $x \ge e$ (we choose $e$ because it's close to three and it's involved with calculus a lot).
Note that $f'(x) = \dfrac{1}{x^2} - \dfrac{\log x}{... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove that:\n\n$\\frac{\\ln (k+1)\\cdot k}{\\ln k \\cdot (k+1)} < 1$\n\nwhere k>3, and k is an integer. \n\n\n[hide=\"My work so far\"]\n\nSince $k+1 > k$\n\nand $\\ln (k+1) > \\ln (k)$\n\nWe get $\\frac... | Prove that
\[
\frac{k\ln(k+1)}{(k+1)\ln k}<1
\]
for integer \(k>3\). | [
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aops_591044 | \[\int_{0}^{\infty}\ln(\tanh(x))dx\]
\[\int_{0}^{\infty}\ln\left(\frac{1-e^{-2x}}{1+e^{-2x}}\right)dx\]
\[\int_{0}^{\infty}\ln(1-e^{-2x})-\int_{0}^{\infty}\ln(1+e^{-2x})dx\]
Use the series for $\ln(1-x)$ for the left one:
\[-\int_{0}^{\infty}\sum_{k=1}^{\infty}\frac{e^{-2kx}}{k}\]
switch sum and integral which is ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate the integral :\n\\[ \\int_{0}^{+\\infty} \\ln{(\\tanh{x})}\\,\\mathrm{d}x \\] \n\n[i]Proposed by Fedor Duzhin[/i]",
"content_html": "Evaluate the integral :<br>\n<img src=\"//latex.artofprob... | Evaluate the integral
\[
\int_{0}^{+\infty} \ln(\tanh x)\,dx.
\] | [
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ours_31026 | Solution: We use integration by parts. Let \( u = \frac{1}{\sqrt{4 x^{3}-x+1}} \) and \( dv = n x^{n} \, dx \). Then
\[
\begin{gathered}
du = -\frac{1}{2} \frac{12 x^{2}-1}{\sqrt{\left(4 x^{3}-x+1\right)^{3}}} \, dx, \\
v = \frac{n x^{n+1}}{n+1}.
\end{gathered}
\]
Thus, we have
\[
\int_{0}^{1} \frac{n x^{n}}{\sqrt{4... | 3 | {
"competition": "bmt",
"dataset": "Ours",
"posts": null,
"source": "CASP18A.md"
} | Compute the following limit:
$$
\lim _{n \rightarrow \infty} \int_{0}^{1} \frac{n x^{n}}{\sqrt{4 x^{3}-x+1}} d x
$$ If the answer is of the form of an irreducible fraction $\frac{a}{b}$, compute the value of $a + b$. | [
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numina_10046714 | ## Solution
$$
\int \operatorname{arctg} \sqrt{4 x-1} d x=
$$
Let:
$$
\begin{aligned}
& u=\operatorname{arctg} \sqrt{4 x-1} ; d u=\frac{1}{1+(\sqrt{4 x-1})^{2}} \cdot \frac{1}{2 \sqrt{4 x-1}} \cdot 4 d x= \\
& =\frac{1}{4 x} \cdot \frac{2}{\sqrt{4 x-1}} \cdot d x=\frac{d x}{2 x \sqrt{4 x-1}} \\
& d v=d x ; v=x
\end{... | x\cdot\operatorname{arctg}\sqrt{4x-1}-\frac{1}{4}\cdot\sqrt{4x-1}+C | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | ## Problem Statement
Calculate the indefinite integral:
$$
\int \operatorname{arctg} \sqrt{4 x-1} \, d x
$$ | [
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0... | null |
aops_227253 | [quote="Hong Quy"]We have : $ x_{n \plus{} 1} < x_n$ and $ 0 < x_n < 1$
Hence exist limit of sequence: $ \lim x_n \equal{} L$ then $ L \equal{} 0$
Apply theorem Stolz's:
$ \lim nx_n \equal{} \lim \frac {n}{\frac {1}{x_n}} \equal{} \lim \frac {1}{\frac {1}{x_{n \plus{} 1}} \minus{} \frac {1}{x_n}} \equal{}... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $ x_n$ be given $ x_1 \\equal{} \\frac {1}{2}$ ,$ x_{n \\plus{} 1} \\equal{} x_n \\minus{} x_n^2$ (for all $ n\\ge 1$\r\nProve that $ lim_{n\\to \\infty}nx_n \\equal{} 1$",
"content_html": "Let <... | Let \(x_1=\tfrac{1}{2}\) and \(x_{n+1}=x_n-x_n^2\) for all \(n\ge1\). Prove that \(\displaystyle\lim_{n\to\infty} n x_n = 1\). | [
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aops_1235753 | [quote=Abhinandan18]$\int (\sin^2 x-\cos^2 x) dx =\int -\cos 2x dx =-\frac{1}{2}\sin 2x +c$[/quote]
Pardon my ignorance, but how do you know integral of $-cos(2x)$ with respect to $x$ is equal to $\frac{-1}{2}*sin(2x)+c$? | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find __________ where $\\int [\\sin^2(x)-\\cos^2(x)] dx = $ _____________ $+ C$ where $C$ is some constant.",
"content_html": "Find __________ where <img src=\"//latex.artofproblemsolving.com/e/3/f/e... | Find __________ where
\[
\int \bigl[\sin^2(x)-\cos^2(x)\bigr]\,dx = \underline{\hspace{3cm}} + C
\]
where \(C\) is some constant. | [
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aops_1107558 | $I_{0}=\int \frac{1}{\sqrt{x^{2} + 2x +3}}\ dx=\int \frac{1}{\sqrt{(x+1)^{2} + 2}}\ dx= \ln(x+1+\sqrt{x^{2} + 2x +3})+C$.
