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Example 2. Prove that the function $y=x^{n} \quad$ ( $n$-integer, positive) is continuous in the interval $(-\infty, \infty)$. | Solution. The function is defined for all $x$, i.e., in the interval $(-\infty, \infty)$. The incremented value of the function for a fixed $x$ is
$$
\begin{aligned}
y+\Delta y=(x+\Delta x)^{n}= & x^{n}+n x^{n-1} \Delta x+\frac{n(n-1)}{1 \cdot 2} x^{n-2} \Delta x^{2}+ \\
& +\ldots+\Delta x^{n}
\end{aligned}
$$
from w... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,617 |
Example 4. Prove that the function $y=\cos x^{2}$ is continuous for any $x$. | Solution. The function $y=\cos x^{2}$ is a composite or function of a function, namely
$$
y=\cos z, z=x^{2}
$$
Each of these functions is continuous (Example 3, Example 2 when $n=2$). By property 4, the function $y=\cos x^{2}$ will also be continuous. | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,618 |
Example 5. Show that for the function $f(x)=\frac{x-2}{|x-2|}$, the point $x=2$ is a point of discontinuity of the 1st kind. | Solution. The function is undefined at the point $x=2$.
By the definition of absolute value (§ 1.1) we have
$$
\begin{gathered}
f(x)=\frac{x-2}{-(x-2)}=-1, \text { when }(x-2)<0 \text { or } x<2 \\
f(x)=\frac{x-2}{x-2}=1, \text { when }(x-2)>0 \text { or } x>2
\end{gathered}
$$
Since
$$
\lim _{x \rightarrow 2-0} f(... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,619 |
Example 7. Show that the function $f(x)=\frac{6}{(x-3)^{2}}$ has a discontinuity at the point $x=3$. | Solution. The function is defined at all points except $x=3$. For $x<3$, $f(x)>0$; for $x>3$, $f(x)>0$.
## We have
$$
\begin{aligned}
& \lim _{x \rightarrow 3-0} f(x)=+\infty \\
& \lim _{x \rightarrow 3+0} f(x)=+\infty
\end{aligned}
$$
The point $x=3$ is a point of discontinuity (Fig. 1.27).
, (2. $6^{\prime \prime \prime}$ ) and (2.8), we find
\[
\begin{gathered}
y^{\prime}=(x \cos x)^{\prime}=x^{\prime} \cos x+x(\cos x)^{\prime}=1 \cdot \cos x+x(-\sin x)= \\
=\cos x-x \sin x
\end{gathered}
\] | \cosx-x\sinx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,621 |
Example 4. Given the function $f(x)=x^{2}-3 x+2$. Calculate $f^{\prime}(-1)$, $f^{\prime}(0), f^{\prime}(1), f^{\prime}(2)$. | Solution. First, we find the derivative of the given function
$$
f^{\prime}(x)=2 x-3
$$
Substituting the values of the argument $x$ into the expression for the derivative, we get:
$$
\begin{gathered}
f^{\prime}(-1)=2(-1)-3=-5 ; f^{\prime}(0)=2 \cdot 0-3=-3 \\
f^{\prime}(1)=2 \cdot 1-3=-1 ; f^{\prime}(2)=2 \cdot 2-3=... | f^{\}(-1)=-5,f^{\}(0)=-3,f^{\}(1)=-1,f^{\}(2)=1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,622 |
Example 5. Find the derivative of the function $f(x)=\frac{x^{2}-2}{x^{2}+2}$. | Solution. By formula (2.5) we find
$$
\begin{gathered}
f^{\prime}(x)=\left(\frac{x^{2}-2}{x^{2}+2}\right)^{\prime}=\frac{\left(x^{2}-2\right)^{\prime}\left(x^{2}+2\right)-\left(x^{2}-2\right)\left(x^{2}+2\right)^{\prime}}{\left(x^{2}+2\right)^{2}}= \\
=\frac{2 x\left(x^{2}+2\right)-\left(x^{2}-2\right) 2 x}{\left(x^{2... | \frac{8x}{(x^{2}+2)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,623 |
Example 1. Find the derivative of the function $y=\sin 5 x$. | Solution. The argument of the sine here is not $x$, but $5 x$. This is a composite trigonometric function, which can be represented as:
$$
y=\sin z, z=5 x
$$
We have:
$$
y_{z}^{\prime}=(\sin z)_{z}^{\prime}=\cos z=\cos 5 x ; z_{x}^{\prime}=(5 x)_{x}^{\prime}=5
$$
Substituting the expressions for $y_{z}^{\prime}$ an... | 5\cos5x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,624 |
Example 8. Find the derivative of the function $y=\frac{\cos ^{2} x}{\sin x}$. | Solution. Differentiating it as a quotient and a composite function, we find
$$
\begin{aligned}
y^{\prime}= & \left(\frac{\cos ^{2} x}{\sin x}\right)^{\prime}=\frac{\left(\cos ^{2} x\right)^{\prime} \sin x-\cos ^{2} x(\sin x)^{\prime}}{\sin ^{2} x}= \\
& =\frac{2 \cos x(-\sin x) \sin x-\cos ^{2} x \cos x}{\sin ^{2} x}... | -\cosx(\operatorname{cosec}^{2}x+1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,625 |
Example 1. Find the derivative of the function $y=4^{x}$. | Solution. We use formula (2.12). In this case $a=4$. We get
$$
\left(4^{x}\right)^{\prime}=4^{x} \ln 4=2 \ln 2 \cdot 4^{x}
$$ | 4^{x}\ln4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,626 |
Example 2. Find the derivative of the function $y=a^{x^{2}}$. | Solution. By formula ( $2.12^{\prime}$ ) we get
$$
\left(a^{x^{2}}\right)^{\prime}=a^{x^{2}} \cdot \ln a \cdot\left(x^{2}\right)^{\prime}=a^{x^{2}} \cdot \ln a \cdot 2 x=2 \ln a \cdot x \cdot a^{x^{2}}
$$ | 2\ln\cdotx\cdot^{x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,627 |
Example 5. Find the derivative of the function $y=\ln \sqrt{\frac{1+2 x}{1-2 x}}$. | ## Solution.
