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class | __index_level_0__ int64 0 742k |
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## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{4 x^{3}+x^{2}+2}{x(x-1)(x-2)} d x
$$ | ## Solution
$$
\int \frac{4 x^{3}+x^{2}+2}{x(x-1)(x-2)} d x=\int \frac{4 x^{3}+x^{2}+2}{x^{3}-3 x^{2}+2 x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
4 x^{3}+x^{2}+2 & \mid x^{3}-3 x^{2}+2 x \\
\frac{4 x^{3}-12 x^{2}+8 x}{13 x^{2}-8 x+2} & 4
\end{a... | 4x+\ln|x|-7\cdot\ln|x-1|+19\cdot\ln|x-2|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,691 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{3 x^{3}-2}{x^{3}-x} d x
$$ | ## Solution
$$
\int \frac{3 x^{3}-2}{x^{3}-x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
3 x^{3}-2 & \mid x^{3}-x \\
\frac{3 x^{3}-3 x}{3 x}-2 & 3
\end{array}
$$
We get:
$$
=\int\left(3+\frac{3 x-2}{x^{3}-x}\right) d x=\int 3 d x+\int \frac{3 x-2... | 3x+2\ln|x|+\frac{1}{2}\ln|x-1|-\frac{5}{2}\ln|x+1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,692 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-3 x^{2}-12}{(x-4)(x-2) x} d x
$$ | ## Solution
$$
\int \frac{x^{3}-3 x^{2}-12}{(x-4)(x-2) x} d x=\int \frac{x^{3}-3 x^{2}-12}{x^{3}-6 x^{2}+8 x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
x^{3}-3 x^{2}-12 & \mid x^{3}-6 x^{2}+8 x \\
\frac{x^{3}-6 x^{2}+8 x}{3 x^{2}-8 x}-12
\end{arra... | x+\frac{1}{2}\cdot\ln|x-4|+4\cdot\ln|x-2|-\frac{3}{2}\cdot\ln|x|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,693 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{5}-x^{3}+1}{x^{2}-x} d x
$$ | ## Solution
$$
\int \frac{x^{5}-x^{3}+1}{x^{2}-x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
\frac{x^{5}-x^{3}+1}{x^{5}-x^{4}} & \frac{\mid x^{2}-x}{x^{3}+x^{2}} \\
\frac{x^{4}-x^{3}+1}{1} &
\end{array}
$$
We get:
$$
=\int\left(x^{3}+x^{2}\right)... | \frac{x^{4}}{4}+\frac{x^{3}}{3}-\ln|x|+\ln|x-1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,694 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{5}+3 x^{3}-1}{x^{2}+x} d x
$$ | ## Solution
$$
\int \frac{x^{5}+3 x^{3}-1}{x^{2}+x} d x=
$$
Under the integral is an improper fraction. Let's separate the integer part:
$$
\begin{aligned}
& x^{5}+3 x^{3}-1 \\
& \frac{x^{5}+x^{4}}{-x^{4}+3 x^{3}-1} \\
& \frac{-x^{4}-x^{3}}{4 x^{3}-1} \\
& \frac{4 x^{3}+4 x^{2}}{-4 x^{2}}-1 \\
& \quad \frac{-4 x^{2}... | \frac{x^{4}}{4}-\frac{x^{3}}{3}+2x^{2}-4x-\ln|x|+5\cdot\ln|x+1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,695 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{5}-8 x^{3}+3}{x^{2}-2 x} d x
$$ | ## Solution
$$
\int \frac{2 x^{5}-8 x^{3}+3}{x^{2}-2 x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{aligned}
& \begin{array}{ll}
2 x^{5}-8 x^{3}+3 & \frac{\mid x^{2}-2 x}{2 x^{3}+4 x^{2}} \\
\frac{2 x^{5}-4 x^{4}}{4 x^{4}-8 x^{3}+3}
\end{array} \\
& \frac{4 x^... | \frac{x^{4}}{2}+\frac{4x^{3}}{3}-\frac{3}{2}\cdot\ln|x|+\frac{3}{2}\cdot\ln|x-2|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,696 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{3 x^{5}-12 x^{3}-7}{x^{2}+2 x} d x
$$ | ## Solution
$$
\int \frac{3 x^{5}-12 x^{3}-7}{x^{2}+2 x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{aligned}
& \begin{array}{ll}
3 x^{5}-12 x^{3}-7 \\
\frac{3 x^{5}+6 x^{4}}{-6 x^{4}}-12 x^{3}-7
\end{array} \quad \frac{\mid x^{2}+2 x}{3 x^{3}-6 x^{2}} \\
& \f... | \frac{3x^{4}}{4}-2x^{3}-\frac{7}{2}\cdot\ln|x|+\frac{7}{2}\cdot\ln|x+2|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,697 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{-x^{5}+9 x^{3}+4}{x^{2}+3 x} d x
$$ | ## Solution
$$
\int \frac{-x^{5}+9 x^{3}+4}{x^{2}+3 x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
-x^{5}+9 x^{3}+4 \\
\frac{-x^{5}-3 x^{4}}{3 x^{4}+9 x^{3}+4} & \frac{\mid x^{2}+3 x}{-x^{3}+3 x^{2}} \\
\frac{3 x^{4}+9 x^{3}}{4} &
\end{array}
$$
We... | -\frac{x^{4}}{4}+x^{3}+\frac{4}{3}\cdot\ln|x|-\frac{4}{3}\cdot\ln|x+3|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,698 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{-x^{5}+25 x^{3}+1}{x^{2}+5 x} d x
$$ | ## Solution
$$
\int \frac{-x^{5}+25 x^{3}+1}{x^{2}+5 x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{aligned}
& \begin{array}{ll}
-x^{5}+25 x^{3}+1 & \frac{\mid x^{2}+5 x}{-x^{3}+5 x^{2}} \\
\frac{-x^{5}-5 x^{4}}{5 x^{4}+25 x^{3}+1} &
\end{array} \\
& 5 x^{4}+2... | -\frac{x^{4}}{4}+\frac{5x^{3}}{3}+\frac{1}{5}\cdot\ln|x|-\frac{1}{5}\cdot\ln|x+5|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,699 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-5 x^{2}+5 x+23}{(x-1)(x+1)(x-5)} d x
$$ | ## Solution
$$
\int \frac{x^{3}-5 x^{2}+5 x+23}{(x-1)(x+1)(x-5)} d x=\int \frac{x^{3}-5 x^{2}+5 x+23}{x^{3}-5 x^{2}-x+5} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{aligned}
& x^{3}-5 x^{2}+5 x+23 \quad \mid x^{3}-5 x^{2}-x+5 \\
& \frac{x^{3}-5 x^{2}-x+5}{6 x+... | x-3\cdot\ln|x-1|+\ln|x+1|+2\cdot\ln|x-5|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,700 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{5}+2 x^{4}-2 x^{3}+5 x^{2}-7 x+9}{(x+3)(x-1) x} d x
$$ | ## Solution
$$
\int \frac{x^{5}+2 x^{4}-2 x^{3}+5 x^{2}-7 x+9}{(x+3)(x-1) x} d x=\int \frac{x^{5}+2 x^{4}-2 x^{3}+5 x^{2}-7 x+9}{x^{3}+2 x^{2}-3 x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{cc}
x^{5}+2 x^{4}-2 x^{3}+5 x^{2}-7 x+9 & \frac{\mid x^{3}+2 ... | \frac{x^{3}}{3}+x+4\cdot\ln|x+3|+2\cdot\ln|x-1|-3\cdot\ln|x|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,701 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{4}-5 x^{2}-8 x-8}{x(x-2)(x+2)} d x
$$ | ## Solution
$$
\int \frac{2 x^{4}-5 x^{2}-8 x-8}{x(x-2)(x+2)} d x=\int \frac{2 x^{4}-5 x^{2}-8 x-8}{x^{3}-4 x} d x=
$$
The integrand is an improper fraction. We will separate the integer part:
$$
\begin{array}{ll}
2 x^{4}-5 x^{2}-8 x-8 & \left\lvert\, \frac{x^{3}-4 x}{2 x}\right. \\
\frac{2 x^{4}-8 x^{2}}{3 x^{2}-8 ... | x^{2}+2\cdot\ln|x|-\frac{3}{2}\cdot\ln|x-2|+\frac{5}{2}\cdot\ln|x+2|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,702 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{4 x^{4}+2 x^{2}-x-3}{x(x-1)(x+1)} d x
