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class | __index_level_0__ int64 0 742k |
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## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow-2} \frac{x^{3}+5 x^{2}+8 x+4}{x^{3}+3 x^{2}-4}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-2} \frac{x^{3}+5 x^{2}+8 x+4}{x^{3}+3 x^{2}-4}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-2} \frac{(x+2)\left(x^{2}+3 x+2\right)}{(x+2)\left(x^{2}+x-2\right)}= \\
& =\lim _{x \rightarrow-2} \frac{x^{2}+3 x+2}{x^{2}+x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rig... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,842 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 3} \frac{\sqrt[3]{9 x}-3}{\sqrt{3+x}-\sqrt{2 x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 5} \frac{\sqrt[3]{9 x}-3}{\sqrt{3+x}-\sqrt{2 x}}=\lim _{x \rightarrow 3} \frac{(\sqrt[3]{9 x}-3)\left(\sqrt[3]{(9 x)^{2}}+\sqrt[3]{9 x} \cdot 3+9\right)}{(\sqrt{3+x}-\sqrt{2 x})\left(\sqrt[3]{(9 x)^{2}}+\sqrt[3]{9 x} \cdot 3+9\right)}= \\
& =\lim _{x \rightarrow 3}... | -\frac{2\sqrt{6}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,843 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1} \frac{3^{5 x-3}-3^{2 x^{2}}}{\tan \pi x}
$$ | ## Solution
$\lim _{x \rightarrow 1} \frac{3^{5 x-3}-3^{2 x^{2}}}{\tan \pi x}=\lim _{x \rightarrow 1} \frac{3^{2 x^{2}}\left(3^{5 x-3-2 x^{2}}-1\right)}{\tan \pi x}=$[^0]$x=y+1 \Rightarrow y=x-1$
$x \rightarrow 1 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{3^{2(y+1)^{2}}... | \frac{9\ln3}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,845 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{\sqrt[3]{1+\ln ^{2} x}-1}{1+\cos \pi x}$ | ## Solution
Substitution:
$x=y+1 \Rightarrow y=x-1$
$x \rightarrow 1 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\sqrt[3]{1+\ln ^{2} x}-1}{1+\cos \pi x}=\lim _{y \rightarrow 0} \frac{\sqrt[3]{1+\ln ^{2}(y+1)}-1}{1+\cos \pi(y+1)}= \\
& =\lim _{y \rightarrow 0} \frac{\sqrt[... | \frac{2}{3\pi^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,846 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{10^{2 x}-7^{-x}}{2 \tan x-\arctan x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{10^{2 x}-7^{-x}}{2 \tan x-\arctan x}=\lim _{x \rightarrow 0} \frac{\left(100^{x}-1\right)-\left(7^{-x}-1\right)}{2 \tan x-\arctan x}=$
$=\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 100}\right)^{x}-1\right)-\left(\left(e^{\ln 7}\right)^{-x}-1\right)}{2 \tan x-\arctan x}=... | \ln700 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,847 |
## Problem Statement
Calculate the limit of the function:
$\lim _{h \rightarrow 0} \frac{\ln (x+h)+\ln (x-h)-2 \ln x}{h^{2}}, x>0$ | ## Solution
$\lim _{h \rightarrow 0} \frac{\ln (x+h)+\ln (x-h)-2 \ln x}{h^{2}}=\lim _{h \rightarrow 0} \frac{\ln ((x+h)(x-h))-\ln x^{2}}{h^{2}}=$
$=\lim _{h \rightarrow 0} \frac{\ln \frac{x^{2}-h^{2}}{x^{2}}}{h^{2}}=\lim _{h \rightarrow 0} \frac{\ln \left(1-\frac{h^{2}}{x^{2}}\right)}{h^{2}}=$
Using the substitution... | -\frac{1}{x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,848 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(2-e^{\sin x}\right)^{\operatorname{ctg} \pi x}$ | ## Solution
$\lim _{x \rightarrow 0}\left(2-e^{\sin x}\right)^{\operatorname{ctg} \pi x}=\lim _{x \rightarrow 0}\left(e^{\ln \left(2-e^{\sin x}\right)}\right)^{\operatorname{ctg} \pi x}=$
$=\lim _{x \rightarrow 0} e^{\operatorname{ctg} \pi x \cdot \ln \left(2-e^{\sin x}\right)}=$
$=\exp \left\{\lim _{x \rightarrow 0... | e^{-\frac{1}{\pi}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,849 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{x^{3}+8}{3 x^{2}+10}\right)^{x+2}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{x^{3}+8}{3 x^{2}+10}\right)^{x+2}=\left(\frac{0^{3}+8}{3 \cdot 0^{2}+10}\right)^{0+2}=$
$=\left(\frac{8}{10}\right)^{2}=0.8^{2}=0.64$
Problem Kuznetsov Limits 18-15 | 0.64 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,850 |
## Condition of the problem
Calculate the limit of the function:
$$
\lim _{x \rightarrow 3}\left(\frac{9-2 x}{3}\right)^{\tan\left(\frac{\pi x}{6}\right)}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 3}\left(\frac{9-2 x}{3}\right)^{\operatorname{tg}\left(\frac{\pi x}{6}\right)}=\lim _{x \rightarrow 3}\left(3-\frac{2 x}{3}\right)^{\operatorname{tg}\left(\frac{\pi x}{6}\right)}= \\
& =\lim _{x \rightarrow 3}\left(e^{\ln \left(3-\frac{2 x}{3}\right)}\right)^{\oper... | e^{\frac{4}{\pi}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,851 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow a}\left(\frac{\sin x-\sin a}{x-a}\right)^{\frac{x^{2}}{a^{2}}}$ | ## Solution
$\lim _{x \rightarrow a}\left(\frac{\sin x-\sin a}{x-a}\right)^{\frac{x^{2}}{a^{2}}}=\left(\lim _{x \rightarrow a} \frac{\sin x-\sin a}{x-a}\right)^{\lim _{x \rightarrow a} \frac{x^{2}}{a^{2}}}=$
$=\left(\lim _{x \rightarrow a} \frac{2 \sin \frac{x-a}{2} \cos \frac{x+a}{2}}{x-a}\right)^{\frac{a^{2}}{a^{2}... | \cos | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,852 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{11 ;-1 ; 4\}$
$p=\{1 ;-1 ; 2\}$
$q=\{3 ; 2 ; 0\}$
$r=\{-1 ; 1 ; 1\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\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-2r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,854 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{2 ;-1 ; 6\}$
$b=\{-1 ; 3 ; 8\}$
$c_{1}=5 a-2 b$
$c_{2}=2 a-5 b$ | ## 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.