$$
\begin{gathered}
y^{\prime}=\left[\ln \left(\frac{1+2 x}{1-2 x}\right)^{\frac{1}{2}}\right]^{\prime}=\left[\frac{1}{2} \ln \left(\frac{1+2 x}{1-2 x}\right)\right]^{\prime}= \\
=\frac{1}{2}[\ln (1+2 x)-\ln (1-2 x)]^{\prime}=\frac{1}{2}\left(\frac{2}{1+2 x}+\frac{2}{1-2 x}\right)= \\
=\left(\frac{1}{1+2 ... | \frac{2}{1-4x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,628 |
Example 2. Find the derivative of the function $y=\sqrt{1-x^{2}} \arccos x$. | Solution. Differentiating using formulas (2.4) and (2.17), we find
$$
\begin{aligned}
& \left(\sqrt{1-x^{2}} \arccos x\right)^{\prime}=\left(\sqrt{1-x^{2}}\right)^{\prime} \arccos x+ \\
& +\sqrt{1-x^{2}}(\arccos x)^{\prime}=-\frac{x}{\sqrt{1-x^{2}}} \arccos x- \\
& -\sqrt{1-x^{2}} \cdot \frac{1}{\sqrt{1-x^{2}}}=-\left... | -(1+\frac{x}{\sqrt{1-x^{2}}}\arccosx) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,629 |
Example 3. Find the derivative of the function
$$
y=\operatorname{arctg} \frac{x}{a}+\frac{1}{2} \ln \left(x^{2}+a^{2}\right)
$$ | Solution. Differentiate term by term and use formula $\left(2.18^{\prime}\right):$
$$
\begin{gathered}
{\left[\operatorname{arctg} \frac{x}{a}+\frac{1}{2} \ln \left(x^{2}+a^{2}\right)\right]^{\prime}=\left(\operatorname{arctg} \frac{x}{a}\right)^{\prime}+\frac{1}{2}\left[\ln \left(x^{2}+a^{2}\right)\right]^{\prime}=} ... | \frac{x+}{x^{2}+^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,630 |
Example 5. Find the derivative of the function $y=\arcsin \frac{1}{x}$. | Solution. Using formula (2.16'), we get
$$
\begin{gathered}
\left(\arcsin \frac{1}{x}\right)^{\prime}=\frac{1}{\sqrt{1-\left(\frac{1}{x}\right)^{2}}}\left(\frac{1}{x}\right)^{\prime}=\frac{1}{\sqrt{\frac{x^{2}-1}{x^{2}}}}\left(-\frac{1}{x^{2}}\right)= \\
=-\frac{1}{|x| \sqrt{x^{2}-1}}
\end{gathered}
$$ | -\frac{1}{|x|\sqrt{x^{2}-1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,631 |
Example 3. Calculate the value of the derivative of the implicit function $x y^{2}=4$ at the point $M(1,2)$. | Solution. First, let's find the derivative:
$$
x^{\prime} y^{2}+x 2 y y^{\prime}=0, y^{\prime}=-\frac{y}{2 x}
$$
Substituting the values \(x=1\), \(y=2\) into the right-hand side of the last equation, we get
$$
y^{\prime}=-\frac{2}{2 \cdot 1}=-1
$$ | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,632 |
Example 4. Find $x_{t}^{\prime}$ at $t=1$, if $t \ln x - x \ln t = 1$. | Solution. This equation defines $x$ as an implicit function of $t$.
Differentiate with respect to $t$:
$$
t_{t}^{\prime} \cdot \ln x + t \cdot \frac{1}{x} x_{t}^{\prime} - \left(x_{t}^{\prime} \ln t + x \cdot \frac{1}{t}\right) = 0
$$
Since $t_{i}^{\prime}=1$, we have
$$
\ln x + \frac{t}{x} x_{t}^{\prime} - x_{t}^{... | e(1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,633 |
Example 1. Find the second-order derivative of the function $y=\sin ^{2} x$. | Solution. Differentiating, we obtain the first derivative
$$
y^{\prime}=2 \sin x \cos x=\sin 2 x
$$
Differentiating again, we find the required second-order derivative:
$$
y^{\prime \prime}=(\sin 2 x)^{\prime}=\cos 2 x \cdot(2 x)^{\prime}=2 \cos 2 x
$$ | 2\cos2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,634 |
Example 2. Find the third-order derivative of the function $y=x^{2}+3 x+2$. | Solution. Differentiating successively, we obtain:
$$
\begin{gathered}
y^{\prime}=\left(x^{2}+3 x+2\right)^{\prime}=2 x+3 ; y^{\prime \prime}=(2 x+3)^{\prime}=2 \\
y^{\prime \prime \prime}=(2)^{\prime}=0
\end{gathered}
$$ | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,635 |
Example 3. Find the fourth-order derivative of the function $y=\sin x$. | Solution. Differentiating successively, we determine:
$$
y^{\prime}=\cos x, y^{\prime \prime}=-\sin x, y^{\prime \prime \prime}=-\cos x, y^{\mathrm{IV}}=\sin x
$$ | y^{\mathrm{IV}}=\sinx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,636 |
Example 4. Find $f(0), f^{\prime}(0), f^{\prime \prime}(0), f^{\prime \prime \prime}(0), f^{\mathrm{IV}}(0)$, if $f(x)=\cos 2 x$. | Solution. We find the first, second, third, and fourth order derivatives:
$$
\begin{gathered}
f^{\prime}(x)=-2 \sin 2 x, f^{\prime \prime}(x)=-4 \cos 2 x, f^{\prime \prime \prime}(x)=8 \sin 2 x \\
f^{\mathrm{IV}}(x)=16 \cos 2 x
\end{gathered}
$$
By assigning the value of zero to \( x \), we find
$$
f(0)=1, f^{\prime... | f(0)=1,f^{\}(0)=0,f^{\\}(0)=-4,f^{\\\}(0)=0,f^{\text{IV}}(0)=16 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,637 |
Example 5. Find the tenth derivative for the function $y=$ $=e^{x}\left(x^{3}-2\right)$. | Solution. Applying the Leibniz formula, we get
$$
\begin{aligned}
y^{(10)}= & {\left[e^{x}\left(x^{3}-2\right)\right]^{(10)}=\left(e^{x}\right)^{(10)}\left(x^{3}-2\right)+10\left(e^{x}\right)^{(9)}\left(x^{3}-2\right)^{\prime}+} \\
& +\frac{10 \cdot 9}{2}\left(e^{x}\right)^{(8)}\left(x^{3}-2\right)^{\prime \prime}+\fr... | y^{(10)}=e^{x}(x^{3}+30x^{2}+270x+718) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,638 |
Example 4. Find the derivative $y_{x}^{\prime}$ of the function $x=e^{t} \cos t$; $y=e^{t} \sin t$ at $t=0$. | Solution. The functions $x$ and $y$ have the following derivatives with respect to $t$:
$$
\begin{aligned}
& x_{i}^{\prime}=e^{t} \cos t-e^{t} \sin t=e^{t}(\cos t-\sin t) \\
& y_{t}^{\prime}=e^{t} \sin t+e^{t} \cos t=e^{t}(\sin t+\cos t)
\end{aligned}
$$
therefore
$$
y_{x}^{\prime}=\frac{\sin t+\cos t}{\cos t-\sin t... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,639 |
Example 1. Find the differential of the function $y=\sin x$. | Solution. By formula (2.25) we find
$$
d y=d(\sin x)=(\sin x)^{\prime} d x=\cos x d x, d y=\cos x d x
$$ | \cosx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,640 |
Example 4. Calculate $y=x^{3}+2 x$, when $x$ varies from 1 to 1.1. | Solution. First, we find the general expression for the differential of this function:
$$
d y=\left(3 x^{2}+2\right) d x
$$
Substituting the values $x=1, d x=\Delta x=1.1-1=0.1$ into the last formula, we obtain the desired value of the differential:
$$
d y=\left(3 \cdot 1^{2}+2\right) 0.1=5 \cdot 0.1=0.5
$$ | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,641 |
Example 1. Find $\lim _{x \rightarrow 0} \frac{\sin x e^{x}-5 x}{4 x^{2}+7 x}$. | Solution. When $x \rightarrow 0$, both the numerator and the denominator of the fraction approach zero, resulting in an indeterminate form of $\frac{0}{0}$.