$$ | ## Solution
$$
\int \frac{4 x^{4}+2 x^{2}-x-3}{x(x-1)(x+1)} d x=\int \frac{4 x^{4}+2 x^{2}-x-3}{x^{3}-x} d x=
$$
The integrand is an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
4 x^{4}+2 x^{2}-x-3 & \left\lvert\, \frac{x^{3}-x}{4 x}\right. \\
\frac{4 x^{4}-4 x^{2}}{6 x^{2}-x-3}
\end{arra... | 2x^{2}+3\cdot\ln|x|+\ln|x-1|+2\cdot\ln|x+1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,703 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{3 x^{4}+3 x^{3}-5 x^{2}+2}{x(x-1)(x+2)} d x
$$ | ## Solution
$$
\int \frac{3 x^{4}+3 x^{3}-5 x^{2}+2}{x(x-1)(x+2)} d x=\int \frac{3 x^{4}+3 x^{3}-5 x^{2}+2}{x^{3}+x^{2}-2 x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
3 x^{4}+3 x^{3}-5 x^{2}+2 & \mid x^{3}+x^{2}-2 x \\
\frac{3 x^{4}+3 x^{3}-6 x^{2... | \frac{3x^{2}}{2}-\ln|x|+\ln|x-1|+\ln|x+2|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,704 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{4}+2 x^{3}-41 x^{2}+20}{x(x-4)(x+5)} d x
$$ | ## Solution
$$
\int \frac{2 x^{4}+2 x^{3}-41 x^{2}+20}{x(x-4)(x+5)} d x=\int \frac{2 x^{4}+2 x^{3}-41 x^{2}+20}{x^{3}+x^{2}-20 x} d x=
$$
The integrand is an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
2 x^{4}+2 x^{3}-41 x^{2}+20 & \mid x^{3}+x^{2}-20 x \\
\frac{2 x^{4}+2 x^{3}-40 x^{2}}... | x^2-\ln|x|+\frac{1}{9}\ln|x-4|-\frac{1}{9}\ln|x+5|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,705 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{5}-x^{4}-6 x^{3}+13 x+6}{x(x-3)(x+2)} d x
$$ | ## Solution
$$
\int \frac{x^{5}-x^{4}-6 x^{3}+13 x+6}{x(x-3)(x+2)} d x=\int \frac{x^{5}-x^{4}-6 x^{3}+13 x+6}{x^{3}-x^{2}-6 x} d x=
$$
The integrand is an improper fraction. Let's separate the integer part:
(x-2)} d x
$$ | ## Solution
$$
\int \frac{3 x^{3}-x^{2}-12 x-2}{x(x+1)(x-2)} d x=\int \frac{3 x^{3}-x^{2}-12 x-2}{x^{3}-x^{2}-2 x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
3 x^{3}-x^{2}-12 x-2 & \mid x^{3}-x^{2}-2 x \\
\frac{3 x^{3}-3 x^{2}-6 x}{2 x^{2}-6 x-2} &... | 3x+\ln|x|+2\cdot\ln|x+1|-\ln|x-2|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,707 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{4}+2 x^{3}-3 x^{2}+2 x-9}{x(x-1)(x+3)} d x
$$ | ## Solution
$$
\int \frac{2 x^{4}+2 x^{3}-3 x^{2}+2 x-9}{x(x-1)(x+3)} d x=\int \frac{2 x^{4}+2 x^{3}-3 x^{2}+2 x-9}{x^{3}+2 x^{2}-3 x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
2 x^{4}+2 x^{3}-3 x^{2}+2 x-9 & \frac{\mid x^{3}+2 x^{2}-3 x}{2 x-2} \... | x^2-2x+3\ln|x|-\frac{3}{2}\ln|x-1|+\frac{11}{2}\ln|x+3|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,708 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}-x^{2}-7 x-12}{x(x-3)(x+1)} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}-x^{2}-7 x-12}{x(x-3)(x+1)} d x=\int \frac{2 x^{3}-x^{2}-7 x-12}{x^{3}-2 x^{2}-3 x} d x=
$$
The integrand is an improper fraction. Let's separate the integer part:
$$
\begin{aligned}
& 2 x^{3}-x^{2}-7 x-12 \\
& \frac{2 x^{3}-4 x^{2}-6 x}{3 x^{2}-x-12}
\end{aligned}
$$
$$
\frac{\mid ... | 2x+4\cdot\ln|x|+\ln|x-3|-2\cdot\ln|x+1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,709 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}-40 x-8}{x(x+4)(x-2)} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}-40 x-8}{x(x+4)(x-2)} d x=\int \frac{2 x^{3}-40 x-8}{x^{3}+2 x^{2}-8 x} d x=
$$
The integrand is an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
2 x^{3}-40 x-8 & \mid x^{3}+2 x^{2}-8 x \\
\frac{2 x^{3}+4 x^{2}-16 x}{-4 x^{2}-24 x-8} & 2
\end{array}
$$
We ... | 2x+\ln|x|+\ln|x+4|-6\cdot\ln|x-2|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,710 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{9-n^{3}}{1+2 n^{3}}, a=-\frac{1}{2}$ | ## Solution
By the definition of the limit:
$$
\begin{aligned}
& \forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{18-2 n^{3}+1+2 n^{3}}{2\left(1+2 n^{3}\right)}\right| \\
& \left|\frac{19}{2\left(1+2 n^{3}\right)}\right| \\
& \frac{19... | N(\varepsilon)=[\sqrt[3]{\frac{1}{2}(\frac{19}{2\varepsilon}-1)}]+1 | Calculus | proof | Yes | Yes | olympiads | false | 47,711 |
Condition of the problem
$$
\lim _{n \rightarrow \infty} \frac{(1+2 n)^{3}-8 n^{5}}{(1+2 n)^{2}+4 n^{2}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow x} \frac{(1+2 n)^{3}-8 n^{5}}{(1+2 n)^{2}+4 n^{2}}=\lim _{n \rightarrow x} \frac{1+6 n+12 n^{2}+8 n^{3}-8 n^{5}}{1+4 n+4 n^{2}+4 n^{2}}= \\
& =\lim _{n \rightarrow x} \frac{1+6 n+12 n^{2}}{1+4 n+8 n^{2}}=\lim _{n \rightarrow x} \frac{n^{2}\left(\frac{1}{n^{2}}+\fra... | 1.5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,712 |
## Problem Statement
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt{n^{2}+2}}{\sqrt[4]{4 n^{4}+1}-\sqrt[3]{n^{4}-1}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt{n^{2}+2}}{\sqrt[4]{4 n^{4}+1}-\sqrt[3]{n^{1}-1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+2}-\sqrt{n^{2}+2}\right)}{\frac{1}{n}\left(\sqrt[4]{4 n^{\frac{1}{4}+1}}-\sqrt[3]{n^{4}-1}\right)}= \\
& =\lim _{n \rightarro... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,713 |
Condition of the problem
Prove, find $\delta(\varepsilon)$ :
$$
\lim _{x \rightarrow-\frac{1}{3}} \frac{9 x^{2}-1}{x+\frac{1}{3}}=-6
$$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). The number
_{\text {is continuous at the point }} x_{0 \text { (find }} \delta(\varepsilon)_{)}$:
$f(x)=-4 x^{2}-7, x_{0}=1$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$. $\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$ when $\left|x-x_{0}\right|<\delt... | proof | Calculus | proof | Yes | Yes | olympiads | false | 47,717 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x+x^{5}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x+x^{5}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{1^{: 5}+3 \cdot 1^{2} \cdot x^{2}+3 \cdot 1 \cdot x^{2}+x^{3}-1-3 x}{x\left(1+x^{4}\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1+3 x+3 x^{2}+x^{3}-1-3 x}{x\left(1+x^{4}\... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,718 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 8} \frac{\sqrt{9+2 x}-5}{\sqrt[3]{x}-2}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 8} \frac{\sqrt{9+2 x}-5}{\sqrt[3]{x}-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 8} \frac{(\sqrt{9+2 x}-5)(\sqrt{9+2 x}+5)}{(\sqrt[3]{x}-2)(\sqrt{9+2 x}+5)}= \\