We find:
$$
\begin{aligned}
& c_{1}=5 a-2 b=\{5 \cdot 2-2 \cdot(-1) ; 5 \cdot(-1)-2 \cdot 3 ; 5 \cdot 6-2 \cdot 8\}=\{12 ;-11 ; 14\} \\
& c_{... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,855 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(3 ; 3 ;-1), B(5 ; 1 ;-2), C(4 ; 1 ;-3)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(5-3 ; 1-3 ;-2-(-1))=(2 ;-2 ;-1)$
$\overrightarrow{A C}=(4-3 ; 1-3 ;-3-(-1))=(1 ;-2 ;-2)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrightar... | \frac{8}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,856 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a_{\text {and }} b$.
$$
\begin{aligned}
& a=5 p-q \\
& b=p+q \\
& |p|=5 \\
& |q|=3 \\
& (\widehat{p, q})=\frac{5 \pi}{6}
\end{aligned}
$$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(5 p-q) \times(p+q)=5 \cdot p \times p+5 \cdot p \times q-q \times p-q \times q=$... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,857 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{1 ;-1 ; 4\}$
$b=\{1 ; 0 ; 3\}$
$c=\{1 ;-3 ; 8\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$(a, b, c)=\left|\begin{array}{ccc}1 & -1 & 4 \\ 1 & 0 & 3 \\ 1 & -3 & 8\end{array}\right|=$
$=1 \cdot\left|\begin{array}{cc}0 & 3 \\ -... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,858 |
## Task Condition
Calculate the volume of the tetrahedron with vertices at points $A_{1}, A_{2}, A_{3}, A_{4_{\text {and }}}$ its height dropped from vertex $A_{4 \text { to the face }} A_{1} A_{2} A_{3}$.
$A_{1}(0 ;-3 ; 1)$
$A_{2}(-4 ; 1 ; 2)$
$A_{3}(2 ;-1 ; 5)$
$A_{4}(3 ; 1 ;-4)$ | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$\overrightarrow{A_{1} A_{2}}=\{-4-0 ; 1-(-3) ; 2-1\}=\{-4 ; 4 ; 1\}$
$\overrightarrow{A_{1} A_{3}}=\{2-0 ;-1-(-3) ; 5-1\}=\{2 ; 2 ; 4\}$
$\overrightarrow{A_{1} A_{4}}=\{3-0 ; 1-(-3) ;-4-1\}=\{3 ; 4 ;-5\}$
According to the geometric meaning of the scalar ... | 32\frac{1}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,859 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ; 2 ; 0)$
$M_{2}(3 ; 0 ;-3)$
$M_{3}(5 ; 2 ; 6)$
$M_{0}(-13 ;-8 ; 16)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-1 & y-2 & z-0 \\
3-1 & 0-2 & -3-0 \\
5-1 & 2-2 & 6-0
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-2 & z \\
2 & -2 & -3 \\
4 & 0 & 6
\... | \frac{134}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,860 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-3, -1, 7)$
$B(0, 2, -6)$
$C(2, 3, -5)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{2-0 ; 3-2 ;-5-(-6)\}=\{2 ; 1 ; 1\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$2 \cdot(x-(-3))+(y-(-1))+(z-7)=0$
$2 ... | 2x+y+0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,861 |
## Task Condition
Find the angle between the planes:
$2 y+z-9=0$
$x-y+2 z-1=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{0 ; 2 ; 1\}$
$\overrightarrow{n_{2}}=\{1 ;-1 ; 2\}$
The angle $\phi$ between the planes is determined by the formula:
$\cos \phi=\frac{\left(\overr... | \frac{\pi}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,862 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(x ; 0 ; 0)$
$B(4 ; 5 ;-2)$
$C(2 ; 3 ; 4)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(4-x)^{2}+(5-0)^{2}+(-2-0)^{2}}=\sqrt{16-8 x+x^{2}+25+4}=\sqrt{x^{2}-8 x+45} \\
& A C=\sqrt{(2-x)^{2}+(3-0)^{2}+(4-0)^{2}}=\sqrt{4-4 x+x^{2}+9+16}=\sqrt{x^{2}-4 x+29}
\end{aligned}
$$
Since according to the problem's condition $A B=... | A(4;0;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,863 |
## 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 $a$?
$A(-3; -2; 4)$
$a: 2x - 3y + z - 5 = 0$
$k = -\frac{4}{5}$ | ## 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 $a$:
$a^{\prime}: 2 x-3 y+z+4=0$
Substitute the coordinates of point $A$ into the equat... | 8\neq0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,864 |
## Task Condition
Write the canonical equations of the line.