$I_{1}=\int \frac{x+1}{\sqrt{x^{2} + 2x +3}}\ dx=\sqrt{x^{2} + 2x +3}+C$.
$I_{2}=\int \frac{x^{2}+x+1}{\sqrt{x^{2} + 2x +3}}\ dx=\frac{1}{2}\int \frac{2x^{2}+2x+2}{\sqrt{x^{2} + 2x +3}}\ dx$
$I_{2... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find \n\n(i) $\\int \\frac{1}{\\sqrt{x^{2} + 2x +3}}\\ dx$,\n\n(ii) $\\int \\frac{x^{2} + x + 1}{\\sqrt{x^{2} + 2x +3}}\\ dx$,\n\n(iii) $\\int \\frac{x^{3} + x^{2} + x +1}{\\sqrt{x^{2} + 2x +3}}\\ dx$.",
... | Find
(i) \(\displaystyle \int \frac{1}{\sqrt{x^{2} + 2x + 3}}\,dx,\)
(ii) \(\displaystyle \int \frac{x^{2} + x + 1}{\sqrt{x^{2} + 2x + 3}}\,dx,\)
(iii) \(\displaystyle \int \frac{x^{3} + x^{2} + x + 1}{\sqrt{x^{2} + 2x + 3}}\,dx.\) | [
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aops_612146 | [quote="Unpredictability"][b]Question.[/b] $\displaystyle \int \frac{\sin x}{\sec^{2} x} \,dx $
[/quote]
Ok, something quicker than that. Notice that :
\[\int \frac{\sin x}{\sec^2 x}\,dx=\int \sin x\cos^2 x \,dx=\int \left [ -\frac{1}{3}\cos^3 x \right ]'\,dx=\]
\[=-\frac{1}{3}\cos^3 x +c, c\in \mathbb{R}\] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "[b]Question.[/b] \\int $ \\frac{\\sin x}{\\sec^{2} x} $ \\,dx = ?\n\n[b]Attempt.[/b] Let $ u=\\sec x $. Then, we get \\int $ \\frac{\\sin x\\cdot u^{-2}}{\\sec x\\tan x}$ \\,du . Remember $ \\frac{\\s... | Question. Evaluate the integral
\[
\int \frac{\sin x}{\sec^{2}x}\,dx.
\]
Answer. \[
-\frac{\cos^{3}x}{3}+C.
\] | [
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aops_3222632 | We have
$$\int_0^{\pi/2}(\sin x -\cos x)^2dx=\frac{\pi}{2}-1=-\int_0^{\frac{\pi}2}(f(x)^2-2f(x)(\sin x-\cos x))dx$$
and so
$$\int_0^{\pi/2}(f(x)-(\sin x -\cos x))^2dx=0,$$
which gives $f(x)=\sin x - \cos x$ hence the result. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Given a continuous function $f:\\left[0,\\frac{\\pi}2\\right]\\to\\mathbb{R}$ that satisfies $$\\int_0^{\\frac{\\pi}2}(f(x)^2-2f(x)(\\sin x-\\cos x))dx=1-\\frac{\\pi}2$$ Show that $$\\int_0^{\\frac{\\pi}2}... | Given a continuous function \(f:\left[0,\frac{\pi}{2}\right]\to\mathbb{R}\) that satisfies
\[
\int_0^{\frac{\pi}{2}}\bigl(f(x)^2-2f(x)(\sin x-\cos x)\bigr)\,dx
=1-\frac{\pi}{2},
\]
show that
\[
\int_0^{\frac{\pi}{2}}f(x)\,dx=0.