& =\lim _{x \rightarrow 8} \frac{9+2 x-25}{(\sqrt[3]{x}-2)(\sqrt{9+2 x}+5)}=\lim _{x \rightarrow ... | \frac{12}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,719 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{1-\cos ^{3} x}{4 x^{2}}
$$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$$
1-\cos x \sim \frac{x^{2}}{2}, \text { as } x \rightarrow 0
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1-\cos ^{3} x}{4 x^{2}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{(1-\cos x)\left(1+\cos x+\cos ^{2} x\ri... | \frac{3}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,720 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\sin ^{2} x-\tan ^{2} x}{(x-\pi)^{4}}$ | ## Solution
$\lim _{x \rightarrow \pi} \frac{\sin ^{2} x-\tan ^{2} x}{(x-\pi)^{4}}=\lim _{x \rightarrow \pi} \frac{\frac{\sin ^{2} x \cdot \cos ^{2} x}{\cos ^{2} x}-\tan ^{2} x}{(x-\pi)^{4}}=$
$=\lim _{x \rightarrow \pi} \frac{\tan ^{2} x \cdot \cos ^{2} x-\tan ^{2} x}{(x-\pi)^{4}}=\lim _{x \rightarrow \pi} \frac{\ta... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,721 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 3} \frac{\sin \left(\sqrt{2 x^{2}-3 x-5}-\sqrt{1+x}\right)}{\ln (x-1)-\ln (x+1)+\ln 2}$ | ## Solution
Substitution:
$x=y+3 \Rightarrow y=x-3$
$x \rightarrow 3 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 3} \frac{\sin \left(\sqrt{2 x^{2}-3 x-5}-\sqrt{1+x}\right)}{\ln (x-1)-\ln (x+1)+\ln 2}= \\
& =\lim _{y \rightarrow 0} \frac{\sin \left(\sqrt{2(y+3)^{2}-3(y+3)-5}-\sqrt... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,722 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{3^{5 x}-2^{x}}{x-\sin 9 x}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{3^{5 x}-2^{x}}{x-\sin 9 x}=\lim _{x \rightarrow 0} \frac{\left(243^{x}-1\right)-\left(2^{x}-1\right)}{x-\sin 9 x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 243}\right)^{x}-1\right)-\left(\left(e^{\ln 2}\right)^{x}-1\right)}{x-\sin 9 x}= \\
& ... | \frac{1}{8}\ln\frac{2}{243} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,723 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-1}{e^{x^{2}}-1}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-1}{e^{x^{2}}-1}= \\
& =\lim _{x \rightarrow 0} \frac{(\sqrt{1+x \sin x}-1)(\sqrt{1+x \sin x}+1)}{\left(e^{x^{2}}-1\right)(\sqrt{1+x \sin x}+1)}= \\
& =\lim _{x \rightarrow 0} \frac{1+x \sin x-1}{\left(e^{x^{2}}-1\right)(\sqrt{1+x \sin x}+... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,724 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}(1-\ln (1+\sqrt[3]{x}))^{\frac{x}{\sin ^{4} \sqrt[3]{x}}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}(1-\ln (1+\sqrt[3]{x}))^{\frac{\sin ^{4} \sqrt[3]{x}}{x}}= \\
& =\lim _{x \rightarrow 0}\left(e^{\ln (1-\ln (1+\sqrt[3]{x})))^{\frac{x}{\sin ^{4} \sqrt[3]{x}}}=}\right. \\
& =\lim _{x \rightarrow 0} e^{\frac{x \ln (1-\ln (1+\sqrt[3]{x}))}{\sin ^{4} \sqrt[3]{x}}}=... | e^{-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,725 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(\frac{\ln (1+x)}{6 x}\right)^{\frac{x}{x+2}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{\ln (1+x)}{6 x}\right)^{\frac{x}{x+2}}=\left(\lim _{x \rightarrow 0} \frac{\ln (1+x)}{6 x}\right)^{\lim _{x \rightarrow 0} \frac{x}{x+2}}= \\
& =\left(\lim _{x \rightarrow 0} \frac{\ln (1+x)}{6 x}\right)^{\frac{0}{0+2}}=\left(\lim _{x \rightarrow 0} \... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,726 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{1 /(\sqrt[5]{x}-1)}
$$ | ## Solution
$\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\frac{1}{\sqrt[5]{x}-1}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{2 x-1}{x}\right)}\right)^{\frac{1}{\sqrt[5]{x}-1}}=$
$=\lim _{x \rightarrow 1} e^{\frac{1}{\sqrt[5]{x}-1} \cdot \ln \left(\frac{2 x-1}{x}\right)}=\exp \left\{\lim _{x \rightarrow ... | e^5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,727 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 3}\left(2-\frac{x}{3}\right)^{\sin (\pi x)}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 3}\left(2-\frac{x}{3}\right)^{\sin (\pi x)}=\lim _{x \rightarrow 3}\left(e^{\ln \left(2-\frac{x}{3}\right)}\right)^{\sin (\pi x)}= \\
& =\lim _{x \rightarrow 3} e^{\sin (\pi x) \cdot \ln \left(2-\frac{x}{3}\right)}=\exp \left\{\lim _{x \rightarrow 3} \sin (\pi x) \... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,728 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{d x}{(3 \operatorname{tg} x+5) \sin 2 x}
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which we get:
$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\operatorname{arctg} 3 \Rightarrow t=\operatorname{tg}(\operatorname{arctg} 3)... | \frac{1}{10}\ln\frac{12}{7} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,730 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\frac{4}{\sqrt{17}}}^{\pi / 4} \frac{2 \operatorname{ctg} x+1}{(2 \sin x+\cos x)^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{\arccos \frac{4}{\sqrt{17}}}^{\frac{\pi}{4}} \frac{2 \operatorname{ctg} x+1}{(2 \sin x+\cos x)^{2}} d x=\int_{\arccos \frac{4}{\sqrt{17}}}^{\frac{\pi}{4}} \frac{2 \operatorname{tg} x+1}{(2 \operatorname{tg} x+1)^{2} \cdot \cos ^{2} x} d x=\left|d(\operatorname{tg} x)=\frac{d x}{\... | 2\ln2-\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,731 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\arccos \frac{4}{\sqrt{17}}} \frac{3+2 \tan x}{2 \sin ^{2} x+3 \cos ^{2} x-1} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\arccos \frac{4}{\sqrt{17}}} \frac{3+2 \tan x}{2 \arccos ^{2} \frac{4}{\sqrt{17}}} d x=\int_{0}^{\arccos \frac{4}{\sqrt{17}}} \frac{3+2 \tan x}{2+\cos ^{2} x-1} d x= \\