Applying L'Hôpital's rule, we find
$$
\lim _{x \rightarrow 0} \frac{\sin x e^{x}-5 x}{4 x^{2}+7 x}=\lim _{x \rightarrow 0} \frac{\left(\sin x e^{x}-5 x\right)^{\... | -\frac{4}{7} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,642 |
Example 2. Find $\lim _{x \rightarrow 0} \frac{\sin x e^{2 x}-x}{5 x^{2}+x^{3}}$. | Solution. In this case, the L'Hôpital-Bernoulli rule needs to be applied twice, as the ratio of the first derivatives again represents an indeterminate form of $\frac{0}{0}$.
Indeed,
$$
\lim _{x \rightarrow 0} \frac{\sin x e^{2 x}-x}{5 x^{2}+x^{3}}=\lim _{x \rightarrow 0} \frac{\left(\sin x e^{2 x}-x\right)^{\prime}}... | \frac{2}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,643 |
Example 3. Find $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan x}{\tan 3 x}$. | Solution. Using the L'Hôpital-Bernoulli rule, we get
$$
\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan x}{\tan 3 x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{1}{\cos ^{2} x}}{\frac{3}{\cos ^{2} 3 x}}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos ^{2} 3 x}{3 \cos ^{2} x}
$$
The limit of the ratio of the first d... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,644 |
Example 4. Find $\lim _{x \rightarrow \infty} \frac{a^{x}}{x^{\alpha}}$ for $a>1, \alpha>0$. | Solution. When calculating this limit, we can assume $x>1$ (since $x \rightarrow \infty$) and if $n$ is the nearest integer greater than $\alpha$, then
$$
\frac{a^{x}}{x^{\alpha}}>\frac{a^{x}}{x^{n}}(n>0)
$$
Therefore,
$$
\lim _{x \rightarrow \infty} \frac{a^{x}}{x^{\alpha}} \geq \lim _{x \rightarrow \infty} \frac{a... | \lim_{xarrow\infty}\frac{^{x}}{x^{\alpha}}=\infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,645 |
Example 5. Find $\lim _{x \rightarrow \infty} \frac{\ln x}{x^{\alpha}}$ for $\alpha>0, x>0$. | Solution. Applying the L'Hôpital-Bernoulli rule, we get
$$
\lim _{x \rightarrow \infty} \frac{\ln x}{x^{\alpha}}=\lim _{x \rightarrow \infty} \frac{1}{x \alpha x^{\alpha-1}}=\lim _{x \rightarrow \infty} \frac{1}{\alpha x^{\alpha}}=0
$$
Thus,
$$
\lim _{x \rightarrow \infty} \frac{\ln x}{x^{\alpha}}=0 \text { for } \a... | \lim_{xarrow\infty}\frac{\lnx}{x^{\alpha}}=0\text{for}\alpha>0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,646 |
Example 6. Find $\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{\sin x}\right)$. | Solution. When $x \rightarrow \mathbf{0}$, we get an indeterminate form of $\infty-\infty$. We will resolve this indeterminacy by converting it to an indeterminate form of $\frac{0}{0}$ and applying L'Hôpital-Bernoulli's rule,
$$
\begin{gathered}
\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{\sin x}\right)=\lim _{... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,647 |
Example 7. Find $\lim _{x \rightarrow 0}[(x-\sin x) \ln x]$. | Solution. Here we have an indeterminate form $0 \cdot \infty$. The given function can be represented as
$$
(x-\sin x) \ln x=\frac{\ln x}{\frac{1}{x-\sin x}}
$$
The resulting indeterminate form $\frac{\infty}{\infty}$ is resolved using L'Hôpital's rule:
$$
\lim _{x \rightarrow 0} \frac{\ln x}{\frac{1}{x-\sin x}}=\lim... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,648 |
Example 8. Find $\lim _{x \rightarrow 1}(x-1)^{\ln x}$. | Solution. When $x \rightarrow 1$, we have the indeterminate form $0^{0}$. We will use the identity
$$
[f(x)]^{\varphi(x)}=e^{\varphi(x) \ln f(x)}
$$
which in this case will be
$$
(x-1)^{\ln x}=e^{\ln x \cdot \ln (x-1)}
$$
We have
$$
\lim _{x \rightarrow 1}(x-1)^{\ln x}=\lim _{x \rightarrow 1} e^{\ln x \cdot \ln (x... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,649 |
Example 9. Find $\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{\frac{1}{x}}$. | Solution. Since $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1, \lim _{x \rightarrow 0} \frac{1}{x}=\infty$, we have an indeterminate form of $1^{\infty}$ here.
Taking into account identity (A) (see example 8), using L'Hôpital-Bernoulli's rule, we find
$$
\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{\frac{1}{x}... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,650 |
Example 11. Find $\lim _{x \rightarrow \infty} \frac{x-\sin x}{x+\sin x}$. | Solution. Apply L'Hôpital's rule:
$$
\lim _{x \rightarrow \infty} \frac{x-\sin x}{x+\sin x}=\lim _{x \rightarrow \infty} \frac{1-\cos x}{1+\cos x}
$$
In the right-hand side of the last equality, the limit does not exist, so L'Hôpital's rule is not applicable here.
The specified limit can be found directly:
$$
\lim ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,652 |
Example 1. Write the equations of the tangent and normal to the curve $f(x)=x^{3}$ at the point $M_{0}(2,8)$. | Solution. First of all, the point $M_{0}$ lies on the curve, since its coordinates satisfy the given equation.
We find the derivative of the given function and its value at $x_{0}=2$:
$$
f^{\prime}(x)=\left(x^{3}\right)^{\prime}=3 x^{2}, f^{\prime}\left(x_{0}\right)=f^{\prime}(2)=3 \cdot 2^{2}=12
$$
Substituting the... | 12x-y-16=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,653 |
Example 2. Derive the equations of the tangent and normal to the ellipse
$$
\frac{x^{2}}{18}+\frac{y^{2}}{8}=1
$$
at the point $M_{0}(3,2)$. | Solution. We find $\frac{x^{2}}{18}+\frac{y^{2}}{8}-1=0$ :
$$
\frac{2 x}{18}+\frac{2 y y^{\prime}}{8}=0,
$$
from which
$$
y^{\prime}(x)=-\frac{4 x}{9 y}, y^{\prime}\left(x_{0}\right)=y^{\prime}(3)=-\frac{4 \cdot 3}{2 \cdot 9}=-\frac{2}{3}
$$
Substituting the values $x_{0}=3, y_{0}=2, y^{\prime}\left(x_{0}\right)=-\... | 2x+3y-12=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,654 |
Example 3. Form the equations of the tangent and normal to the curve given parametrically,
$$
x=2(t-\sin t) ; y=2(1-\cos t)
$$
1) at an arbitrary point; 2) at $t=\frac{\pi}{2}$. | Solution. The equation of the tangent (3.1) can be written as
$$
Y-y=y_{x}^{\prime}(x)(X-x)
$$
where $(x, y)$ are the coordinates of a point on the curve, and $(X, Y)$ are the coordinates of a point on the tangent.