\] | [
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aops_265177 | Both series converge. Just consider series $ \frac{1}{n}$ and $ \frac{1}{n^2}$, which clearly converge to $ 0$. Clearly, in the first case $ |a_n|\leq\frac{1}{n}$ and in the second case $ |a_n|\leq\frac{1}{n^2}$. Hence both $ \{a_n\}$ series converge to $ 0$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "a. $ a_n\\equal{}\\frac{\\sin{n}}{n}$\r\n\r\nb. $ a_n\\equal{}\\frac{\\cos{\\pi}n}{n^2}$\r\n\r\nIf it converges, find its limit.",
"content_html": "a. <img src=\"//latex.artofproblemsolving.com/d/f/6... | a. \(a_n=\dfrac{\sin n}{n}\)
b. \(a_n=\dfrac{\cos \pi n}{n^2}\)
If it converges, find its limit. | [
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numina_10198158 | Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$, using the L'Hôpital's rule, we have $\lim \limits_{x\to 1}\frac{x^2-1}{x-1} = \lim \limits_{x\to 1}\frac{2x}{1} = 2$.
Thus, the answer is $\boxed{\textbf{(D)}\ 2}$.
~ MATH__is__FUN | 2 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "amc_aime"
} | The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$ | [
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aops_365898 | Actually keep using Stolze theorem we have
$\lim_{n\to \infty} \ln \sqrt[n(n+1)]{a_n} = \lim_{n\to \infty} \frac{\ln a_n}{n(n+1)} = \lim_{n\to \infty} \frac{\ln \frac{a_{n+1}}{a_n}}{2(n+1)}=\lim_{n\to \infty} \frac{\ln \frac{a_{n+2}a_n}{a_{n+1}^2}}{2}=\frac{\ln \lambda}{2}$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "[color=darkred]Let $(a_n)_{n\\ge 1}$ be a sequence of positive real numbers . [b]IF[/b] $\\lim_{n\\to\\infty}\\ \\frac {a_n\\cdot a_{n+2}}{a_{n+1}^2}=\\lambda$ , [b]THEN[/b] show that : $\\lim_{n\\to\\inft... | Let \((a_n)_{n\ge 1}\) be a sequence of positive real numbers. If
\[
\lim_{n\to\infty}\frac{a_n a_{n+2}}{a_{n+1}^2}=\lambda,
\]
then show that
\[
\lim_{n\to\infty}\sqrt[n(n+1)]{a_n}=\sqrt{\lambda}.
\] | [
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aops_1376479 | Do you mean $\lim_{n\to\infty} \left[\left(\sum_{k=1}^n k^{\frac1n}\right)-n-\ln n\right]$? | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Compute the limit of the sequence an= (sum from k=1 to n from k^(1/n))-n-ln(n) ",
"content_html": "Compute the limit of the sequence an= (sum from k=1 to n from k^(1/n))-n-ln(n)",
"post_id": 76... | Compute the limit of the sequence
\[
a_n=\sum_{k=1}^n k^{1/n}-n-\ln n.
\] | [
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numina_10148715 | (1) 1/√(n 2 + i 2 ) = (1/n) / √(1 + (i/n) 2 ). So the sum is just a Riemann sum for the integral ∫ 0 1 dx/√(1 + x 2 ) = sinh -1 1 = ln(1 + √2) = 0.8814. (2) 1/√(n 2 + i) = (1/n) / √(1 + i/n 2 ). Each term is less than 1/n, so the (finite) sum is less than 1. But each term is at least (1/n) / √(1 + 1/n). So the sum is a... | (1)\ln(1+\sqrt{2})\approx0.8814,(2)1,(3)\infty | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 4th Putnam 1941 Problem B2 Find: (1) lim n→∞ ∑ 1≤i≤n 1/√(n 2 + i 2 ); (2) lim n→∞ ∑ 1≤i≤n 1/√(n 2 + i); (3) lim n→∞ ∑ 1≤i≤n 2 1/√(n 2 + i); | [
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aops_258682 | [hide]$ \lim \frac{n(n^{1/n}\minus{}1)}{\ln n} \equal{} \lim \frac{n(e^{\ln n/n}\minus{}1)}{\ln n} \equal{} \lim \equal{} \frac{n(1 \plus{} \frac{\ln n}{n} \plus{} O[(\frac{\ln n}{n})^2] \minus{} 1)}{\ln n}$
$ \equal{} \lim \left(1 \plus{} O \left(\frac{\ln n}{n}\right)\right) \equal{} 1$.[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Calculate $ \\lim_{n\\to\\infty}\\frac{n(n^{1/n}\\minus{}1)}{ln(n)}$",
"content_html": "Calculate <img src=\"//latex.artofproblemsolving.com/b/a/c/bacf067751aeb0d0d9a49ee946ab4883c7b7fc16.png\" class... | Calculate \[
\lim_{n\to\infty}\frac{n\bigl(n^{1/n}-1\bigr)}{\ln n}.