$x-3 y+z+2=0$
$x+3 y+2 z+14=0$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- are coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction ... | \frac{x+8}{-9}=\frac{y+2}{-1}=\frac{z}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,865 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-1}{6}=\frac{y-3}{1}=\frac{z+5}{3}$
$3 x-2 y+5 z-3=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-1}{6}=\frac{y-3}{1}=\frac{z+5}{3}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+6 t \\
y=3+t \\
z=-5+3 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$3(1+6 t)-2(3+t)+5(-5+3 t)-3=0$
$3+18 t... | (7;4;-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,866 |
## problem statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(3 ;-3 ;-1)$
$2 x-4 y-4 z-13=0$ | ## 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:
$$
\vec{s}=\vec{n}=\{2 ;-4 ;-4\}
$$
Then the equation of the desired line is:
... | M^{\}(2;-1;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,867 |
## Problem Statement
Calculate the definite integral:
$$
\int_{e+1}^{e^{2}+1} \frac{1+\ln (x-1)}{x-1} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{e+1}^{e^{2}+1} \frac{1+\ln (x-1)}{x-1} d x=\int_{e+1}^{e^{2}+1} \frac{1}{x-1} d x+\int_{e+1}^{e^{2}+1} \frac{\ln (x-1)}{x-1} d x= \\
& =\left.\ln |x-1|\right|_{e+1} ^{e^{2}+1}+\int_{e+1}^{e^{2}+1} \ln (x-1) d(\ln (x-1))= \\
& =\ln \left|e^{2}+1-1\right|-\ln |e+1-1|+\left.\frac{1}... | 2\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,868 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} \frac{\left(x^{2}+1\right) d x}{\left(x^{3}+3 x+1\right)^{2}}
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{1} \frac{\left(x^{2}+1\right) d x}{\left(x^{3}+3 x+1\right)^{2}}=\int_{0}^{1} \frac{\frac{1}{3} \cdot\left(3 x^{2}+3\right) d x}{\left(x^{3}+3 x+1\right)^{2}}=\int_{0}^{1} \frac{\frac{1}{3} \cdot d\left(x^{3}+3 x+1\right)}{\left(x^{3}+3 x+1\right)^{2}}= \\
& =\left.\frac{1}{3... | \frac{4}{15} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,869 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} \frac{4 \operatorname{arctg} x - x}{1 + x^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{1} \frac{4 \operatorname{arctg} x - x}{1 + x^{2}} d x = 4 \cdot \int_{0}^{1} \frac{\operatorname{arctg} x}{1 + x^{2}} d x - \int_{0}^{1} \frac{x}{1 + x^{2}} d x = \\
& = 4 \cdot \int_{0}^{1} \operatorname{arctg} x \cdot d(\operatorname{arctg} x) - \frac{1}{2} \cdot \int_{0}^{... | \frac{\pi^{2}-4\cdot\ln2}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,870 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2} \frac{x^{3}}{x^{2}+4} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{2} \frac{x^{3}}{x^{2}+4} d x=\frac{1}{2} \cdot \int_{0}^{2} \frac{x^{2}}{x^{2}+4} d\left(x^{2}\right)=\frac{1}{2} \cdot \int_{0}^{2} \frac{x^{2}+4-4}{x^{2}+4} d\left(x^{2}\right)= \\
& =\frac{1}{2} \cdot \int_{0}^{2} d\left(x^{2}\right)-\frac{1}{2} \cdot \int_{0}^{2} \frac{4}... | 2-2\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,871 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\pi}^{2 \pi} \frac{x+\cos x}{x^{2}+2 \sin x} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{\pi}^{2 \pi} \frac{x+\cos x}{x^{2}+2 \sin x} d x=\int_{\pi}^{2 \pi} \frac{\frac{1}{2}(2 x+2 \cos x)}{x^{2}+2 \sin x} d x=\frac{1}{2} \cdot \int_{\pi}^{2 \pi} \frac{d\left(x^{2}+2 \sin x\right)}{x^{2}+2 \sin x}= \\
& =\left.\frac{1}{2} \cdot \ln \left|x^{2}+2 \sin x\right|\right|_... | \ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,872 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{4}} \frac{2 \cos x + 3 \sin x}{(2 \sin x - 3 \cos x)^{3}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{4}} \frac{2 \cos x+3 \sin x}{(2 \sin x-3 \cos x)^{3}} d x=\int_{0}^{\frac{\pi}{4}} \frac{d(2 \sin x-3 \cos x)}{(2 \sin x-3 \cos x)^{3}}=-\left.\frac{1}{2 \cdot(2 \sin x-3 \cos x)^{2}}\right|_{0} ^{\frac{\pi}{4}}= \\
& =-\frac{1}{2 \cdot\left(2 \sin \frac{\pi}{4}-3 ... | -\frac{17}{18} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,873 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{1}{2}} \frac{8 x-\operatorname{arctg} 2 x}{1+4 x^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\frac{1}{2}} \frac{8 x-\operatorname{arctg} 2 x}{1+4 x^{2}} d x=\int_{0}^{\frac{1}{2}} \frac{8 x}{1+4 x^{2}} d x-\int_{0}^{\frac{1}{2}} \frac{\operatorname{arctg} 2 x}{1+4 x^{2}} d x= \\
& =\int_{0}^{\frac{1}{2}} \frac{d\left(1+4 x^{2}\right)}{1+4 x^{2}}-\frac{1}{2} \cdot \in... | \ln2-\frac{\pi^2}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,874 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{4} \frac{\frac{1}{2 \sqrt{x}}+1}{(\sqrt{x}+x)^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{1}^{4} \frac{\frac{1}{2 \sqrt{x}}+1}{(\sqrt{x}+x)^{2}} d x=\int_{1}^{4} \frac{d(\sqrt{x}+x)}{(\sqrt{x}+x)^{2}}=-\left.\frac{1}{\sqrt{x}+x}\right|_{1} ^{4}= \\