We find the derivative $y_{x}^{\prime}$ using the formula
$$
y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{... | X-Y-\pi+4=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,655 |
Example 4. Prove that the curvature of the line $x^{2}+y^{2}=25$ is constant. | Solution. We will calculate the required curvature using formula (3.3). First, we find the first and second derivatives of the implicit function $x^{2}+y^{2}-25=0$. Differentiating, we get:
$$
\begin{gathered}
2 x+2 y y^{\prime}=0, y^{\prime}=-\frac{x}{y} \\
y^{\prime \prime}=\left(-\frac{x}{y}\right)^{\prime}=-\frac{... | \frac{1}{5} | Calculus | proof | Yes | Yes | olympiads | false | 32,656 |
Example 5. Find the radius of curvature of the cycloid $x=a(t-\sin t), y=$ $=a(1-\cos t)$ at any point of it. | Solution. From formulas (3.4) and (3.5), it follows that the radius of curvature of a curve given by parametric equations is determined by the formula
$$
R=\frac{\left(x^{\prime 2}+y^{\prime 2}\right)^{\frac{3}{2}}}{\left|x^{\prime} y^{\prime \prime}-y^{\prime} x^{\prime \prime}\right|}
$$
Let's find the first and se... | 4|\sin\frac{}{2}| | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,657 |
Example 7. Write the equation of the evolute of the cycloid
$$
x=a(t-\sin t), y=a(1-\cos t)
$$ | Solution. Since
$$
x^{\prime 2}+y^{\prime 2}=2 a^{2}(1-\cos t), x^{\prime} y^{\prime \prime}-y^{\prime} x^{\prime \prime}=-a^{2}(1-\cos t)
$$
(see example 5), then according to formulas (3.7) we get
$$
\begin{gathered}
X=x-y^{\prime} \cdot \frac{x^{\prime 2}+y^{\prime 2}}{x^{\prime} y^{\prime \prime}-y^{\prime} x^{\... | X=(\tau-\sin\tau)+\pi;\quadY=(1-\cos\tau)-2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,658 |
Example 9. At what angles do the curves $y=x^{2}$, $y^{2}=x$ intersect? | Solution. The angle between two curves at their point of intersection is called the angle between the tangents to the curves at this point.
Let's find the points of intersection of the given lines. Solving the system of their equations
$$
\left.\begin{array}{l}
y=x^{2} \\
y^{2}=x,
\end{array}\right\}
$$
we get two p... | 90,\operatorname{arctg}\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,659 |
Example 1. Find the intervals of increase and decrease of the function
$$
f(x)=x^{3}-\frac{3}{2} x^{2}-6 x+4
$$ | Solution. We find the first derivative:
$$
f^{\prime}(x)=3 x^{2}-3 x-6=3\left(x^{2}-x-2\right)
$$
The derivative is zero when $x^{2}-x-2=0$, from which $x_{1}=-1, x_{2}=2$. These points divide the infinite interval into three intervals: $(-\infty,-1),(-1,2),(2, \infty)$. We investigate the sign of the derivative in e... | (-\infty,-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,660 |
Example 2. Investigate the function for maximum and minimum
$$
f(x)=\frac{1}{3} x^{3}-\frac{1}{2} x^{2}-6 x+2 \frac{2}{3}
$$ | Solution. We find the first derivative:
$$
f^{\prime}(x)=x^{2}-x-6
$$
The derivative is continuous everywhere. Let's find the points where it is zero: $x^{2}-x-6=0$.
The critical points of the function argument are $x_{1}=-2, x_{2}=3$.
We will investigate the critical points using the first derivative.
Since
$$
f... | -10\frac{5}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,661 |
Example 3. Investigate the function for extremum
$$
f(x)=\frac{1}{4} x^{4}-2 x^{3}+\frac{11}{2} x^{2}-6 x+\frac{9}{4}
$$ | Solution. We will use the second rule of the sufficient condition for an extremum. We find the first and second derivatives:
$$
f^{\prime}(x)=x^{3}-6 x^{2}+11 x-6, f^{\prime \prime}(x)=3 x^{2}-12 x+11
$$
We look for points of extremum among the critical points, i.e., points for which $f^{\prime}(x)=0$. The first deri... | \f(x)=0,\maxf(x)=\frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,662 |
Example 4. Investigate the function for extremum
$$
f(x)=(x-2) \sqrt[3]{x^{2}}
$$ | Solution. We find the first derivative:
$$
f^{\prime}(x)=\sqrt[3]{x^{2}}+\frac{2(x-2)}{3 \sqrt[3]{x}}=\frac{5 x-4}{3 \sqrt[3]{x}}
$$
The derivative is zero at the point $x_{1}=\frac{4}{5}$, and it does not exist at the point $x_{2}=0$ (it goes to infinity). Therefore, the critical points are $x_{1}=\frac{4}{5}$ and $... | \f(x)=-\frac{6}{5}\sqrt[3]{\frac{16}{25}},\maxf(x)=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,663 |
Example 5. Find the maximum and minimum values of the function $f(x)=2 x^{3}-6 x+5$ on the interval $\left[-\frac{5}{2}, \frac{3}{2}\right]$. | Solution. We find the extrema of the function:
\[
\begin{gathered}
f^{\prime}(x)=6 x^{2}-6=6\left(x^{2}-1\right) \\
f^{\prime}(x)=0,6\left(x^{2}-1\right)=0, x_{1}=-1, x_{2}=1 \\
f^{\prime \prime}(x)=12 x, f^{\prime \prime}(-1)0
\end{gathered}
\]
Therefore, at the point \( x_{1}=-1 \) the function has a maximum, and
... | f(-1)=9,f(-\frac{5}{2})=-11\frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,664 |
Example 6. Given a positive number a, decompose it into two addends so that their product is the largest possible. | Solution. Let $x$ be the first addend, then the second is $a-x$. Their product represents a function of $x$:
$$
f(x)=x(a-x)=a x-x^{2}
$$
We will investigate this function. Let's find the derivatives:
$$
f^{\prime}(x)=a-2 x, f^{\prime \prime}(x)=-2
$$
The first derivative is zero when $a-2 x=0$ or $x=\frac{a}{2}$. S... | \frac{}{2},-\frac{}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,665 |
Example 7. What dimensions of a box (without a lid) made from a square sheet of cardboard with side $a$ will have the maximum capacity | Solution. To manufacture a box, it is necessary to cut out squares from the corners of the sheet and fold the protrusions of the resulting cross-shaped figure. Let the side of the cut-out square be denoted by $x$, then the side of the box base will be $a-2 x$. The volume of the box can be expressed by the function $V=(... | \frac{}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,666 |
Example 1. Find the intervals of concavity and convexity, as well as the inflection points of the Gaussian curve $y=e^{-x^{2}}$. | Solution. We find the first and second derivatives:
$$
y^{\prime}=-2 x e^{-x^{2}} ; y^{\prime \prime}=\left(4 x^{2}-2\right) e^{-x^{2}}
$$
Setting the second derivative to zero, we obtain the critical points of the second kind:
$$
x_{1}=-\frac{1}{\sqrt{2}} ; x_{2}=\frac{1}{\sqrt{2}}
$$
These points divide the numbe... | M_{1}(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{e}})M_{2}(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{e}}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,667 |
Example 4. Find the asymptotes of the curve $f(x)=\frac{x^{2}+1}{x}$. | Solution. Since
$$
\lim _{x \rightarrow 0} \frac{x^{2}+1}{x}=\infty
$$
the line $x=0$ is a vertical asymptote (see formula (3.12)).
Further,
$$
f(x)=\frac{x^{2}+1}{x}=x+\frac{1}{x}=x+\alpha(x)
$$
where $\alpha(x) \rightarrow 0$ as $x \rightarrow \pm \infty$.