= & \int_{0}^{2} \frac{3+2 \tan x}{1+\cos ^{2} x} d x=
\end{aligned}
$$
We will use the substitution:
$$
... | \frac{3}{\sqrt{2}}\arctan\frac{1}{4\sqrt{2}}+\ln\frac{33}{32} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,732 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{0}^{\arccos \frac{1}{\sqrt{17}}} \frac{3+2 \operatorname{tg} x}{2 \sin ^{2} x+3 \cos ^{2} x-1} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\arccos \frac{1}{\sqrt{17}}} \frac{3+2 \tan x}{2 \sin ^{2} x+3 \cos ^{2} x-1} d x=\int_{0}^{\arccos } \frac{1}{\sqrt{17}} \frac{3+2 \tan x}{2+\cos ^{2} x-1} d x= \\
= & \int_{0}^{\frac{\arccos }{\sqrt{17}}} \frac{3+2 \tan x}{1+\cos ^{2} x} d x=
\end{aligned}
$$
We will use t... | \frac{3}{\sqrt{2}}\arctan(2\sqrt{2})+\ln9 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,733 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{4 \operatorname{tg} x-5}{1-\sin 2 x+4 \cos ^{2} x} d x
$$ | ## Solution
$$
\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{4 \operatorname{tg} x-5}{1-\sin 2 x+4 \cos ^{2} x} d x
$$
We will use the substitution:
$$
t=\operatorname{tg} x
$$
From which:
$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}} \\
& \cos ^{2} x=\frac{1}{t^{2}+1} \\
& d x=\frac{d t}{1+t^{2}} \\
& x=\frac... | 2\ln2-\frac{\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,734 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{0}^{\operatorname{arctg} \frac{1}{3}} \frac{8+\operatorname{tg} x}{18 \sin ^{2} x+2 \cos ^{2} x} d x
$$ | ## Solution
$$
\int_{0}^{\operatorname{arctg} \frac{1}{3}} \frac{8+\operatorname{tg} x}{18 \sin ^{2} x+2 \cos ^{2} x} d x=
$$
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which we get:
$$
\begin{aligned}
& \sin ^{2} x=\frac{t^{2}}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
... | \frac{\pi}{3}+\frac{\ln2}{36} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,735 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\arccos \sqrt{2 / 3}} \frac{\tan x + 2}{\sin^2 x + 2 \cos^2 x - 3} \, dx
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\arccos \sqrt{2 / 3}} \frac{\operatorname{tg} x+2}{\sin ^{2} x+2 \cos ^{2} x-3} d x=\int_{0}^{\arccos \sqrt{2 / 3}} \frac{\operatorname{tg} x+2}{\cos ^{2} x-2} d x= \\
& =\int_{0}^{\arccos \sqrt{2 / 3}} \frac{\operatorname{tg} x+2}{\frac{1}{2} \cdot \cos 2 x-\frac{3}{2}} d x=... | -\frac{\ln2+\sqrt{2}\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,736 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\arcsin (1 / \sqrt{37})}^{\pi / 4} \frac{6 \operatorname{tg} x d x}{3 \sin 2 x+5 \cos ^{2} x}
$$ | ## Solution
$\int_{\arcsin \frac{1}{\sqrt{37}}}^{\frac{\pi}{4}} \frac{6 \operatorname{tg} x d x}{3 \sin 2 x+5 \cos ^{2} x}=\int_{\arcsin \frac{1}{\sqrt{37}}}^{\frac{\pi}{4}} \frac{6 \operatorname{tg} x d x}{6 \sin x \cdot \cos x+5 \cos ^{2} x}=$
$=\int_{\arcsin \frac{1}{\sqrt{37}}}^{\frac{\pi}{4}} \frac{6 \operatorna... | \frac{5}{6}\ln\frac{6\cdot\mathrm{e}}{11} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,737 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\operatorname{arctan}(1 / 3)}^{0} \frac{3 \tan x + 1}{2 \sin 2x - 5 \cos 2x + 1} d x
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which we get:
$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, \cos 2 x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=-\operatorname{arctg} \frac{1}{3} \Rightarrow t=\operatorname{tg}\left(-\operatorname{arctg} \frac{1}{3}\right)=-... | \frac{1}{4}\ln\frac{6}{7} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,739 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{1+\operatorname{ctg} x}{(\sin x+2 \cos x)^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{\pi / 4}^{\operatorname{arctg} 3} \frac{1+\operatorname{ctg} x}{(\sin x+2 \cos x)^{2}} d x=\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{1+\operatorname{ctg} x}{\sin ^{2} x+4 \sin x \cos x+4 \cos ^{2} x} d x= \\
& =\int_{\pi / 4}^{1+2 \sin 2 x+3 \cos ^{2} x} \frac{1+\operatorname... | \frac{1}{4}\ln\frac{9}{5}+\frac{1}{15} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,740 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\pi / 4}^{\arccos (1 / \sqrt{3})} \frac{\tan x}{\sin ^{2} x-5 \cos ^{2} x+4} d x
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which we get:
$$
\begin{aligned}
& \sin ^{2} x=\frac{t^{2}}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\arccos \frac{1}{\sqrt{3}} \Rightarrow t=\... | \frac{1}{10}\cdot\ln\frac{9}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,741 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi / 4} \frac{6 \sin ^{2} x}{3 \cos 2 x-4} d x
$$ | ## Solution
## Solution №1
We will prove the lemma:
$$
\begin{aligned}
& \int \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}=\frac{1}{a b} \operatorname{arctg}\left(\frac{b}{a} \operatorname{tg} x\right)+C \\
& \int \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}=\int \frac{\frac{1}{\cos ^{2} x} d x}{a^{2}+b^{2} \o... | -\frac{\pi}{4}+\frac{1}{\sqrt{7}}\cdot\operatorname{arctg}\sqrt{7} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,742 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\operatorname{arctan} 3} \frac{4+\operatorname{tan} x}{2 \sin ^{2} x+18 \cos ^{2} x} d x
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which we get:
$$
\begin{aligned}
& \sin ^{2} x=\frac{t^{2}}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\operatorname{arctg} 3 \Rightarrow t=\operatorname{tg}(\operatorna... | \frac{\pi}{6}+\frac{\ln2}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,743 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\operatorname{arctan} 2} \frac{12+\operatorname{tan} x}{3 \sin ^{2} x+12 \cos ^{2} x} d x
$$ | ## Solution
We will use the substitution:
$$
t=\operatorname{tg} x
$$
From which:
$$
\begin{aligned}
& \sin ^{2} x=\frac{t^{2}}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\operatorname{arctg} 2 \Rightarrow t=\operatorname{tg}(\operatorname{ar... | \frac{\pi}{2}+\frac{\ln2}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,744 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\operatorname{arctan}(2 / 3)} \frac{6+\operatorname{tan} x}{9 \sin ^{2} x+4 \cos ^{2} x} d x