& =-\frac{1}{\sqrt{4}+4}+\frac{1}{\sqrt{1}+1}=-\frac{1}{6}+\frac{1}{2}=\frac{2}{6}=\frac{1}{3}
\end{aligned}
$$
Source —... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,875 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} \frac{x}{x^{4}+1} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{1} \frac{x}{x^{4}+1} d x=\frac{1}{2} \cdot \int_{0}^{1} \frac{d\left(x^{2}\right)}{x^{4}+1}=\frac{1}{2} \cdot \int_{0}^{1} \frac{d\left(x^{2}\right)}{x^{4}+1}=\left.\frac{1}{2} \cdot \operatorname{arctg} x^{2}\right|_{0} ^{1}= \\
& =\frac{1}{2} \cdot \operatorname{arctg} 1^{2... | \frac{\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,876 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\sqrt{3}}^{\sqrt{8}} \frac{x+\frac{1}{x}}{\sqrt{x^{2}+1}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{\sqrt{3}}^{\sqrt{8}} \frac{x+\frac{1}{x}}{\sqrt{x^{2}+1}} d x=\int_{\sqrt{3}}^{\sqrt{8}} \frac{x}{\sqrt{x^{2}+1}} d x+\int_{\sqrt{3}}^{\sqrt{8}} \frac{1}{x \sqrt{x^{2}+1}} d x= \\
& =\frac{1}{2} \cdot \int_{\sqrt{3}}^{\sqrt{8}} \frac{d\left(x^{2}+1\right)}{\sqrt{x^{2}+1}}+\int_{\... | 1+\ln\sqrt{\frac{3}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,877 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\sqrt{3}}^{\sqrt{8}} \frac{x-\frac{1}{x}}{\sqrt{x^{2}+1}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{\sqrt{3}}^{\sqrt{8}} \frac{x-\frac{1}{x}}{\sqrt{x^{2}+1}} d x=\int_{\sqrt{3}}^{\sqrt{8}} \frac{x}{\sqrt{x^{2}+1}} d x-\int \frac{1}{x \sqrt{x^{2}+1}} d x= \\
& =\frac{1}{2} \cdot \int_{\sqrt{3}}^{\sqrt{8}} \frac{d\left(x^{2}+1\right)}{\sqrt{x^{2}+1}}-\int_{\sqrt{3}}^{\sqrt{8}} \f... | 1+\ln\sqrt{\frac{2}{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,878 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\sqrt{3}} \frac{\operatorname{arctg} x + x}{1 + x^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\sqrt{3}} \frac{\operatorname{arctg} x + x}{1 + x^{2}} d x = \int_{0}^{\sqrt{3}} \frac{\operatorname{arctg} x}{1 + x^{2}} d x + \int_{0}^{\sqrt{3}} \frac{x}{1 + x^{2}} d x = \\
& = \frac{1}{2} \int_{0}^{\sqrt{3}} \operatorname{arctg} x d(\operatorname{arctg} x) + \frac{1}{2} ... | \frac{\pi^{2}}{18}+\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,879 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\sqrt{3}} \frac{x-(\operatorname{arctg} x)^{4}}{1+x^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\sqrt{3}} \frac{x-(\operatorname{arctg} x)^{4}}{1+x^{2}} d x=\int_{0}^{\sqrt{3}} \frac{x d x}{1+x^{2}}-\int_{0}^{\sqrt{3}} \frac{(\operatorname{arctg} x)^{4}}{1+x^{2}}= \\
& =\frac{1}{2} \int_{0}^{\sqrt{3}} \frac{d\left(1+x^{2}\right)}{1+x^{2}}-\int_{0}^{\sqrt{3}}(\operatorna... | \ln2-\frac{\pi^5}{5\cdot3^5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,880 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} \frac{x^{3}}{x^{2}+1} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{1} \frac{x^{3}}{x^{2}+1} d x=\int_{0}^{1} \frac{x\left(x^{2}+1\right)-x}{x^{2}+1} d x=\int_{0}^{1} x d x-\int \frac{x}{x^{2}+1}= \\
& =\left.\frac{1}{2} x^{2}\right|_{0} ^{1}-\frac{1}{2} \int_{0}^{1} \frac{d\left(x^{2}+1\right)}{x^{2}+1}=\frac{1}{2} \cdot 1^{2}-\frac{1}{2} \c... | \frac{1-\ln2}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,881 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\sin 1} \frac{(\arcsin x)^{2}+1}{\sqrt{1-x^{2}}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\sin 1} \frac{(\arcsin x)^{2}+1}{\sqrt{1-x^{2}}} d x=\int_{0}^{\sin 1} \frac{\arcsin ^{2} x}{\sqrt{1-x^{2}}} d x+\int_{0}^{\sin 1} \frac{d x}{\sqrt{1-x^{2}}}= \\
& =\int_{0}^{\sin 1}(\arcsin x)^{2} d(\arcsin x)+\left.\arcsin x\right|_{0} ^{\sin 1}= \\
& =\left.\frac{1}{3} \cd... | \frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,882 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{3} \frac{1-\sqrt{x}}{\sqrt{x} \cdot(x+1)} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{1}^{3} \frac{1-\sqrt{x}}{\sqrt{x} \cdot(x+1)} d x=\int_{1}^{3} \frac{1}{\sqrt{x} \cdot(x+1)} d x-\int_{1}^{3} \frac{1}{x+1} d x= \\
& =2 \cdot \int_{1}^{3} \frac{1}{x+1} d(\sqrt{x})-\ln |x+1|_{1}^{3}=\left.2 \operatorname{arctg} \sqrt{x}\right|_{1} ^{3}-\ln |3+1|+\ln |1+1|= \\
& ... | \frac{\pi}{6}-\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,883 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\sqrt{3}}^{\sqrt{8}} \frac{d x}{x \sqrt{x^{2}+1}}
$$ | ## Solution
$$
\begin{aligned}
& \int_{\sqrt{3}}^{\sqrt{8}} \frac{1}{x \sqrt{x^{2}+1}} d x=\int_{\sqrt{3}}^{\sqrt{8}} \frac{1}{x^{2} \cdot \frac{1}{x} \sqrt{x^{2}+1}} d x=\int_{\sqrt{3}}^{\sqrt{8}} \frac{1}{x^{2} \sqrt{1+\frac{1}{x^{2}}}} d x= \\
& =-\int_{\sqrt{3}}^{\sqrt{3}} \frac{1}{\sqrt{1+\frac{1}{x^{2}}}} d\left... | \ln\sqrt{\frac{3}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,884 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{e} \frac{1+\ln x}{x} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{1}^{e} \frac{1+\ln x}{x} d x=\int_{1}^{e}(1+\ln x) d(\ln x)=\left.\left(\ln x+\frac{1}{2} \ln ^{2} x\right)\right|_{1} ^{e}= \\