Thus, the line $y=x$ (see formula (3.15)) will be an as... | x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,668 |
Example 5. Find the asymptotes of the curve $y=\frac{x^{3}}{1-x^{2}}$. | Solution. The graph of this function has two vertical asymptotes $x=-1, x=1$, since
$$
\lim _{x \rightarrow-1} \frac{x^{3}}{1-x^{2}}=\lim _{x \rightarrow 1} \frac{x^{3}}{1-x^{2}}=\infty
$$
By isolating the integer part of the function through direct division or using the following simple transformations, we get
$$
y... | -x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,669 |
Example 6. Find the asymptotes of the curve $y=\frac{x^{2}}{\sqrt{x^{2}-1}}$. | Solution. By equating the denominator to zero, we obtain two vertical asymptotes: $x=-1, x=1$.
Using formulas (3.13) and (3.14), we find the oblique asymptotes.
We have:
$$
\begin{gathered}
k=\lim _{x \rightarrow+\infty} \frac{y}{x}=\lim _{x \rightarrow+\infty} \frac{x^{2}}{x \sqrt{x^{2}-1}}=1 \\
b=\lim _{x \rightar... | x,-x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,670 |
Example 1. Investigate the function $f(x)=\frac{x^{3}}{x^{2}-3}$ and plot its graph. | Solution. We use the function investigation scheme.
1. The function is undefined only at points where the denominator is zero, i.e., at $x_{1}=-\sqrt{3}, x_{2}=\sqrt{3}$. Therefore, the domain consists of three intervals: $(-\infty,-\sqrt{3}),(-\sqrt{3}, \sqrt{3})$, $(\sqrt{3}, \infty)$
2. We investigate the change in... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,671 | |
Example 4. Investigate the function $x^{2}=y^{2}+x^{4}$ and plot its graph. | Solution. Solving the given equation for $y$, we get
$$
y= \pm x \sqrt{1-x^{2}}
$$
Let's investigate the function $y=x \sqrt{1-x^{2}}$. This function is defined for $1-x^{2} \geq 0$ or $x^{2} \leq 1$, i.e., for $-1 \leq x \leq 1$. The domain of the function is the segment $[-1,1]$. Therefore, the graph of the functio... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,673 |
Example 5. Investigate the function $y=\ln \left|\frac{x-1}{x+1}\right|$ and plot its graph. | Solution. The function is undefined at $x=-1$ and $x=1$. The domain of the function consists of three intervals: $(-\infty,-1),(-1,1)$, $(1,+\infty)$.
At the ends of the intervals of the domain of existence, we have:
$$
\begin{aligned}
& \lim _{x \rightarrow-\infty} \ln \left|\frac{x-1}{x+1}\right|=0, \lim _{x \right... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,674 |
Example 7. Plot the graph of the function $y=\sqrt[3]{6 x^{2}-x^{3}}$. | Solution. The function is defined and continuous for all $x$. The first derivative
$$
y^{\prime}=\frac{12 x-3 x^{2}}{3 \sqrt[3]{\left(6 x^{2}-x^{3}\right)^{2}}}=\frac{4-x}{\sqrt[3]{x(6-x)^{2}}}
$$
exists everywhere except at the points $x_{1}=0, x_{2}=6$.
We investigate the limit values of the derivative as $x$ appr... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,676 |
Example 1. Find the integral $\int x^{3} d x$. | Solution. We apply formula (4.5) for the case $m=3$. According to this formula, we get
$$
\int x^{3} d x=\frac{x^{3+1}}{3+1}+C=\frac{x^{4}}{4}+C
$$ | \frac{x^{4}}{4}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,677 |
Example 2. Find the integral $\int \frac{x^{4}-2 x^{3}+3 x^{2}}{x^{2}} d x$. | Solution. Dividing term by term the numerator by the denominator, using properties 3 and 4 and formula (4.5), we find
$$
\begin{gathered}
\int \frac{x^{4}-2 x^{3}+3 x^{2}}{x^{2}} d x=\int\left(x^{2}-2 x+3\right) d x=\int x^{2} d x-2 \int x d x+3 \int d x= \\
=\frac{x^{2+1}}{2+1}-2 \frac{x^{1+1}}{1+1}+3 x+C=\frac{x^{3}... | \frac{x^{3}}{3}-x^{2}+3x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,678 |
Example 3. Find the integral $\int\left(1-\frac{1}{x^{2}}\right)^{2} d x$. | Solution. Expanding the brackets and using formula (4.5) for the case when $m$ is a negative number, we find
$$
\begin{aligned}
& \int\left(1-\frac{1}{x^{2}}\right)^{2} d x=\int\left(1-\frac{2}{x^{2}}+\frac{1}{x^{4}}\right) d x=\int d x-2 \int \frac{d x}{x^{2}}+\int \frac{d x}{x^{4}}= \\
& =\int d x-2 \int x^{-2} d x+... | x+\frac{2}{x}-\frac{1}{3x^{3}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,679 |
Example 4. Find the integral $\int \sqrt{x} d x$. | Solution.
$$
\int \sqrt{x} d x=\int x^{\frac{1}{2}} d x=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{2}{3} x^{\frac{3}{2}}+C=\frac{2}{3} x \sqrt{x}+C
$$ | \frac{2}{3}x\sqrt{x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,680 |
Example 5. Find the integral $\int \operatorname{ctg}^{2} x d x$. | $$
\begin{gathered}
\int \operatorname{ctg}^{2} x d x=\int\left(\operatorname{cosec}^{2} x-1\right) d x=\int \operatorname{cosec}^{2} x d x-\int d x= \\
=\int \frac{1}{\sin ^{2} x} d x-\int d x=-\operatorname{ctg} x-x+C
\end{gathered}
$$
Translation:
$$
\begin{gathered}
\int \cot^{2} x \, dx = \int \left(\csc^{2} x -... | -\cotx-x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,681 |
Example 6. Find the integral $\int\left(a x^{3}+b x^{2}+c x+p\right) d x$. | Solution.
$$
\begin{gathered}
\int\left(a x^{3}+b x^{2}+c x+p\right) d x=a \int x^{3} d x+b \int x^{2} d x+c \int x d x+ \\
+p \int d x=\frac{a}{4} x^{4}+\frac{b}{3} x^{3}+\frac{c}{2} x^{2}+p x+C
\end{gathered}
$$ | \frac{}{4}x^{4}+\frac{b}{3}x^{3}+\frac{}{2}x^{2}+px+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,682 |
Example 7. Find the integral $\int \frac{d x}{\sin ^{2} x \cos ^{2} x}$. | Solution.