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which we get:
$$
\begin{aligned}
& \sin ^{2} x=\frac{t^{2}}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\operatorname{arctg} \frac{2}{3} \Rightarrow t=\operatorname{tg}\l... | \frac{\pi}{4}+\frac{\ln2}{18} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,745 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\arcsin \sqrt{3 / 7}} \frac{\tan^{2} x \, dx}{3 \sin ^{2} x + 4 \cos ^{2} x - 7}
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which we get:
$$
\begin{aligned}
& \sin ^{2} x=\frac{t^{2}}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\arcsin \sqrt{\frac{3}{7}} \Rightarrow t=\operatorname{tg}\left(\a... | -\frac{\sqrt{3}}{8}+\frac{\sqrt{3}\pi}{32} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,746 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{0}^{\pi / 4} \frac{7+3 \tan x}{(\sin x+2 \cos x)^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\pi / 4} \frac{7+3 \tan x}{(\sin x+2 \cos x)^{2}} d x=\int_{0}^{\pi / 4} \frac{7+3 \tan x}{\sin ^{2} x+4 \sin x \cos x+4 \cos ^{2} x} d x= \\
& =\int_{0}^{\pi / 4} \frac{7+3 \tan x}{1+2 \sin 2 x+3 \cos ^{2} x} d x=
\end{aligned}
$$
We will use the substitution:
$$
t=\tan x
... | 3\ln\frac{3}{2}+\frac{1}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,747 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\arcsin (2 / \sqrt{5})}^{\arcsin (3 / \sqrt{10})} \frac{2 \tan x + 5}{(5 - \tan x) \sin 2x} \, dx
$$ | ## Solution
We will use the substitution:
$$
t=\operatorname{tg} x
$$
From which:
$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\arcsin \frac{2}{\sqrt{5}} \Rightarrow t=\operatorname{tg}\left(\arcsin \frac{2}{\sqrt{5}}\right)=\sqrt{\frac{1}{1-\sin ^{2}\left(\arcsin \frac{2}{\sqrt... | 2\ln\frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,748 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{0}^{\pi / 4} \frac{5 \operatorname{tg} x+2}{2 \sin 2 x+5} d x
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which:
$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1
\end{aligned}
$$
Substitute:
$$
\begin{aligne... | \frac{1}{2}\ln\frac{14}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,750 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{\pi / 4}^{\arcsin (2 / \sqrt{5})} \frac{4 \operatorname{tg} x-5}{4 \cos ^{2} x-\sin 2 x+1} d x
$$ | ## Solution
We will use the substitution:
$$
t=\operatorname{tg} x
$$
From which:
$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\arcsin \frac{2}{\sqrt{5}} \Rightarrow t=\operatorna... | 2\ln\frac{5}{4}-\frac{1}{2}\operatorname{arctg}\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,751 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\arcsin \sqrt{7 / 8}} \frac{6 \sin ^{2} x}{4+3 \cos 2 x} d x
$$ | ## Solution
We will use the substitution:
$$
t=\operatorname{tg} x
$$
From which:
$$
\begin{aligned}
& \sin ^{2} x=\frac{t^{2}}{1+t^{2}}, \cos 2 x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\arcsin \sqrt{\frac{7}{8}} \Rightarrow t=\operatorname{tg}\left(\arc... | \frac{\sqrt{7}\pi}{4}-\operatorname{arctg}\sqrt{7} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,752 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\arccos (1 / \sqrt{5})}^{0} \frac{11-3 \tan x}{\tan x+3} d x
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which we get:
$$
\begin{aligned}
& d x=\frac{d t}{1+t^{2}} \\
& x=-\arccos \frac{1}{\sqrt{5}} \Rightarrow t=\operatorname{tg}\left(-\arccos \frac{1}{\sqrt{5}}\right)=-\operatorname{tg}\left(\arccos \frac{1}{\sqrt{5}}\right)= \\
& =-\sqrt{\frac... | \ln45+3\operatorname{arctg}2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,753 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\arcsin (3 / \sqrt{10})} \frac{2 \tan x-5}{(4 \cos x-\sin x)^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \arcsin (3 / \sqrt{10}) \quad \arcsin (3 / \sqrt{10}) \\
& \int_{0} \frac{2 \operatorname{tg} x-5}{(4 \cos x-\sin x)^{2}} d x=\quad \int_{0} \frac{2 \operatorname{tg} x-5}{16 \cos ^{2} x-8 \cos x \sin x+\sin ^{2} x} d x= \\
& \arcsin (3 / \sqrt{10}) \\
& =\int_{0} \frac{2 \operatorname... | \frac{9}{4}-\ln16 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,754 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\pi / 4}^{\arccos (1 / \sqrt{26})} \frac{d x}{(6-\operatorname{tg} x) \sin 2 x}
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which:
$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\arccos \frac{1}{\sqrt{26}} \Rightarrow t=\operatorname{tg}\left(\arccos \frac{1}{\sq... | \frac{\ln5}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,755 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\pi / 4}^{\arccos (1 / \sqrt{26})} \frac{36 d x}{(6-\operatorname{tg} x) \sin 2 x}
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which we get:
$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\arccos \frac{1}{\sqrt{26}} \Rightarrow t=\operatorname{tg}\left(\arccos \frac... | 6\ln5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,756 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi / 4} \frac{4-7 \tan x}{2+3 \tan x} d x
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which we get:
$$
\begin{aligned}
& d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1
\end{aligned}
$$
Substitute:
$$
\int_{0}^{\pi / 4} \frac{4-7 \operat... | \ln\frac{25}{8}-\frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,757 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\arcsin (2 / \sqrt{5})}^{\pi / 4} \frac{2-\operatorname{tg} x}{(\sin x+3 \cos x)^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{-\arcsin (2 / \sqrt{5})}^{\pi / 4} \frac{2-\operatorname{tg} x}{(\sin x+3 \cos x)^{2}} d x=\int_{-\arcsin (2 / \sqrt{5})}^{\pi / 4} \frac{2-\operatorname{tg} x}{\sin ^{2} x+6 \sin x \cos x+9 \cos ^{2} x} d x= \\