& =\left(\ln e+\frac{1}{2} \ln ^{2} e\right)-\left(\ln 1+\frac{1}{2} \ln ^{2} 1\right)=\left(1+\frac{1}{2} \cdot 1^{2}\right)-\left(0+\frac{1}{2} \cdot ... | \frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,885 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}}
$$ | ## Solution
$$
\begin{aligned}
& \int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}}=\int_{\sqrt{2}}^{2} \frac{d x}{x^{2} \cdot \frac{1}{x} \cdot \sqrt{x^{2}-1}}=\int_{\sqrt{2}}^{2} \frac{d x}{x^{2} \cdot \sqrt{1-\frac{1}{x^{2}}}}=-\int_{\sqrt{2}}^{2} \frac{d\left(\frac{1}{x}\right)}{\sqrt{1-\frac{1}{x^{2}}}}= \\
= & \in... | \frac{\pi}{12} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,886 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{e} \frac{x^{2}+\ln x^{2}}{x} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{1}^{e} \frac{x^{2}+\ln x^{2}}{x} d x=\int_{1}^{e} \frac{x^{2}}{x} d x+\int_{1}^{e} \frac{\ln x^{2}}{x} d x= \\
& =\int_{1}^{e} x \cdot d x+\int_{1}^{e} \ln x^{2} \cdot d(\ln x)=\left.\frac{x^{2}}{2}\right|_{1} ^{e}+\int_{1}^{e} 2 \ln x \cdot d(\ln x)= \\
& =\frac{e^{2}}{2}-\frac{... | \frac{e^{2}+1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,887 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} \frac{x}{\sqrt{x^{4}+x^{2}+1}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{1} \frac{x}{\sqrt{x^{4}+x^{2}+1}} d x=\int_{0}^{1} \frac{\frac{1}{2} \cdot d\left(x^{2}+\frac{1}{2}\right)}{\sqrt{\left(x^{2}+\frac{1}{2}\right)^{2}+\frac{3}{4}}}=\frac{1}{2} \cdot \ln \left|x^{2}+\frac{1}{2}+\sqrt{\left(x^{2}+\frac{1}{2}\right)^{2}+\frac{3}{4}}\right|_{0}^{1... | \ln\sqrt{\frac{3+2\sqrt{3}}{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,888 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} \frac{x^{3} d x}{\left(x^{2}+1\right)^{2}}
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{1} \frac{x^{3}}{\left(x^{2}+1\right)^{2}} d x=\int_{0}^{1} \frac{x\left(x^{2}+1\right)-x}{\left(x^{2}+1\right)^{2}} d x=\int_{0}^{1} \frac{x}{x^{2}+1} d x-\int_{0}^{1} \frac{x}{\left(x^{2}+1\right)^{2}} d x= \\
& =\frac{1}{2} \cdot \int_{0}^{1} \frac{d\left(x^{2}+1\right)}{x^... | \frac{\ln4-1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,889 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{4}} \operatorname{tg} x \cdot \ln (\cos x) d x
$$ | ## Solution
$$
\int_{0}^{\frac{\pi}{4}} \operatorname{tg} x \cdot \ln (\cos x) d x=
$$
Substitution:
$$
\begin{aligned}
& y=\ln \cos x \\
& d y=\frac{1}{\cos x} \cdot(-\sin x) \cdot d x=-\operatorname{tg} x \cdot d x \\
& x=0 \Rightarrow y=\ln \cos 0=\ln 1=0 \\
& x=\frac{\pi}{4} \Rightarrow y=\ln \cos \frac{\pi}{4}=... | -\frac{1}{2}\cdot\ln^{2}\frac{\sqrt{2}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,890 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-1}^{0} \frac{\tan(x+1)}{\cos^2(x+1)} \, dx
$$ | ## Solution
$$
\begin{aligned}
& \int_{-1}^{0} \frac{\tan(x+1)}{\cos^2(x+1)} \, dx = \int_{-1}^{0} \tan(x+1) \cdot d(\tan(x+1)) = \left. \frac{\tan^2(x+1)}{2} \right|_{-1}^{0} = \\
& = \frac{\tan^2(0+1)}{2} - \frac{\tan^2(-1+1)}{2} = \frac{\tan^2 1}{2} - \frac{\tan^2 0}{2} = \frac{\tan^2 1}{2} - \frac{0^2}{2} = \frac{... | \frac{\tan^21}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,891 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{1}{\sqrt{2}}} \frac{(\arccos x)^{3}-1}{\sqrt{1-x^{2}}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\frac{1}{\sqrt{2}}} \frac{\arccos ^{3} x-1}{\sqrt{1-x^{2}}} d x=\int_{0}^{\frac{1}{\sqrt{2}}} \frac{\arccos ^{3} x}{\sqrt{1-x^{2}}} d x+\int_{0}^{\frac{1}{\sqrt{2}}} \frac{-d x}{\sqrt{1-x^{2}}}= \\
& =\int_{0}^{\frac{1}{\sqrt{2}}}-(\arccos x)^{3} d(\arccos x)+\left.\arccos x\... | \frac{15\pi^{4}}{2^{10}}-\frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,892 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\pi}^{2 \pi} \frac{1-\cos x}{(x-\sin x)^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{\pi}^{2 \pi} \frac{1-\cos x}{(x-\sin x)^{2}} d x=\int_{\pi}^{2 \pi} \frac{d(x-\sin x)}{(x-\sin x)^{2}}=-\left.\frac{1}{x-\sin x}\right|_{\pi} ^{2 \pi}= \\
& =-\frac{1}{2 \pi-\sin (2 \pi)}+\frac{1}{\pi-\sin \pi}=-\frac{1}{2 \pi-0}+\frac{1}{\pi-0}=-\frac{1}{2 \pi}+\frac{2}{2 \pi}=\... | \frac{1}{2\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,893 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{4}} \frac{\sin x - \cos x}{(\cos x + \sin x)^{5}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{4}} \frac{\sin x-\cos x}{(\cos x+\sin x)^{5}} d x=-\int_{0}^{\frac{\pi}{4}} \frac{d(\cos x+\sin x)}{(\cos x+\sin x)^{5}}=\left.\frac{1}{4} \cdot \frac{1}{(\cos x+\sin x)^{4}}\right|_{0} ^{\frac{\pi}{4}}= \\
& =\frac{1}{4} \cdot \frac{1}{\left(\cos \frac{\pi}{4}+\si... | -\frac{3}{16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,894 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{x \cdot \cos x+\sin x}{(x \cdot \sin x)^{2}} d x