$$
\begin{gathered}
\int \frac{d x}{\sin ^{2} x \cos ^{2} x}=\int \frac{\cos ^{2} x+\sin ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x} d x=\int \frac{d x}{\sin ^{2} x}+\int \frac{d x}{\cos ^{2} x}= \\
=-\operatorname{ctg} x+\operatorname{tg} x+C
\end{gathered}
$$ | -\operatorname{ctg}x+\operatorname{tg}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,683 |
Example 9. Find the integral $\int \sin ^{2} \frac{x}{2} d x$. | Solution. Since
$$
\sin ^{2} \frac{x}{2}=\frac{1}{2}(1-\cos x)
$$
then
$$
\int \sin ^{2} \frac{x}{2} d x=\frac{1}{2} \int(1-\cos x) d x=\frac{1}{2} \int d x-\frac{1}{2} \int \cos x d x=\frac{1}{2} x-\frac{1}{2} \sin x+C
$$
## Problems
Find the indefinite integrals:
1. $\int \frac{x^{4}-3 x^{3}+4 x^{2}+6 x-8}{x^{2... | \frac{1}{2}x-\frac{1}{2}\sinx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,684 |
E x a m p l e 1. Find the indefinite integral $\int(2 x+3)^{2} d x$. | Solution. Based on the transformation of 3 differentials, we have
$$
d x=\frac{1}{2} d(2 x+3)
$$
Applying formula (9.5') for the case where $u=2 x+3, \alpha=2$, we find
$$
\int(2 x+3)^{2} d x=\int(2 x+3)^{2} \frac{1}{2} d(2 x+3)=\frac{1}{2} \int(2 x+3)^{2} d(2 x+3)=
$$
$$
=\frac{1}{2} \frac{(2 x+3)^{3}}{3}+C=\frac{... | \frac{1}{6}(2x+3)^{3}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,685 |
Example 2. Find the integral $\int \sqrt{x+4} d x$. | Solution. According to transformation 1,
$$
d x=d(x+4)
$$
Applying formula (9.5') for the case when $u=x+4$ and $\alpha=\frac{1}{2}$, we get
$$
\begin{gathered}
\int \sqrt{x+4} d x=\int(x+4)^{\frac{1}{2}} d(x+4)=\frac{(x+4)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C= \\
=\frac{2}{3}(x+4)^{\frac{3}{2}}+C=\frac{2}{3}(x+4) \sqr... | \frac{2}{3}(x+4)\sqrt{x+4}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,686 |
Example 3. Find the integral $\int \frac{d x}{a x+b}$. | Solution.
$$
\int \frac{d x}{a x+b}=\int \frac{\frac{1}{a} d(a x+b)}{(a x+b)}=\frac{1}{a} \int \frac{d(a x+b)}{(a x+b)}=\frac{1}{a} \ln |a x+b|+C
$$ | \frac{1}{}\ln|+b|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,687 |
Example 4. Find the integral $\int \frac{x d x}{x^{2}+2}$. | Solution.
$$
\int \frac{x d x}{x^{2}+2}=\int \frac{\frac{1}{2} d\left(x^{2}+2\right)}{x^{2}+2}=\frac{1}{2} \int \frac{d\left(x^{2}+2\right)}{x^{2}+2}=\frac{1}{2} \ln \left(x^{2}+2\right)+C
$$ | \frac{1}{2}\ln(x^{2}+2)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,688 |
Example 5. Find the integral $\int \operatorname{tg} x d x$.
Note: In English, the tangent function is usually denoted as $\tan x$ rather than $\operatorname{tg} x$. So the integral would typically be written as $\int \tan x \, dx$. | ## Solution.
$$
\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx = -\int \frac{d(\cos x)}{\cos x} = -\ln |\cos x| + C
$$ | -\ln|\cosx|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,689 |
Example 7. Find the integral $\int e^{\frac{x}{2}} d x$. | Solution.
$$
\int e^{\frac{x}{2}} d x=\int e^{\frac{x}{2}} 2 d\left(\frac{x}{2}\right)=2 \int e^{\frac{x}{2}} d\left(\frac{x}{2}\right)=2 e^{\frac{x}{2}}+C
$$ | 2e^{\frac{x}{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,690 |
Example 8. Find the integral $\int \cos \frac{x}{4} d x$. | Solution.
$$
\int \cos \frac{x}{4} d x=\int \cos \frac{x}{4} 4 d\left(\frac{x}{4}\right)=4 \int \cos \frac{x}{4} d\left(\frac{x}{4}\right)=4 \sin \frac{x}{4}+C
$$ | 4\sin\frac{x}{4}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,691 |
Example 9. Find the integral $\int \frac{d x}{\sin ^{2} 3 x}$. | Solution.
$$
\int \frac{d x}{\sin ^{2} 3 x}=\int \frac{\frac{1}{3} d(3 x)}{\sin ^{2} 3 x}=\frac{1}{3} \int \frac{d(3 x)}{\sin ^{2} 3 x}=-\frac{1}{3} \operatorname{ctg}^{2} 3 x+C
$$ | -\frac{1}{3}\operatorname{ctg}^{2}3x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,692 |
Example 10. Find the integral $\int \frac{d x}{1+4 x^{2}}$. | Solution.
$$
\int \frac{d x}{1+4 x^{2}}=\int \frac{d x}{1+(2 x)^{2}}=\frac{1}{2} \int \frac{d(2 x)}{1+(2 x)^{2}}=\frac{1}{2} \operatorname{arctg} 2 x+C
$$ | \frac{1}{2}\operatorname{arctg}2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,693 |
Example 11. Find the integral $\int \frac{d x}{\sqrt{1-9 x^{2}}}$. | Solution.
$$
\int \frac{d x}{\sqrt{1-9 x^{2}}}=\int \frac{d x}{\sqrt{1-(3 x)^{2}}}=\frac{1}{3} \int \frac{d(3 x)}{\sqrt{1-(3 x)^{2}}}=\frac{1}{3} \arcsin 3 x+C
$$
## Problems
Using the simplest transformations of the differential and the table of integrals $\left(9.5^{\prime}\right)-\left(9.15^{\prime}\right)$, find... | \frac{1}{3}\arcsin3x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,694 |
Example 1. Find the integral $\int x e^{x^{2}} d x$. | Solution. Let $x^{2}=t$, then $2 x d x=d t, x d x=\frac{d t}{2}$. Substituting the obtained values into the integrand, we get
$$
\int x e^{x^{2}} d x=\int e^{x^{2}} x d x=\int e^{t} \frac{d t}{2}=\frac{1}{2} \int e^{t} d t=\frac{1}{2} e^{t}+C=\frac{1}{2} e^{x^{2}}+C
$$
This example can also be solved in another way (... | \frac{1}{2}e^{x^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,695 |
Example 2. Find the integral $\int x \sqrt{x-2} d x$. | Solution. To get rid of the root, let
$$
\sqrt{x-2}=t
$$
Squaring this equality, we find $x$ :
$$
x=t^{2}+2
$$
from which
$$
d x=2 t d t .