& =\int_{\arcsin (\rho / \sqrt{5})}^{\pi / 4} \frac{2-\operatorname{... | \frac{15}{4}-\ln4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,758 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\pi / 4}^{\arcsin \sqrt{2 / 3}} \frac{8 \operatorname{tg} x d x}{3 \cos ^{2} x+8 \sin 2 x-7}
$$ | ## Solution
We will use the substitution:
$$
t=\operatorname{tg} x
$$
From which:
$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\arcsin \sqrt{\frac{2}{3}} \Rightarrow t=\operatorna... | \frac{4}{21}\cdot\ln|\frac{7\sqrt{2}-2}{5}|-\frac{4}{3}\cdot\ln|2-\sqrt{2}| | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,759 |
## Task Condition
Calculate the definite integral:
$$
\int_{\arccos (1 / \sqrt{10})}^{\arccos (1 / \sqrt{26})} \frac{12 d x}{(6+5 \tan x) \sin 2 x}
$$ | ## Solution
Let's use the substitution:
$$
t=\operatorname{tg} x
$$
From which:
$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\arccos \frac{1}{\sqrt{10}} \Rightarrow t=\operatorname{tg}\left(\arccos \frac{1}{\sqrt{10}}\right)=\sqrt{\frac{1}{\cos ^{2}\left(\arccos \frac{1}{\sqrt{1... | \ln\frac{105}{93} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,760 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\arccos (1 / \sqrt{6})} \frac{3 \tan^{2} x-1}{\tan^{2} x+5}
$$ | ## Solution
We will use the substitution:
$$
t=\operatorname{tg} x
$$
From which:
$$
\begin{aligned}
& d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\arccos \frac{1}{\sqrt{6}} \Rightarrow t=\operatorname{tg}\left(\arccos \frac{1}{\sqrt{6}}\right)=\sqrt{\frac{1}{\cos ^{2}\left(\arccos \... | \frac{\pi}{\sqrt{5}}-\operatorname{arctg}\sqrt{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,762 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{-9 ;-8 ;-3\}$
$p=\{1 ; 4 ; 1\}$
$q=\{-3 ; 2 ; 0\}$
$r=\{1 ;-1 ; 2\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | -3p+2q | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,763 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{3 ;-1 ; 6\}$
$b=\{5 ; 7 ; 10\}$
$c_{1}=4 a-2 b$
$c_{2}=b-2 a$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
It is not hard to notice that $c_{1}=-2(b-2 a)=-2 c_{2}$ for any $a$ and $b$. That is, $c_{1}=-\frac{1}{2} \cdot c_{2}$, which means the vecto... | c_{1}=-2c_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,764 |
## problem statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-4 ; 3 ; 0), B(0 ; 1 ; 3), C(-2 ; 4 ;-2)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-(-4) ; 1-3 ; 3-0)=(4 ;-2 ; 3)$
$\overrightarrow{A C}=(-2-(-4) ; 4-3 ;-2-0)=(2 ; 1 ;-2)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overr... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,765 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=6 p-q$
\[
\begin{aligned}
& b=p+q \\
& |p|=3 \\
& |q|=4 \\
& (\widehat{p, q})=\frac{\pi}{4}
\end{aligned}
\] | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(6 p-q) \times(p+q)=6 \cdot p \times p+6 \cdot p \times q-q \times p-q \times q=$... | 42\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,766 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{4 ; 2 ; 2\}$
$b=\{-3 ;-3 ;-3\}$
$c=\{2 ; 1 ; 2\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$(a, b, c)=\left|\begin{array}{ccc}4 & 2 & 2 \\ -3 & -3 & -3 \\ 2 & 1 & 2\end{array}\right|=$
$$
\begin{aligned}
& =4 \cdot\left|\begi... | -6\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,767 |
## problem statement
Calculate the volume of the tetrahedron with vertices at points $A_{1}, A_{2}, A_{3}, A_{4_{\text {and }}}$ its height dropped from vertex $A_{4 \text { to the face }} A_{1} A_{2} A_{3}$.
$A_{1}(4 ;-1 ; 3)$
$A_{2}(-2 ; 1 ; 0)$
$A_{3}(0 ;-5 ; 1)$
$A_{4}(3 ; 2 ;-6)$ | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-2-4 ; 1-(-1) ; 0-3\}=\{-6 ; 2 ;-3\} \\
& A_{1} A_{3}=\{0-4 ;-5-(-1) ; 1-3\}=\{-4 ;-4 ;-2\} \\
& \overrightarrow{A_{1} A_{4}}=\{3-4 ; 2-(-1) ;-6-3\}=\{-1 ; 3 ;-9\}
\end{aligned}
$$
According to the geometr... | \frac{17}{\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,768 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(2 ; 1 ; 4)$
$M_{2}(3 ; 5 ;-2)$
$M_{3}(-7 ;-3 ; 2)$
$M_{0}(-3 ; 1 ; 8)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-2 & y-1 & z-4 \\
3-2 & 5-1 & -2-4 \\
-7-2 & -3-1 & 2-4
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-1 & z-4 \\
1 & 4 & -6 \\
-9 & -4 ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,769 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-7 ; 1 ;-4)$
$B(8 ; 11 ;-3)$
$C(9 ; 9 ;-1)$ | ## Solution
Let's find the vector $\overrightarrow{B C}:$
$$
\overrightarrow{B C}=\{9-8 ; 9-11 ;-1-(-3)\}=\{1 ;-2 ; 2\}
$$
Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$(x-(-7))-2 \cdot(y-1)+2 \cdot(... | x-2y+2z+17=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,770 |
## Problem Statement
Find the angle between the planes:
\[
\begin{aligned}
& x+y+3z-7=0 \\
& y+z-1=0
\end{aligned}
\] | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{1 ; 1 ; 3\}$
$\overrightarrow{n_{2}}=\{0 ; 1 ; 1\}$
The angle $\phi_{\text {between the planes is determined by the formula: }}$
$\cos \phi=\frac{\left... | \arccos2\sqrt{\frac{2}{11}}\approx3128'56'' | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,771 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; y ; 0)$
$B(-2 ; 4 ;-6)$
$C(8 ; 5 ; 1)$ | ## Solution
Let's find the distances $A B$ and $A C$:
\[
\begin{aligned}
& A B=\sqrt{(-2-0)^{2}+(4-y)^{2}+(-6-0)^{2}}=\sqrt{4+16-8 y+y^{2}+36}=\sqrt{y^{2}-8 y+56} \\
& A C=\sqrt{(8-0)^{2}+(5-y)^{2}+(1-0)^{2}}=\sqrt{64+25-10 y+y^{2}+1}=\sqrt{y^{2}-10 y+90}
\end{aligned}
\]
Since by the condition of the problem $A B=A... | A(0;17;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,772 |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(2 ; 5 ; 1)$
$a: 5 x-2 y+z-3=0$
$k=\frac{1}{3}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 5 x-2 y+z-1=0$
Substitute the coordinates of point $A$ into the equat... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,773 |
## problem statement
Write the canonical equations of the line.