\] | [
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numina_10268757 | 1. We start by considering the integral \(\int \frac{e^{x/2} \cos x}{\sqrt[3]{3 \cos x + 4 \sin x}} \, dx\).
2. Let \( y = 3 \cos x + 4 \sin x \). Then, we have:
\[
y^{2/3} = \frac{3 \cos x + 4 \sin x}{y^{1/3}}
\]
and
\[
(y^{2/3})' = \frac{2}{3} \frac{4 \cos x - 3 \sin x}{y^{1/3}}.
\]
3. The integr... | null | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "aops_forum"
} | Final Round (Time limit: 5 mins per integral)
Q1. $\int\frac{e^{x/2}\cos x}{\sqrt[3]{3\cos x+4\sin x}}\,dx$
Q2. $\int_0^\infty\frac{\log(2e^x-1)}{e^x-1}\,dx$
Q3. $\int_{-\infty}^\infty\frac{dx}{x^4+x^3+x^2+x+1}$
Q4. $\int_{-1/3}^1\left(\sqrt[3]{1+\sqrt{1-x^3}}+\sqrt[3]{1-\sqrt{1-x^3}}\right)\,dx$
Q5. $\int_0^1\max... | [
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aops_2652932 | Since the inequality is homogeneous, we can assume $\sum_{i=1}^{n}x_i=1$ ( If $x_1=\ldots=x_n=0$ the inequality is trivial).
Denote $y_k=x_k+x_{k+1}$ for brevity and let $f(x)=x-x^p$. Note that $f''(x)=-p(p-1)x^{p-2} \ge 0$ for $x \ge 0$, hence $f(x)$ is convex. We need to prove
\[ \sum_{k=1}^{n} x_k - \sum_{k=1}^{n} x... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let$n \\ge 4$ be an integer, and let $x_1,x_2,\\cdots,x_n \\ge 0; p\\ge1$. Prove that\n$$\\sum_{i=1}^n (x_i+x_{i+1})^p \\le \\sum_{i=1}^n x_i^p+\\left(\\sum_{i=1}^n x_i\\right)^p$$\nPS: $x_{n+1}=x_1$.",
... | Let \(n\ge 4\) be an integer, and let \(x_1,x_2,\ldots,x_n\ge 0\) and \(p\ge 1\). Prove that
\[
\sum_{i=1}^n (x_i+x_{i+1})^p \le \sum_{i=1}^n x_i^p+\Bigl(\sum_{i=1}^n x_i\Bigr)^p,
\]
where \(x_{n+1}=x_1\). | [
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aops_543514 | $\Re(z)\ ,\ \Im(z)$ denotes to the real part or the imaginary part respectively of the complex number $z$.
Using Euler's identity we get that : $\sin x=\Im \left(e^{i x}\right)$ for any real $x$, then your definite would be : \[\Im \int (x e^{ix })^2 \ \mathrm{d}x.\] Which is easier to do by parts. Also, you may want ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Integrate:\n\n$\\int{(x\\sin x)^2}$\n\n[hide=\"Note/hint\"]\n\nI am pretty sure there is a solution with Taylor Polynomials, but is there another solution? Feel free to post any non-computerized solution y... | Integrate:
\[
\int (x\sin x)^2\,dx.
\] | [
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aops_3366230 | [quote=Alphaamss]First, we prove $$\lim_{n \to \infty} \frac{n}{u^2_n}=0,$$ by Stolz theorem, we have
$$\lim_{n \to \infty} \frac{n}{u^2_n}=\lim_{n \to \infty} \frac{1}{u^2_{n+1}-u^2_n}
=\lim_{n \to \infty}\frac1{2\ln n+\frac{\ln^2n}{u^2_n}}=0,$$
then$$\lim_{n \to \infty}\frac{\ln^2n}{u^2_n}=\lim_{n \to \infty}\frac{n}... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $(u_n)$ be a set satisfying $u_1 = 1$ and $u_{n+1} = u_n + \\frac{\\ln n}{u_n}$ $\\forall n \\geq 1$.\n\nProve that $u_{2023} > \\sqrt{2023 \\cdot \\ln 2023}$.\n\nFind \n$$\n\\lim_{n \\to \\infty} \\fr... | Let \( (u_n) \) be a sequence satisfying \(u_1 = 1\) and
\[
u_{n+1} = u_n + \frac{\ln n}{u_n}\quad\text{for all } n\ge 1.
\]
1. Prove that \(u_{2023} > \sqrt{2023\cdot\ln 2023}\).
2. Find
\[
\lim_{n\to\infty} \frac{u_n\cdot\ln n}{n}.