$$
Substituting the obtained equalities into the integrand, we find
$$
\begin{aligned}
\int x \sqrt{x-2} d x & =\int\left(t^{2}+2\right) t 2 t d t=\int\left(2 t^{4}+4 t^{2}\r... | \frac{2}{5}(x-2)^{\frac{5}{2}}+\frac{4}{3}(x-2)^{\frac{3}{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,696 |
Example 4. Find the integral $\int \cos ^{2} x \sin x d x$. | Solution. Let $\cos x=u$, then
$$
-\sin x d x=d u, \sin x d x=-d u
$$
Thus,
$$
\begin{gathered}
\int \cos ^{2} x \sin x d x=\int u^{2}(-d u)=-\int u^{2} d u=-\frac{u^{3}}{3}+C= \\
=-\frac{\cos ^{3} x}{3}+C
\end{gathered}
$$
The same result can be obtained directly (see § 4.2):
$$
\begin{gathered}
\int \cos ^{2} x ... | -\frac{\cos^{3}x}{3}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,698 |
Example 6. Find the integral $\int \frac{\ln x}{x} d x$. | Solution. Let
$$
\ln x=t
$$
then
$$
\frac{1}{x} d x=d t
$$
Therefore,
$$
\int \frac{\ln x}{x} d x=\int \ln x \frac{1}{x} d x=\int t d t=\frac{t^{2}}{2}+C=\frac{1}{2} \ln ^{2} x+C
$$ | \frac{1}{2}\ln^{2}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,699 |
Example 7. Find the integral $\int \frac{d x}{\sin x \cos x}$. | Solution. Dividing the numerator and the denominator by $\cos ^{2} x$, we get
$$
\frac{1}{\sin x \cos x}=\frac{\frac{1}{\cos ^{2} x}}{\tan x}
$$
Let
$$
\tan x=t
$$
then
$$
\frac{1}{\cos ^{2} x} d x=d t
$$
Thus,
$$
\int \frac{d x}{\sin x \cos x}=\int \frac{\frac{d x}{\cos ^{2} x}}{\tan x}=\int \frac{d t}{t}=\ln |... | \ln|\tanx|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,700 |
Example 8. Find the integral $\int \frac{d x}{\sin x}$. | Solution. Let $\frac{x}{2}=t$, we get
$$
\begin{gathered}
\int \frac{d x}{\sin x}=\int \frac{d x}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\int \frac{d\left(\frac{x}{2}\right)}{\sin \frac{x}{2} \cos \frac{x}{2}}=\int \frac{d t}{\sin t \cos t}= \\
=\ln |\operatorname{tg} t|+C=\ln \left|\operatorname{tg} \frac{x}{2}\right|+... | \ln|\operatorname{tg}\frac{x}{2}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,701 |
Example 9. Find the integral $\int \sqrt{a^{2}-x^{2}} d x(a>0)$. | Solution. Apply the trigonometric substitution
$$
x=a \cos t
$$
## We have
$$
d x=-a \sin t d t
$$
Therefore,
$$
\begin{gathered}
\int \sqrt{a^{2}-x^{2}} d x=\int \sqrt{a^{2}-a^{2} \cos ^{2} t}(-a \sin t) d t= \\
=-a^{2} \int \sqrt{1-\cos ^{2} t} \sin t d t=-a^{2} \int \sin ^{2} t d t=-a^{2} \int \frac{1-\cos 2 t}... | -\frac{^{2}}{2}\arccos\frac{x}{}+\frac{x}{2}\sqrt{^{2}-x^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,702 |
Example 10. Find the integral $\int \frac{d x}{\sqrt{\left(x^{2}-a^{2}\right)^{3}}}$. | Solution. Apply the trigonometric substitution
$$
x=a \sec t
$$
We have:
$$
\begin{gathered}
d x=\frac{a \sin t d t}{\cos ^{2} t}=\frac{a \operatorname{tg} t d t}{\cos t} \\
\sqrt{\left(x^{2}-a^{2}\right)^{3}}=\sqrt{\left(a^{2} \sec ^{2} t-a^{2}\right)^{3}}=\sqrt{a^{6}\left(\frac{1-\cos ^{2} t}{\cos ^{2} t}\right)^{... | -\frac{1}{^{2}}\frac{x}{\sqrt{x^{2}-^{2}}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,703 |
Example 11. Show that
$$
\int \frac{d x}{\sqrt{x^{2}+a}}=\ln \left|x+\sqrt{x^{2}+a}\right|+C, \text { where } a=\text { const. }
$$ | Solution. Let's apply the so-called Euler substitution
$$
\sqrt{x^{2}+a}=t-x
$$
where $t$ is the new variable. Squaring both sides of this equation, we get
160
$$
x^{2}+a=t^{2}-2 t x+x^{2}
$$
or
$$
a=t^{2}-2 t x
$$
Finding the differentials of both sides of the last equation:
$$
0=2 t d t-(2 x d t+2 t d x)
$$
... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,704 |
Example 1. Find $\int x \sin x d x$. | Solution. Let: $x=u, \sin x d x=d v$.
To apply formula (4.18), we need to know $v$ and $d u$ as well. Differentiating the equation $x=u$, we get $d x=d u$. Integrating the equation $d v=\sin x d x=d(-\cos x)$, we determine $v=-\cos x$.
Substituting the values of $u, v, d u, d v$ into formula (4.18), we find
$$
\begi... | \sinx-x\cosx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,705 |
Example 2. Find $\int x \ln x d x$. | Solution. Let
$$
\ln x=u, x d x=d v
$$
we get
$$
\frac{1}{x} d x=d u, \frac{x^{2}}{2}=v
$$
By formula (4.18) we find
$$
\int x \ln x d x=\frac{x^{2}}{2} \ln x-\int \frac{x^{2}}{2} \cdot \frac{1}{x} d x=\frac{x^{2}}{2} \ln x-\frac{1}{4} x^{2}+C
$$ | \frac{x^{2}}{2}\lnx-\frac{1}{4}x^{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,706 |
Example 3. Find $\int \arcsin x d x$. | Solution. Let
$$
u=\arcsin x, d v=d x
$$
we determine
$$
d u=\frac{1}{\sqrt{1-x^{2}}} d x, v=x
$$
Therefore,
$$
\begin{gathered}
\int \arcsin x d x=x \arcsin x-\int x \frac{1}{\sqrt{1-x^{2}}} d x= \\
=x \arcsin x+\frac{1}{2} \int\left(1-x^{2}\right)^{-\frac{1}{2}} d\left(1-x^{2}\right)=x \arcsin x+ \\
+\frac{1}{2}... | x\arcsinx+\sqrt{1-x^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,707 |
Example 4. Find $\int x^{2} \cos x d x$. | Solution. Setting
$$
u=x^{2}, d v=\cos x d x=d(\sin x)
$$
we get
$$
d u=2 x d x, v=\sin x
$$
Therefore,
$$
\int x^{2} \cos x d x=x^{2} \sin x-\int \sin x \cdot 2 x d x=x^{2} \sin x-2 \int x \sin x d x
$$
The obtained integral can again be found by integration by parts (Example 1). It can also be found without exp... | x^{2}\sinx+2(x\cosx-\sinx)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,708 |
Example 5. Find $\int e^{x} \sin x d x$. | Solution. Let
$$
u=e^{x}, d v=\sin x d x
$$
from this
$$
d u=e^{x} d x, v=-\cos x
$$
Applying formula (4.18), we get
$$
\begin{gathered}
\int e^{x} \sin x d x=e^{x}(-\cos x)-\int(-\cos x) e^{x} d x= \\
=-e^{x} \cos x+\int e^{x} \cos x d x
\end{gathered}
$$