$$
\begin{aligned}
& 4 x+y+z+2=0 \\
& 2 x-y-3 z-8=0
\end{aligned}
$$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vect... | \frac{x-1}{-2}=\frac{y+6}{14}=\frac{z}{-6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,774 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-5}{-2}=\frac{y-2}{0}=\frac{z+4}{-1}$
$2 x-5 y+4 z+24=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-5}{-2}=\frac{y-2}{0}=\frac{z+4}{-1}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=5-2 t \\
y=2 \\
z=-4-t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$2(5-2 t)-5 \cdot 2+4(-4-t)+24=0$
$10-4 t... | (3;2;-5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,775 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(2 ; 1 ; 0)$
$y+z+2=0$ | ## Solution
Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$. Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:
$\vec{s}=\vec{n}=\{0 ; 1 ; 1\}$
Then the equation of the desired line is:
$\f... | M^{\}(2;-2;-3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,776 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
e^{\sin \left(x^{\frac{3}{2}} \sin \frac{2}{x}\right)}-1+x^{2}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,777 |
## Condition of the problem
To derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=14 \sqrt{x}-15 \sqrt[3]{x}+2, x_{0}=1$ | ## Solution
Let's find $y^{\prime}:$
$y^{\prime}=(14 \sqrt{x}-15 \sqrt[3]{x}+2)^{\prime}=\frac{14}{2 \sqrt{x}}-15 \cdot \frac{1}{3} \cdot x^{-\frac{2}{3}}=\frac{7}{\sqrt{x}}-\frac{5}{\sqrt[3]{x^{2}}}$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{7}{\sqrt{1}}-\frac{5}{\sqrt[3]{1^{2}}}=7-5=2$
 e^{2 \sqrt{x-1}}
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\left(\sqrt{x-1}-\frac{1}{2}\right) e^{2 \sqrt{x-1}}\right)^{\prime} d x= \\
& =\left(\left(\sqrt{x-1}-\frac{1}{2}\right)^{\prime} e^{2 \sqrt{x-1}}+\left(\sqrt{x-1}-\frac{1}{2}\right) \cdot\left(e^{2 \sqrt{x-1}}\right)^{\prime}\right) d x= \\
& =\left(\fr... | e^{2\sqrt{x-1}}\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,779 |
## Task Condition
Find the derivative.
$$
y=\frac{x+7}{6 \sqrt{x^{2}+2 x+7}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x+7}{6 \sqrt{x^{2}+2 x+7}}\right)^{\prime}=\left(\frac{(x+7)^{\prime} \cdot 6 \sqrt{x^{2}+2 x+7}-(x+7)\left(6 \sqrt{x^{2}+2 x+7}\right)^{\prime}}{36\left(x^{2}+2 x+7\right)}\right)^{\prime}= \\
& =\frac{6 \sqrt{x^{2}+2 x+7}-(x+7) \cdot 6 \cdot \frac{1}{2 \sqrt{x^... | -\frac{x}{(x^{2}+2x+7)\sqrt{x^{2}+2x+7}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,781 |
## Task Condition
Find the derivative.
$y=\operatorname{arctg}\left(e^{x}-e^{-x}\right)$ | ## Solution
$y^{\prime}=\left(\operatorname{arctg}\left(e^{x}-e^{-x}\right)\right)^{\prime}=\frac{1}{1+\left(e^{x}-e^{-x}\right)^{2}} \cdot\left(e^{x}-e^{-x}\right)^{\prime}=$
$$
\begin{aligned}
& =\frac{1}{1+\left(e^{x}-e^{-x}\right)^{2}} \cdot\left(e^{x}+e^{-x}\right)=\frac{e^{x}+e^{-x}}{1+e^{2 x}-2+e^{-2 x}}= \\
&... | \frac{e^{x}+e^{-x}}{e^{2x}+e^{-2x}-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,782 |
## Task Condition
Find the derivative.
$$
y=\ln \left(e^{x}+\sqrt{1+e^{2 x}}\right)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \left(e^{x}+\sqrt{1+e^{2 x}}\right)\right)^{\prime}=\frac{1}{e^{x}+\sqrt{1+e^{2 x}}} \cdot\left(e^{x}+\sqrt{1+e^{2 x}}\right)^{\prime}= \\
& =\frac{1}{e^{x}+\sqrt{1+e^{2 x}}} \cdot\left(e^{x}+\frac{1}{2 \sqrt{1+e^{2 x}}} \cdot 2 e^{2 x}\right)= \\
& =\frac{1}{e^{x}... | \frac{e^{x}}{\sqrt{1+e^{2x}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,783 |
## Condition of the problem
Find the derivative.
$$
y=\sqrt[3]{\cos \sqrt{2}}-\frac{1}{52} \cdot \frac{\cos ^{2} 26 x}{\sin 52 x}
$$ | ## Solution
$y^{\prime}=\left(\sqrt[3]{\cos \sqrt{2}}-\frac{1}{52} \cdot \frac{\cos ^{2} 26 x}{\sin 52 x}\right)^{\prime}=0-\frac{1}{52}\left(\frac{\cos ^{2} 26 x}{\sin 52 x}\right)^{\prime}=$
$=-\frac{1}{52} \cdot \frac{\left(\cos ^{2} 26 x\right)^{\prime} \cdot \sin 52 x-\cos ^{2} 26 x \cdot(\sin 52 x)^{\prime}}{\s... | \frac{1}{2\sin^{2}26x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,784 |
## Task Condition
Find the derivative.
$y=\left(2 x^{2}+6 x+5\right) \operatorname{arctg} \frac{x+1}{x+2}-x$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\left(2 x^{2}+6 x+5\right) \operatorname{arctg} \frac{x+1}{x+2}-x\right)^{\prime}= \\
& =\left(2 x^{2}+6 x+5\right)^{\prime} \operatorname{arctg} \frac{x+1}{x+2}+\left(2 x^{2}+6 x+5\right)\left(\operatorname{arctg} \frac{x+1}{x+2}\right)^{\prime}-1= \\
& =(4 x+6) \oper... | (4x+6)\operatorname{arctg}\frac{x+1}{x+2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,785 |
## Task Condition
Find the derivative.
$y=\frac{3}{2} \cdot \ln \left(\tanh \frac{x}{2}\right)+\cosh x-\frac{\cosh x}{2 \sinh^{2} x}$ | ## Solution
$y^{\prime}=\left(\frac{3}{2} \cdot \ln \left(\operatorname{th} \frac{x}{2}\right)+\operatorname{ch} x-\frac{\operatorname{ch} x}{2 \operatorname{sh}^{2} x}\right)^{\prime}=$
$$
\begin{aligned}
& =\frac{3}{2} \cdot \frac{1}{\operatorname{th} \frac{x}{2}} \cdot \frac{1}{\operatorname{ch}^{2} \frac{x}{2}} \... | \frac{\operatorname{ch}^{4}x}{\operatorname{sh}^{3}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,786 |
## Problem Statement
Find the derivative.
$$
y=(\tan x)^{\ln (\tan x) / 4}
$$ | ## Solution
$$
\begin{aligned}
& y=(\tan x)^{\ln (\tan x) / 4} \\
& \ln y=\frac{\ln (\tan x)}{4} \cdot \ln (\tan x) \\
& \ln y=\frac{\ln ^{2}(\tan x)}{4} \\
& \frac{y^{\prime}}{y}=\left(\frac{\ln ^{2}(\tan x)}{4}\right)^{\prime}=\frac{2 \ln (\tan x)}{4} \cdot \frac{1}{\tan x} \cdot \frac{1}{\cos ^{2} x}=\frac{\ln (\ta... | (\tanx)^{(\ln(\tanx)/4)}\cdot\frac{\ln(\tanx)}{\sin2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,787 |
## Task Condition
Find the derivative.