\] | [
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aops_371097 | [quote="medjool"]This is similar to one i posted earlier, and as i found while attempting a solution for the first, much easier =D
$ \int_{0}^{\frac{\pi}{2}}{\tan^{-1}({\sin{x}})dx + \int_{0}^{\frac{\pi}{4}}{\sin^{-1}({\tan{x}})dx }}$[/quote]
$\frac{\pi^2}{8}$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "This is similar to one i posted earlier, and as i found while attempting a solution for the first, much easier =D\n\n$ \\int_{0}^{\\frac{\\pi}{2}}{\\tan^{-1}({\\sin{x}})dx + \\int_{0}^{\\frac{\\pi}{4}}{\\s... | \[
\int_{0}^{\frac{\pi}{2}}\tan^{-1}(\sin x)\,dx \;+\; \int_{0}^{\frac{\pi}{4}}\sin^{-1}(\tan x)\,dx
\] | [
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aops_437209 | So, that's one clever trick. I haven't tested whether it works.
Alternately, we can complete the square in the numerator and make a trigonometric substitution. That leads to a rational function in $\sin\theta$ and $\cos\theta$, which we can follow with a $t=\tan\frac{\theta}{2}$ substitution to get a rational function... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Hello, can someone help me? Please, solve the indefinite integral following\n\n\\[ \\frac{\\sqrt{6-x-x^{2}}}{x^2}. \\]",
"content_html": "Hello, can someone help me? Please, solve the indefinite inte... | Solve the indefinite integral
\[
\int \frac{\sqrt{6-x-x^{2}}}{x^{2}}\,dx.
\] | [
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aops_168860 | [hide="Number 1"]$ [\ln(\cos x \plus{} \sqrt{\cos 2x})]' \equal{} \frac{\minus{} \sin x \minus{} \frac{\sin 2x}{\sqrt{\cos 2x}}}{\cos x \plus{} \sqrt{\cos 2x}} \equal{} \frac{\cos x}{\sin x} \minus{} \frac{1}{\sin x \sqrt{\cos 2x}}$.
On the other hand, $ 1 \minus{} \cos 2x \equal{} 2 \sin^2x$. So:
$ P \equal{} \frac{... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find:\r\n1. $ \\int \\frac {ln(cosx \\plus{} \\sqrt {cos2x})}{1 \\minus{} cos2x} \\,dx$\r\n2. $ \\int \\frac {1}{sin^{n}x \\plus{} cos^{n}x} \\,dx$\r\n\r\n--\r\nmath10.com",
"content_html": "Find:<br... | Find:
1. \[
\int \frac{\ln\big(\cos x+\sqrt{\cos 2x}\big)}{1-\cos 2x}\,dx
\]
2. \[
\int \frac{1}{\sin^{n}x+\cos^{n}x}\,dx
\] | [
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aops_246408 | hello
[hide="Probably the worst way but nothing else is striking in my mind at this moment.Even then click on this to see a stupid way "]
$ \int sin^{5/2} \theta cos ^{3/2} \theta d \theta$
=$ \int \frac {cos^{3/2}\theta}{sin^{3/2}\theta sin^{ \minus{} 4}\theta}d\theta$
=$ \int cot^{3/2}\theta cosec^{4}\theta d\theta... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$ \\int sin^{3/2}2\\theta sin \\theta d \\theta \\equal{} 2 \\sqrt{2} \\int sin^{5/2} \\theta cos ^{3/2} \\theta d \\theta$\r\n\r\nAfter that ...",
"content_html": "<img src=\"//latex.artofproblemsol... | \[
\int \sin^{3/2}(2\theta)\sin\theta\,d\theta = 2\sqrt{2}\int \sin^{5/2}\theta\cos^{3/2}\theta\,d\theta
\] | [
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aops_465803 | hello, for c) we get with integration by parts $\int x\tan(x)^2\,dx=x\tan(x)-\frac{1}{2}x^2-\frac{1}{2}\ln(1+\tan(x)^2)$ and the value of our integral is $1/4\,\pi -1/32\,{\pi }^{2}-1/2\,\ln \left( 2 \right) $
Sonnhard. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "(a) $\\int_{\\frac{\\pi}{4}}^{\\frac{\\pi}{3}} \\cot^{4} x\\ dx$;\n\n(b) $\\int_{0}^{\\frac{\\pi}{2}} \\frac{\\cos x}{\\sqrt{1+\\sin x}}\\ dx$;\n\n(c) $\\int_{0}^{\\frac{\\pi}{4}} x \\tan^{2} x\\ dx$;\n\n(... | (a) \[
\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot^{4} x\,dx
\]
(b) \[
\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sqrt{1+\sin x}}\,dx
\]
(c) \[
\int_{0}^{\frac{\pi}{4}} x\,\tan^{2} x\,dx
\]
(d) \[
\int_{0}^{\frac{\pi}{2}} e^{x}\cos 4x\,dx
\] | [
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aops_312314 | hello
$ I_{m,n}\equal{} \int_{0}^{\infty}\frac{x^{m}}{(x^{2}\plus{}a^{2})^{n}}\ dx$
using integration by parts
$ I_{m,n}\equal{}[\frac{x^{m\plus{}1}}{(m\plus{}1)(x^{2}\plus{}a^{2})^{n}}]_{0}^{\infty}\minus{}\int_{0}^{\infty}\frac{2x}{(m\plus{}1)(x^2\plus{}a^2)^{n\plus{}1}}.(\minus{}n)x^{m\plus{}1}dx$
replacing $ m... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $ I_{m,n}$ denotes $ \\int_{0}^{\\infty} \\frac{x^{m}}{(x^{2}\\plus{}a^{2})^{n}}\\ dx$, find a relation between $ I_{m,n}$ and $ I_{m\\minus{}2, n\\minus{}1}$, and state the values of $ m$ and $ n$ for ... | Let
\[
I_{m,n}=\int_{0}^{\infty}\frac{x^{m}}{(x^{2}+a^{2})^{n}}\,dx.