To the integral on the right side, we again apply the int... | \frac{e^{x}}{2}(\sinx-\cosx)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,709 |
Example 6. Show that
$$
\int \sqrt{x^{2}+a} d x=\frac{1}{2}\left[x \sqrt{x^{2}+a}+a \ln \left|x+\sqrt{x^{2}+a}\right|\right]+C
$$ | ## Solution. Let
$$
\sqrt{x^{2}+a}=u, d x=d v
$$
from this
$$
\frac{x}{\sqrt{x^{2}+a}} d x=d u, v=x
$$
By formula (4.18) we get
$$
\int \sqrt{x^{2}+a} d x=x \sqrt{x^{2}+a}-\int x \frac{x}{\sqrt{x^{2}+a}} d x
$$
Transform the integral on the right side:
$$
\begin{gathered}
\int \frac{x^{2}}{\sqrt{x^{2}+a}} d x=\i... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,710 |
Example 2. Find the integral $\int \frac{d x}{x^{2}-6 x-16}$. | Solution. Completing the square trinomial to a perfect square and integrating based on formula (4.20) for the case when $u=x-3$, $a=5$, we get
$$
\begin{aligned}
& \int \frac{d x}{x^{2}-6 x-16}=\int \frac{d x}{\left(x^{2}-6 x+9\right)-9-16}=\int \frac{d x}{(x-3)^{2}-25}= \\
= & \int \frac{d(x-3)}{(x-3)^{2}-5^{2}}=\fra... | \frac{1}{10}\ln|\frac{x-8}{x+2}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,711 |
Example 3. Find the integral $\int \frac{d x}{3 x^{2}+4 x+1}$. | Solution. As in Example 2, we get
$$
\int \frac{d x}{3 x^{2}+4 x+1}=\int \frac{d x}{3\left(x^{2}+\frac{4}{3} x+\frac{1}{3}\right)}=
$$
$$
\begin{gathered}
=\int \frac{d x}{3\left[\left(x^{2}+2 \cdot \frac{2}{3} x+\frac{4}{9}\right)-\frac{4}{9}+\frac{1}{3}\right]}=\int \frac{d x}{3\left[\left(x+\frac{2}{3}\right)^{2}-... | \frac{1}{2}\ln|\frac{3x+1}{3x+3}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,712 |
Example 4. Find the integral $\int \frac{x+2}{x^{2}+2 x+5} d x$. | Solution. This is an integral of the form (4.21). By transforming the quadratic trinomial and changing the variable to $x+1=t$, we find
$$
\begin{gathered}
\int \frac{x+2}{x^{2}+2 x+5} d x=\int \frac{x+2}{\left(x^{2}+2 x+1\right)+4} d x=\int \frac{(x+1)+1}{(x+1)^{2}+4} d x= \\
=\int \frac{x+1}{(x+1)^{2}+2^{2}} d(x+1)+... | \frac{1}{2}\ln(x^{2}+2x+5)+\frac{1}{2}\operatorname{arctg}\frac{x+1}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,713 |
Example 5. Find the integral $\int \frac{d x}{\sqrt{5-4 x-x^{2}}}$. | Solution. Completing the square trinomial to a perfect square and using formula (4.23), we find
$$
\begin{aligned}
\int \frac{d x}{\sqrt{5-4 x-x^{2}}} & =\int \frac{d x}{\sqrt{-\left(x^{2}+4 x+4-4-5\right)}}=\int \frac{d x}{\sqrt{-(x+2)^{2}+9}}= \\
& =\int \frac{d(x+2)}{\sqrt{3^{2}-(x+2)^{2}}}=\arcsin \frac{x+2}{3}+C
... | \arcsin\frac{x+2}{3}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,714 |
Example 6. Find the integral $\int \frac{d x}{\sqrt{6-4 x-2 x^{2}}}$. | Solution. As in example 5, we get
$$
\begin{gathered}
\int \frac{d x}{\sqrt{6-4 x-2 x^{2}}}=\int \frac{d x}{\sqrt{-2\left(x^{2}+2 x-3\right)}}= \\
=\int \frac{d x}{\sqrt{-2\left[\left(x^{2}+2 x+1\right)-1-3\right]}}=\frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{-\left[(x+1)^{2}-4\right]}}= \\
=\frac{1}{\sqrt{2}} \int \frac... | \frac{1}{\sqrt{2}}\arcsin\frac{x+1}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,715 |
Example 7. Find the integral $\int \frac{d x}{\sqrt{x^{2}-6 x+3}}$. | Solution. By transforming the quadratic trinomial and applying formula (4.24), we find
$$
\begin{aligned}
\int \frac{d x}{\sqrt{x^{2}-6 x+3}} & =\int \frac{d x}{\sqrt{\left(x^{2}-6 x+9\right)-9+3}}=\int \frac{d(x-3)}{\sqrt{(x-3)^{2}-6}}= \\
= & \ln \left|(x-3)+\sqrt{(x-3)^{2}-6}\right|+C= \\
& =\ln \left|x-3+\sqrt{x^{... | \ln|x-3+\sqrt{x^{2}-6x+3}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,716 |
Example 8. Find the integral $\int \frac{d x}{\sqrt{3 x^{2}-6 x+9}}$. | Solution. As in example 7, we get
$$
\begin{gathered}
\int \frac{d x}{\sqrt{3 x^{2}-6 x+9}}=\int \frac{d x}{\sqrt{3\left(x^{2}-2 x+3\right)}}=\frac{1}{\sqrt{3}} \int \frac{d x}{\sqrt{\left(x^{2}-2 x+1\right)+2}}= \\
=\frac{1}{\sqrt{3}} \int \frac{d(x-1)}{\sqrt{(x-1)^{2}+2}}=\frac{1}{\sqrt{3}} \ln \left|(x-1)+\sqrt{(x-... | \frac{1}{\sqrt{3}}\ln|x-1+\sqrt{x^{2}-2x+3}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,717 |
Example 9. Find the integral $\int \sqrt{x^{2}+8 x+25} d x$. | Solution. This is an integral of the form (4.25). By transforming the quadratic trinomial and applying formula (4.26), we find
$$
\begin{aligned}
& \int \sqrt{x^{2}+8 x+25} d x=\int \sqrt{\left(x^{2}+8 x+16\right)-16+25} d x= \\
& =\int \sqrt{(x+4)^{2}+9} d(x+4)=\frac{x+4}{2} \sqrt{(x+4)^{2}+9}+ \\
& \quad+\frac{9}{2}... | \frac{x+4}{2}\sqrt{x^{2}+8x+25}+\frac{9}{2}\ln|x+4+\sqrt{x^{2}+8x+25}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,718 |
Example 10. Find the integral $\int \sqrt{8+2 x-x^{2}} d x$. | Solution. This is also an integral of the form (4.25). By transforming the quadratic trinomial and applying formula (4.27), we get
$$
\begin{gathered}
\int \sqrt{8+2 x-x^{2}} d x=\int \sqrt{-\left(x^{2}-2 x-8\right)}= \\
=\int \sqrt{-\left(x^{2}-2 x+1-1-8\right)} d x=\int \sqrt{-\left(x^{2}-2 x+1\right)+9} d x= \\
=\i... | \frac{x-1}{2}\sqrt{8+2x-x^{2}}+\frac{9}{2}\arcsin\frac{x-1}{3}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,719 |
Example 1. Find the integral $\int \frac{9-5 x}{x^{3}-6 x^{2}+11 x-6} d x$. | Solution. First, we decompose the proper rational fraction
$$
\frac{9-5 x}{x^{3}-6 x^{2}+11 x-6}
$$
into partial fractions, for which we need to find the roots of its denominator, i.e., the roots of the equation
$$
x^{3}-6 x^{2}+11 x-6=0 .
$$
One root is immediately apparent, which is $x_{1}=1$. Since
$$
\left(x^{... | \ln|\frac{(x-1)^{2}(x-2)}{(x-3)^{3}}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,720 |
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