$y=(3 x+1)^{4} \arcsin \frac{1}{3 x+1}+\left(3 x^{2}+2 x+1\right) \sqrt{9 x^{2}+6 x}, 3 x+1>0$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left((3 x+1)^{4} \arcsin \frac{1}{3 x+1}+\left(3 x^{2}+2 x+1\right) \sqrt{9 x^{2}+6 x}\right)^{\prime}= \\
& =\left((3 x+1)^{4}\right)^{\prime} \arcsin \frac{1}{3 x+1}+(3 x+1)^{4}\left(\arcsin \frac{1}{3 x+1}\right)^{\prime}+ \\
& +\left(3 x^{2}+2 x+1\right)^{\prime} \sqrt{... | 12(3x+1)^{3}\cdot\arcsin\frac{1}{3x+1}+(3x+1)\cdot\frac{18x^{2}}{\sqrt{9x^{2}+6x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,788 |
## Problem Statement
Find the derivative.
$$
y=x^{2} \arccos x-\frac{x^{2}+2}{3} \sqrt{1-x^{2}}
$$ | ## Solution
$y^{\prime}=\left(x^{2} \arccos x-\frac{x^{2}+2}{3} \sqrt{1-x^{2}}\right)^{\prime}=$
$=2 x \cdot \arccos x+x^{2} \cdot \frac{-1}{\sqrt{1-x^{2}}}-\frac{2 x}{3} \sqrt{1-x^{2}}-\frac{x^{2}+2}{3} \frac{1}{2 \sqrt{1-x^{2}}} \cdot(-2 x)=$
$=2 x \cdot \arccos x-\frac{x^{2}}{\sqrt{1-x^{2}}}-\frac{2 x\left(1-x^{2}... | 2x\cdot\arccosx-x^{2}\cdot\sqrt{\frac{1-x}{1+x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,789 |
## Problem Statement
Find the derivative.
$y=\frac{\ln (\operatorname{ctg} x+\operatorname{ctg} \alpha)}{\sin \alpha}$ | ## Solution
$y^{\prime}=\left(\frac{\ln (\operatorname{ctg} x+\operatorname{ctg} \alpha)}{\sin \alpha}\right)^{\prime}=\frac{1}{\sin \alpha \cdot(\operatorname{ctg} x+\operatorname{ctg} \alpha)} \cdot \frac{-1}{\sin ^{2} x}=$
$=-\frac{1}{\sin \alpha \cdot \cos x+\cos \alpha \cdot \sin x}=-\frac{1}{\sin (\alpha+x)}$
#... | -\frac{1}{\sin(\alpha+x)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,790 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\sqrt{t-t^{2}}-\operatorname{arctg} \sqrt{\frac{1-t}{t}} \\
y=\sqrt{t}-\sqrt{1-t} \cdot \arcsin \sqrt{t}
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\sqrt{t-t^{2}}-\operatorname{arctg} \sqrt{\frac{1-t}{t}}\right)^{\prime}=\frac{1}{2 \sqrt{t-t^{2}}} \cdot(1-2 t)-\frac{1}{1+\left(\sqrt{\frac{1-t}{t}}\right)^{2}} \cdot\left(\sqrt{\frac{1-t}{t}}\right)^{\prime}= \\
& =\frac{1-2 t}{2 \sqrt{t-t^{2}}}-\frac{t}{t+(1-t)... | \frac{\sqrt{}\cdot\arcsin\sqrt{}}{2(1-)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,791 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=2 \cos t \\
y=\sin t
\end{array}\right. \\
& t_{0}=-\frac{\pi}{3}
\end{aligned}
$$ | ## Solution
Since $t_{0}=-\frac{\pi}{3}$, then
$x_{0}=2 \cos \left(-\frac{\pi}{3}\right)=2 \cdot 0.5=1$
$y_{0}=\sin \left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}$
Let's find the derivatives:
$x_{t}^{\prime}=(2 \cos t)^{\prime}=-2 \sin t$
$y_{t}^{\prime}=(\sin t)^{\prime}=\cos t$
$y_{x}^{\prime}=\frac{y_{t}^{\p... | \frac{\sqrt{3}}{6}\cdotx-\frac{2\sqrt{3}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,792 |
Condition of the problem
Find the $n$-th order derivative.
$y=2^{k x}$ | ## Solution
$y=2^{k x}=\left(e^{\ln 2}\right)^{k x}=e^{k \ln 2 \cdot x}$
$y^{\prime}=\left(e^{k \ln 2 \cdot x}\right)^{\prime}=e^{k \ln 2 \cdot x} \cdot k \ln 2$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(e^{k \ln 2 \cdot x} \cdot k \ln 2\right)^{\prime}=e^{k \ln 2 \cdot x} \cdot k^{2} \ln ^{2} 2$
$\c... | 2^{kx}\cdotk^{n}\ln^{n}2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,793 |
## Task Condition
Find the derivative of the specified order.
$y=(5 x-8) \cdot 2^{-x}, y^{(I V)}=?$ | ## Solution
$y^{\prime}=\left((5 x-8) \cdot 2^{-x}\right)^{\prime}=5 \cdot 2^{-x}+(5 x-8) \cdot 2^{-x} \cdot(-\ln 2)=2^{-x} \cdot(-5 \ln 2 \cdot x+8 \ln 2+5)$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(2^{-x} \cdot(-5 \ln 2 \cdot x+8 \ln 2+5)\right)^{\prime}=$
$=2^{-x} \cdot(-\ln 2) \cdot(-5 \ln 2 \cd... | 2^{-x}\cdot\ln^{3}2\cdot(5\ln2\cdotx-8\ln2-20) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,794 |
## Task Condition
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\operatorname{arctg} t \\
y=\frac{t^{2}}{2}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\operatorname{arctg} t)^{\prime}=\frac{1}{1+t^{2}}$
$y_{t}^{\prime}=\left(\frac{t^{2}}{2}\right)^{\prime}=t$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{t}{\left(\frac{1}{1+t^{2}}\right)}=t\left(1+t^{2}\right) \\
& \left(y_{x}^{\prime}\righ... | y_{xx}^{\\}=(1+3t^2)(1+^2) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,795 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
$y=\frac{1}{\sqrt{\sin x+x}}$
$2 \sin x \cdot y^{\prime}+y \cdot \cos x=y^{3}(x \cdot \cos x-\sin x)$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{\sqrt{\sin x+x}}\right)^{\prime}=\left((\sin x+x)^{-\frac{1}{2}}\right)^{\prime}= \\
& =-\frac{1}{2} \cdot(\sin x+x)^{-\frac{3}{2}} \cdot(\cos x+1)=-\frac{\cos x+1}{2 \sqrt{(\sin x+x)^{3}}}
\end{aligned}
$$
Substitute into equation (1):
$2 \sin x \cdot\left(... | proof | Calculus | proof | Yes | Yes | olympiads | false | 47,796 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$f(x)=\left\{\begin{array}{c}\sqrt{1+\ln \left(1+x^{2} \sin \frac{1}{x}\right)}-1, x \neq 0 ; \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$$
f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}
$$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} ... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,797 |
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