\]
(a) Find a relation between \(I_{m,n}\) and \(I_{m-2,n-1}\), and state the values of \(m\) and \(n\) for which this relation is valid.
(b) Hence evaluate
\[
\text{(i)}\quad \int_{0}^{\infty}\left(\frac{x}{x^{2}+a^{2}}\right)^{7}dx,\qquad
\text{(ii... | [
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aops_2284192 | The problem is straightforward: Let $t = - x$. Hence:
$$ \int^1_{-1} \frac{1}{(1+e^x)(3+x^2)}\,dx = \int^0_{-1} \frac{1}{(1+e^x)(3+x^2)}\,dx+\int^0_{1} \frac{1}{(1+e^x)(3+x^2)}\,dx$$$$= \int^1_{0} \frac{1}{(1+e^{-t})(3+t^2)}\,dt+\int^1_{0} \frac{1}{(1+e^x)(3+x^2)}\,dx= \int_{0}^{1} \frac{1}{x^2+3} dx = \frac{\pi}{6\sq... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Calculate \n$\\int^1_{-1} \\frac{1}{(1+e^x)(3+x^2)}\\,dx$",
"content_html": "Calculate<br>\n<img src=\"//latex.artofproblemsolving.com/c/4/e/c4e4a41ce06e48bcf003114face35274151ac9de.png\" class=\"lat... | Calculate
\[
\int_{-1}^{1}\frac{1}{(1+e^{x})(3+x^{2})}\,dx.
\] | [
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aops_248243 | [quote="kan"]How to integrate this.
$ \int x^2 \frac {(a \minus{} x)}{(a \plus{} x)} dx$.[/quote]
subsitute x = x-a and break the fraction in the integral into two parts. Expand-divide and then integrate. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How to integrate this.\r\n\r\n$ \\int x^2 \\frac{(a\\minus{}x)}{(a\\plus{}x)} dx$.",
"content_html": "How to integrate this.<br>\n<br>\n<span style=\"white-space:nowrap;\"><img src=\"//latex.artofpro... | How to integrate this:
\[
\int x^2 \frac{(a-x)}{(a+x)}\,dx.
\] | [
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aops_1079542 | With $t=\dfrac{\sqrt{x}}{1+x}$, we obtain $\int \frac{x-1}{x+1} \frac{1}{\sqrt{x(x^{2}+x+1)}} \ dx=C-2arcsin\left(\dfrac{\sqrt{x}}{1+x}\right)$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find $\\int \\frac{x-1}{x+1} \\frac{1}{\\sqrt{x(x^{2}+x+1)}} \\ dx$.",
"content_html": "Find <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolving.com/5/c/a/5ca0f5898eac6bf733a2d... | Find
\[
\int \frac{x-1}{x+1}\cdot\frac{1}{\sqrt{x(x^{2}+x+1)}}\,dx.
\] | [
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aops_2794917 | [hide=Wolfram alpha gives]
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"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the value of\n\n$\\int_{0}^{1} (1- x^2) ^{2017} dx$",
"content_html": "Find the value of<br>\n<br>\n<img src=\"//latex.artofproblemsolving.com/7/b/0/7b048f1891c91993c737f21f1c02a5cd4f66aebd.png\... | Find the value of
\[
\int_{0}^{1} (1-x^2)^{2017}\,dx.