$$ | ## Solution
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{x \cdot \cos x+\sin x}{(x \cdot \sin x)^{2}} d x=
$$
Substitution:
$$
\begin{aligned}
& y=x \cdot \sin x \\
& d y=(\sin x+x \cdot \cos x) d x \\
& x=\frac{\pi}{4} \Rightarrow y=\frac{\pi}{4} \cdot \sin \frac{\pi}{4}=\frac{\sqrt{2} \pi}{8} \\
& x=\frac{\pi}{2}... | \frac{4\sqrt{2}-2}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,895 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} \frac{x^{3}+x}{x^{4}+1} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{0}^{1} \frac{x^{3}+x}{x^{4}+1} d x=\int_{0}^{1} \frac{x^{3}}{x^{4}+1} d x+\int \frac{x}{x^{4}+1} d x= \\
& =\frac{1}{4} \cdot \int_{0}^{1} \frac{1}{x^{4}+1} d\left(x^{4}+1\right)+\frac{1}{2} \cdot \int_{0}^{1} \frac{1}{x^{4}+1} d\left(x^{2}\right)= \\
& =\frac{1}{4} \cdot \ln \le... | \frac{\ln4+\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,896 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\sqrt{2}}^{\sqrt{3}} \frac{x}{\sqrt{x^{4}-x^{2}-1}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{\sqrt{2}}^{\sqrt{3}} \frac{x}{\sqrt{x^{4}-x^{2}-1}} d x=\int_{\sqrt{2}}^{\sqrt{3}} \frac{\frac{1}{2} \cdot d\left(x^{2}-\frac{1}{2}\right)}{\sqrt{\left(x^{2}-\frac{1}{2}\right)^{2}-\frac{5}{4}}}=\frac{1}{2} \cdot \ln \left|x^{2}-\frac{1}{2}+\sqrt{\left(x^{2}-\frac{1}{2}\right)^{2... | \ln\sqrt{\frac{5+2\sqrt{5}}{5}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,897 |
## Problem Statement
Calculate the definite integral:
$$
\int_{2}^{9} \frac{x}{\sqrt[3]{x-1}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int_{2}^{9} \frac{x}{\sqrt[3]{x-1}} d x=\int_{2}^{9} \frac{x-1+1}{\sqrt[3]{x-1}} d x=\int_{2}^{9} \sqrt[3]{(x-1)^{2}} d x+\int_{2}^{9} \frac{1}{\sqrt[3]{x-1}} d x= \\
& =\left.\frac{3}{5} \cdot \sqrt[3]{(x-1)^{5}}\right|_{2} ^{9}+\left.\frac{3}{2} \cdot \sqrt[3]{(x-1)^{2}}\right|_{2} ... | 23.1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,898 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{2 n-1}{2-3 n}, a=-\frac{2}{3}$ | ## Solution
By the definition of the limit:
$\forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{6 n-3+4-6 n}{3(2-3 n)}\right| \\
& \left.\frac{1}{3(2-3 n)} \right\rvert\, \\
& \left|\frac{1}{3(3 n-2)}\right| \\
& \frac{1}{3(3 n-2)} \\
&... | N(\varepsilon)=[\frac{1+15\varepsilon}{9\varepsilon}] | Calculus | proof | Yes | Yes | olympiads | false | 46,899 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(n+7)^{3}-(n+2)^{3}}{(3 n+2)^{2}+(4 n+1)^{2}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+7)^{3}-(n+2)^{3}}{(3 n+2)^{2}+(4 n+1)^{2}}=\lim _{n \rightarrow \infty} \frac{n^{3}+3 \cdot 7 \cdot n^{2}+3 \cdot 7^{2} \cdot n+7^{3}-n^{3}-3 \cdot 2 \cdot n^{2}-3 \cdot 2^{2} \cdot n-2^{3}}{(3 n+2)^{2}+(4 n+1)^{2}}= \\
& =\lim _{n \rightarrow \inf... | \frac{3}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,900 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}\right)}{\frac{1}{n}\left(\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}\right)}= \\
& =\lim _{n \rightarrow \infty} \f... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,901 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{\left(n^{3}+1\right)\left(n^{2}+3\right)}-\sqrt{n\left(n^{4}+2\right)}}{2 \sqrt{n}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{\left(n^{3}+1\right)\left(n^{2}+3\right)}-\sqrt{n\left(n^{4}+2\right)}}{2 \sqrt{n}}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\sqrt{\left(n^{3}+1\right)\left(n^{2}+3\right)}-\sqrt{n\left(n^{4}+2\right)}\right)\left(\sqrt{\left(n^{3}+1\righ... | \frac{3}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,902 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!-(2 n+2)!}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!-(2 n+2)!}=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3) \cdot(2 n+2)!-(2 n+2)!}=$
$$
=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+2)!((2 n+3)-1)}=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+2)!\cd... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,903 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{10 n-3}{10 n-1}\right)^{5 n}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{10 n-3}{10 n-1}\right)^{5 n}=\lim _{n \rightarrow \infty}\left(\frac{10 n-1}{10 n-3}\right)^{-5 n}= \\
& =\lim _{n \rightarrow \infty}\left(\frac{10 n-3+2}{10 n-3}\right)^{-5 n}=\lim _{n \rightarrow \infty}\left(1+\frac{2}{10 n-3}\right)^{-5 n}= ... | \frac{1}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,904 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$$
\lim _{x \rightarrow 5} \frac{5 x^{2}-24 x-5}{x-5}=26
$$ | ## 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
=\frac{\varepsilon}{5} | Calculus | proof | Yes | Yes | olympiads | false | 46,905 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$f(x)=2 x^{2}-3, x_{0}=4$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$.