\] | [
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aops_254513 | I think there is only one thing one could possibly do:
[hide]Let $ x = 1/y$, so $ dx = -\frac{dy}{y^2}$. Then we have
\begin{align*}
I & : = \int_{\frac{1}{a}}^{a}{\frac{(x+1)f(x)}{x\sqrt{x^{2}+1}}} dx \\
& = \int_{a}^{\frac{1}{a}}{\frac{(\frac{1}{y} + 1)f(\frac{1}{y})}{\frac{1}{y}\sqrt{\frac{1}{y^{2}}+1}}} \frac{-d... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "hi,\r\nlet $ a>1$ and a function $ f: [\\frac{1}{a},a] \\rightarrow \\mathbb{R}$ s.t. \r\n\\[ f(x) \\plus{} f(\\frac{1}{x}) \\equal{} k , \\forall x \\in [\\frac{1}{a},a].\\]\r\nCalculate:\r\n\\[ \\int_{\\... | Let a > 1 and let f: [1/a, a] → ℝ satisfy
\[ f(x)+f\!\left(\frac{1}{x}\right)=k \quad\text{for all } x\in\left[\frac{1}{a},a\right]. \]
Calculate
\[
\int_{1/a}^{a}\frac{(x+1)f(x)}{x\sqrt{x^{2}+1}}\,dx.
\] | [
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aops_1457090 | Just use that
\[\int_0^{\pi} xf(\sin x) dx=\int_0^{\pi} (\pi-x) f(\sin (\pi-x)) dx=\int_0^{\pi} (\pi-x) f(\sin x) dx=\frac{1}{2}\left(\int_0^{\pi} xf(\sin x) dx+\int_0^{\pi} (\pi-x) f(\sin x) dx\right)=\frac{\pi}{2} \int_0^{\pi} f(\sin x) dx\] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $f$ be continuous function from set of reals to reals.Prove the following:$$\\int_{0}^{\\pi}xf(sin x)dx=\\frac{\\pi}{2}\\int_{0}^{\\pi}f(sinx)dx$$",
"content_html": "Let <img src=\"//latex.artofp... | Let \(f:\mathbb{R}\to\mathbb{R}\) be continuous. Prove that
\[
\int_{0}^{\pi} x\,f(\sin x)\,dx=\frac{\pi}{2}\int_{0}^{\pi} f(\sin x)\,dx.
\] | [
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aops_1370925 | [quote=jonny]$\int^{\frac{\pi}{2}}_{0}\frac{1}{\left(\sqrt[n]{\sin x}+\sqrt[n]{\cos x}\right)^{2n}}dx,n \in \mathbb{W}$[/quote]
\begin{align*}
\int_{0}^{\pi/2} \frac{{\rm d}x}{\left ( \sqrt[n]{\cos x} + \sqrt[n]{\sin x} \right )^{2n}} &\overset{1+\sqrt[n]{\tan x} = \frac{1}{y}}{=\!=\!=\!=\!=\!=\!=\!=\! } n \int_{0}^{... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$\\int^{\\frac{\\pi}{2}}_{0}\\frac{1}{\\left(\\sqrt[n]{\\sin x}+\\sqrt[n]{\\cos x}\\right)^{2n}}dx,n \\in \\mathbb{W}$",
"content_html": "<img src=\"//latex.artofproblemsolving.com/8/2/5/82525eec73c2... | \[
\text{Evaluate }\int_{0}^{\frac{\pi}{2}}\frac{1}{\bigl(\sin^{1/n}x+\cos^{1/n}x\bigr)^{2n}}\,dx,\qquad n\in\mathbb{W}.
\] | [
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aops_504677 | $\int{\frac{\sqrt{\ln \left( x+\sqrt{1+{{x}^{2}}} \right)+5}}{\sqrt{1+{{x}^{2}}}}}dx$
$=\int{\left( \sqrt{\ln \left( x+\sqrt{1+{{x}^{2}}} \right)+5} \right)}d\left[ \ln \left( x+\sqrt{1+{{x}^{2}}} \right)+5 \right]$
$=\frac{2}{3}{{\left[ \ln \left( x+\sqrt{1+{{x}^{2}}} \right)+5 \right]}^{\frac{3}{2}}}+C$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate : $\\int {\\frac{\\sqrt{\\ln \\left( x+\\sqrt{1+{{x}^{2}}} \\right)+5}}{\\sqrt{1+{{x}^{2}}}}dx}$\nI try to subsitute $x=\\tan t$ and $t=\\sqrt{1+x^2}$. But it seems not working....",
"conten... | Evaluate:
\[
\int \frac{\sqrt{\ln\!\bigl(x+\sqrt{1+x^{2}}\bigr)+5}}{\sqrt{1+x^{2}}}\,dx.