$$
\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, such that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$ when $\left|x-x_{0}\righ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,906 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2} \frac{x^{3}-3 x-2}{x-2}$ | ## Solution
$\lim _{x \rightarrow 2} \frac{x^{3}-3 x-2}{x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^{2}+2 x+1\right)}{x-2}=$
$=\lim _{x \rightarrow 2}\left(x^{2}+2 x+1\right)=2^{2}+2 \cdot 2+1=4+4+1=9$
## Problem Kuznetsov Limits 10-20 | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,907 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{1}{3}} \frac{\sqrt[3]{\frac{x}{9}}-\frac{1}{3}}{\sqrt{\frac{1}{3}+x}-\sqrt{2 x}}$ | ## Solution
$\lim _{x \rightarrow \frac{1}{3}} \frac{\sqrt[3]{\frac{x}{9}}-\frac{1}{3}}{\sqrt{\frac{1}{3}+x}-\sqrt{2 x}}=\lim _{x \rightarrow \frac{1}{3}} \frac{\left(\sqrt[3]{\frac{x}{9}}-\frac{1}{3}\right)\left(\sqrt[3]{\left(\frac{x}{9}\right)^{2}}+\sqrt[3]{\frac{x}{9}} \cdot \frac{1}{3}+\left(\frac{1}{3}\right)^{2... | -\frac{2}{3}\sqrt{\frac{2}{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,908 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{\sin (5(x+\pi))}{e^{3 x}-1}
$$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$e^{3 x}-1 \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$
$\sin 5 x \sim 5 x$, as $x \rightarrow 0(5 x \rightarrow 0)$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin (5(x+\pi))}{e^{3 x}-1}=\left\{\frac{0}{0}\right\}=\lim _{... | -\frac{5}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,909 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2} \frac{\ln \left(9-2 x^{2}\right)}{\sin 2 \pi x}$ | ## Solution
Substitution:
$x=y+2 \Rightarrow y=x-2$
$x \rightarrow 2 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{\ln \left(9-2 x^{2}\right)}{\sin 2 \pi x}=\lim _{y \rightarrow 0} \frac{\ln \left(9-2(y+2)^{2}\right)}{\sin 2 \pi(y+2)}= \\
& =\lim _{y \rightarrow 0} \frac{\... | -\frac{4}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,910 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-2} \frac{\tan \left(e^{x+2}-e^{x^{2}-4}\right)}{\tan x+\tan 2}$ | ## Solution
Substitution:
$x=y-2 \Rightarrow y=x+2$
$x \rightarrow-2 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow-2} \frac{\tan\left(e^{x+2}-e^{x^{2}-4}\right)}{\tan x+\tan 2}=\lim _{y \rightarrow 0} \frac{\tan\left(e^{(y-2)+2}-e^{(y-2)^{2}-4}\right)}{\tan(y-2)+\tan 2}= \\
& =\lim ... | 5\cos^{2}2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,911 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)}{2 \sin x-\tan x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)\right)}{\frac{1}{x}(2 \sin x-\tan x)}=$
$=\frac{... | 7 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,912 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt{x+2}-\sqrt{2}}{\sin 3 x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{\sqrt{x+2}-\sqrt{2}}{\sin 3 x}=$
$\sin 3 x \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$
We get:
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\sqrt{x+2}-\sqrt{2}}{3 x}= \\
& =\lim _{x \rightarrow 0} \frac{(\sqrt{x+2}-\sqrt{2})(\sqrt{x+2}+\sqrt{2})}{3 x(\sqrt{x+2... | \frac{1}{6\sqrt{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,913 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \sqrt[x^{2}]{2-\cos x}$ | ## Solution
$\lim _{x \rightarrow 0} \sqrt[x^{2}]{2-\cos x}=\lim _{x \rightarrow 0}(2-\cos x)^{\frac{1}{x^{2}}}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln (2-\cos x)}\right)^{\frac{1}{x^{2}}}=\lim _{x \rightarrow 0} e^{\frac{\ln (2-\cos x)}{x^{2}}}=$
$=\exp \left\{\lim _{x \rightarrow 0} \frac{\ln (1+(1-\cos x))}{x^{2}... | \sqrt{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,914 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}(1+\cos 3 x)^{\sec x}$ | ## Solution
$\lim _{x \rightarrow \frac{\pi}{2}}(1+\cos 3 x)^{\sec x}=\lim _{x \rightarrow \frac{\pi}{2}}(1+\cos 3 x)^{\frac{1}{\cos x}}=$
Substitution:
$x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2}$
$x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0$
We get:
$=\lim _{y \rightarrow 0}\left(1+\cos 3\left(... | e^{-3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,916 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi}(x+\sin x)^{\sin x+x}$ | ## Solution
$\lim _{x \rightarrow \pi}(x+\sin x)^{\sin x+x}=(\pi+\sin \pi)^{\sin \pi+\pi}=(\pi+0)^{0+\pi}=\pi^{\pi}$
## Problem Kuznetsov Limits 20-20 | \pi^{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,917 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{2+\ln \left(e+x \sin \left(\frac{1}{x}\right)\right)}{\cos x+\sin x}$ | ## Solution
Since $\sin \left(\frac{1}{x}\right)_{\text {- is bounded as }} x \rightarrow 0$, then
$x \sin \left(\frac{1}{x}\right) \rightarrow 0 \quad$ as $x \rightarrow 0$
Then:
$\lim _{x \rightarrow 0} \frac{2+\ln \left(e+x \sin \left(\frac{1}{x}\right)\right)}{\cos x+\sin x}=\frac{2+\ln (e+0)}{\cos 0+\sin 0}=\f... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,918 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\frac{\pi}{2}}^{2 \operatorname{arctan} 2} \frac{d x}{\sin ^{2} x(1-\cos x)}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left(\frac{\pi}{2}\right)}{2}=\operatorname{tg}... | \frac{55}{96} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,919 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{\cos x \, dx}{2+\cos x}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\... | \frac{(9-4\sqrt{3})\pi}{18} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,920 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\frac{\pi}{2}}^{2 \operatorname{arctan} 2} \frac{d x}{\sin ^{2} x(1+\cos x)}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left(\frac{\pi}{2}\right)}{2}=\operatorname{tg}... | \frac{29}{24} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,921 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{2 \operatorname{arctg} \frac{1}{2}}^{\frac{\pi}{2}} \frac{\cos x d x}{(1-\cos x)^{3}}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=2 \operatorname{arctg} \frac{1}{2} \Rightarrow t=\operatorname{tg} \frac{2 \operatorname{arctg} \frac{1}{2}}{2}=\operatorname{tg... | 1.3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,922 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{\cos x - \sin x}{(1 + \sin x)^{2}} d x