\] | [
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aops_15410 | The answer can be expressed as $\displaystyle \frac{\Gamma(5/3)\Gamma(1/2)}{2\Gamma(13/6)}$
Of course, by the usual Calculus II-Calculus III understanding of the idea of "closed form", only $\Gamma(1/2) = \frac{\sqrt{\pi}}2$ would count as "closed form", not $\Gamma$(5/3) or $\Gamma$(13/6).
In the original integr... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Hi. Does any one can solve this problem by hand.? I try many ways but can't solve. :D \r\n\r\nintegrate</sup> (1 - x<sup>2</sup>)<sup>2/3</sup> dx\r\nintegrate from x = 0 to x = 1 :maybe:",
"content_... | Compute the definite integral
\[
\int_{0}^{1} (1 - x^{2})^{2/3}\,dx.
\] | [
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aops_3253917 | We have $\int_0^1 x\ln ^2 x (1-2x+3x^2-\cdots)dx=a_1-2a_2+3a_3-\cdots$ where $a_n=\int_0^1 x^n \ln ^2 dx$.
Partial integration, twice, yields $a_n={2\over {(n+1)^3}}$.
Hence we get $2(1/2^3-2/3^3+3/4^3-\cdots)=2( {{2-1}\over {2^3}}-{{3-1}\over {3^3}}+\dots)=2(1/2^2-1/3^2+\cdots)-2(1/2^3-1/3^3+\cdots)$.
We have $a=\pi^2... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$$\\int_{0}^{1} \\frac{x \\ln^2(x)}{(1+x)^2} \\, dx$$",
"content_html": "<img src=\"//latex.artofproblemsolving.com/c/5/3/c53a5d951ddf1c6da77f5f38eddd862ef2c760b3.png\" class=\"latexcenter\" alt=\"$$... | \[
\int_{0}^{1} \frac{x\ln^2(x)}{(1+x)^2}\,dx
\] | [
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aops_288287 | [quote]NORA.91 wrote
Simple Integration :)
2 questions
(1) $ \int \frac {x^2}{(x \sin x \plus{} \cos x)^2} \ dx$
(2) $ \int \frac {dx}{2 \sin x \plus{} \sec x }$
[/quote]
1)
$ \int \frac {x^2}{(x \sin x \plus{} \cos x)^2} \ dx \equal{} \int \frac {x^2 cosx }{ ( x \sin x \plus{} \cos x )^{2} \cos x }$
$ (... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "[b](1)[/b] $ \\int \\frac{x^2}{(x \\sin x \\plus{} \\cos x)^2} \\ dx$\r\n\r\n[b](2)[/b] $ \\int \\frac{dx}{2 \\sin x \\plus{} \\sec x }$",
"content_html": "<b>(1)</b> <img src=\"//latex.artofproblems... | 1. \[
\int \frac{x^2}{\bigl(x\sin x+\cos x\bigr)^2}\,dx
\]
2. \[
\int \frac{dx}{2\sin x+\sec x}
\] | [
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aops_485241 | The original integral is \[ I = \int_{x=0}^{\pi/2} \frac{\cos x + 4}{3 \cos x + 4 \sin x + 25} \, dx. \] The technique that comes to mind is the substitution $u = \tan \frac{x}{2}$, which turns the integrand into a rational function, via \[ \sin x = \frac{2u}{1+u^2}, \quad \cos x = \frac{1-u^2}{1+u^2}, \quad dx = \fra... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "This is my first time at AoPS and here is my problem:\n\nSolve D= Integral from 0 to pi/2 of [cosx + 4] dx / [3cosx + 4sinx + 25]\n\nI tried dividing everything by cosx, using partial integrals and the sub... | Solve
\[
D=\int_{0}^{\pi/2}\frac{\cos x+4}{3\cos x+4\sin x+25}\,dx.
\] | [
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aops_133311 | You might as well write the integral as
$\int\frac{dx}{(x^{2}+3)^{3/2}}$
That substitution makes no sense to me. The "system" substitution would be $x=\sqrt{3}\sec \theta.$ Since we eventually get an answer with no trig functions, in hindsight there would have been a non-trig method. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Consider the substitution $x+t=\\sqrt{x^{2}+3}$\r\n\r\n$\\int_{0}^{1}\\frac{1}{(x^{2}+3)\\sqrt{x^{2}+3}}\\,dx$",
"content_html": "Consider the substitution <img src=\"//latex.artofproblemsolving.com/... | Consider the substitution \(x+t=\sqrt{x^{2}+3}\).
\[
\int_{0}^{1}\frac{1}{(x^{2}+3)\sqrt{x^{2}+3}}\,dx
\] | [
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