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow... | \frac{1}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,923 |
## Problem Statement
Calculate the definite integral:
$$
\int_{2 \operatorname{arctan} 2}^{2 \operatorname{arctan} 3} \frac{d x}{\cos x(1-\cos x)}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=2 \operatorname{arctg} 2 \Rightarrow t=\operatorname{tg} \frac{2 \operatorname{arctg} 2}{2}=\operatorname{tg}(\operatorname{arct... | \frac{1}{6}+\ln2-\ln3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,924 |
## Problem Statement
Calculate the definite integral:
$$
\int_{2 \operatorname{arctg} \frac{1}{3}}^{2 \operatorname{arctg} \frac{1}{2}} \frac{d x}{\sin x(1-\sin x)}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=2 \operatorname{arctg} \frac{1}{3} \Rightarrow t=\operatorname{tg} \frac{2 \operatorname{arctg} \frac{1}{3}}{2}=\operatorname{tg}\le... | \ln3-\ln2+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,925 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{2 \operatorname{arctg} \frac{1}{2}}^{\frac{\pi}{2}} \frac{d x}{(1+\sin x-\cos x)^{2}}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=2 \operatorname{arctg} \frac{1}{2} \Rightarrow t=\operatorname{tg} \frac{2 \operatorname{arctg} \fra... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,926 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{\cos x d x}{5+4 \cos x}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\... | \frac{\pi}{8}-\frac{5}{6}\cdot\operatorname{arctg}\frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,927 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{2 \pi}{3}} \frac{1+\sin x}{1+\cos x+\sin x} d x
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{2 \pi}{3} \Rightarr... | \frac{\pi}{3}+\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,928 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\cos x \, dx}{1+\sin x-\cos x}
$$ | ## Solution
Let's use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=\frac{\pi}{3} \Rightarrow t=\operatorname{tg} \frac{\left(\frac{\pi}{3}\right)}{2}=\operatorname{tg} \... | \frac{1}{2}\cdot\ln2-\frac{\pi}{12} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,929 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{(1+\cos x) d x}{1+\sin x+\cos x}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow... | \frac{1}{2}\cdot\ln2+\frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,930 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin x \, dx}{1+\sin x+\cos x}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow... | -\frac{1}{2}\cdot\ln2+\frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,931 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{0}^{2 \operatorname{arctan} \frac{1}{2}} \frac{1+\sin x}{(1-\sin x)^{2}} d x
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=2 \operatorname{arctg} \frac{1}{2} \Rightarrow t=\operato... | \frac{26}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,932 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{\cos x \, dx}{1+\sin x+\cos x}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow... | \frac{\pi}{4}-\frac{1}{2}\cdot\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,933 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \operatorname{arctan} \frac{1}{3}} \frac{\cos x d x}{(1-\sin x)(1+\cos x)}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=2 \operatorname{arctg} \f... | -\frac{1}{3}-2\ln\frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,934 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\frac{2 \pi}{3}}^{0} \frac{\cos x \, dx}{1+\cos x-\sin x}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=-\frac{2 \pi}{3} \Rightarrow t=\operatorname{tg} \frac{\left(-\frac{2 \pi}{3}\right)}{2}=\operatorna... | \frac{\pi}{3}-\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,935 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\frac{\pi}{2}}^{0} \frac{\cos x \, dx}{(1+\cos x-\sin x)^{2}}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=-\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left(-\frac{\pi}{2}\right)}{2}=\operatorname{t... | -\frac{1}{2}+\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,936 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{\cos x \, dx}{(1+\cos x+\sin x)^{2}}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow... | -\frac{1}{2}+\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,937 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \operatorname{arctan} \frac{1}{2}} \frac{(1-\sin x) d x}{\cos x(1+\cos x)}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=2 \operatorname{arctg} \f... | -\frac{1}{2}+2\ln\frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,938 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin x \, dx}{(1+\sin x)^{2}}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,939 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin x d x}{(1+\cos x+\sin x)^{2}}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow... | \ln2-\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,940 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\frac{\pi}{2}}^{0} \frac{\sin x d x}{(1+\cos x-\sin x)^{2}}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=-\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left(-\frac{\pi}{2}\right)}{2}=\operatorname{t... | \frac{1}{2}-\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,941 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-\frac{2 \pi}{3}}^{0} \frac{\cos ^{2} x d x}{(1+\cos x-\sin x)^{2}}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=-\frac{2 \pi}{3} \Rightarrow t=\operatorname{tg} \frac{\left(-\frac{2 \pi}{3}\right)}{2}=\operatorna... | \frac{\sqrt{3}}{2}-\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,942 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x d x}{(1+\cos x+\sin x)^{2}}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow... | \frac{1}{2}-\frac{1}{2}\cdot\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,943 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{2 \pi}{3}} \frac{\cos ^{2} x d x}{(1+\cos x+\sin x)^{2}}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{2 \pi}{3} \Rightarr... | \frac{\sqrt{3}}{2}-\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,944 |
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