Example 5. Find the integral $\int \operatorname{ch}^{3} x \operatorname{sh} x d x$.
Solution.

$$
\int \operatorname{ch}^{3} x \operatorname{sh} x d x=\int \operatorname{ch}^{3} x d(\operatorname{ch} x)=\frac{\operatorname{ch}^{4} x}{4}+C
$$

## Problems

Find the integrals of hyperbolic functions:

1. $\int \operatorname{ch}^{2} x d x$.
2. $\int \operatorname{ch}^{3} x d x$.
3. $\int \operatorname{sh}^{3} 4 x \operatorname{ch} 4 x d x$.
4. $\int \operatorname{th}^{2} x d x$.
5. $\int \operatorname{ch}^{4} x \operatorname{sh} x d x$.
6. $\int \operatorname{sh}^{5} x \operatorname{ch} x d x$.
7. $\int \frac{1+2 \operatorname{ch} x}{\operatorname{sh}^{2} x} d x$.
8. $\int \frac{1+\operatorname{sh} x}{\operatorname{ch}^{2} x} d x$.

## Answers

1. $\frac{1}{4} \operatorname{sh} 2 x+\frac{1}{2} x+C$. 2. $\operatorname{sh} x+\frac{\operatorname{sh}^{3} x}{3}+C$. 3. $\frac{\operatorname{sh}^{4} 4 x}{16}+C$. 4. $\operatorname{th} x+x+C$.
5. $\frac{\operatorname{ch}^{5} x}{5}+C$. 6. $\frac{\operatorname{sh}^{6} x}{6}+C$. 7. $-\frac{\operatorname{ch} x+2}{\operatorname{sh} x}+C$. 8. $\frac{\operatorname{sh} x-1}{\operatorname{ch} x}+C$.

## Chapter 5. The Definite Integral and Its Applications

Let the function $y=f(x)$ be defined on the interval $[a, b]$. Divide the interval $[a, b]$ into $n$ parts by points $a=x_{0}<x_{1}<\cdots<x_{n}=b$. In each of the elementary intervals of length $\Delta x_{i}=x_{i}-x_{i-1} .(i=1,2, \ldots, n)$, arbitrarily choose a point $z_{i}$ and form the sum

$$
\sum_{i=1}^{n} f\left(z_{i}\right) \Delta x_{i}=f\left(z_{1}\right) \Delta x_{1}+f\left(z_{2}\right) \Delta x_{2}+\cdots+f\left(z_{n}\right) \Delta x_{n}
$$

This sum is called the integral sum of the function $y=f(x)$ over the interval $[a, b]$.

The definite integral of the function $y=f(x)$ over the interval $[a, b]$ is the limit of its integral sum as the number of elementary intervals increases without bound and the length of the largest of them tends to zero:

$$
\int_{a}^{b} f(x) d x=\lim _{\substack{n \rightarrow \infty \\ \max \Delta x_{i} \rightarrow 0}} \sum_{i=1}^{n} f\left(z_{i}\right) \Delta x_{i}
$$

If the function $y=f(x)$ is continuous, then the limit (5.2) exists and is finite.

Properties of the definite integral:

1. $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(t) d t$.
2. $\int_{a}^{a} f(x) d x=0$.
3. $\int_{a}^{b} f(x) d x=-\int_{b}^{a} f(x) d x$.
4. $\int_{a}^{b} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x$.
5. $\int_{a}^{b}\left[f_{1}(x)-f_{2}(x)+f_{3}(x)\right] d x=\int_{a}^{b} f_{1}(x) d x-\int_{a}^{b} f_{2}(x) d x+\int_{a}^{b} f_{3}(x) d x$.
6. $\int_{a}^{b} c f(x) d x=c \int_{a}^{b} f(x) d x$.
7. $\left[\int_{a}^{x} f(x) d x\right]^{\prime}=f(x)$.

## § 5.1. Calculation of the Definite Integral

1. Theorem of Newton-Leibniz. The definite integral of a continuous function over a given interval is equal to the difference of the values of any antiderivative of the function at the upper and lower limits of integration:

$$
\int_{a}^{b} f(x) d x=\left.F(x)\right|_{a} ^{b}=F(b)-F(a)
$$

where

$$
F^{\prime}(x)=f(x)
$$

2. Substitution of variables in the definite integral is performed according to the formula

$$
\int_{a}^{b} f(x) d x=\int_{\alpha}^{\beta} f[\varphi(t)] \varphi^{\prime}(t) d t
$$

where $x=\varphi(t), a=\varphi(\alpha), b=\varphi(\beta) ; t$ - the new variable; $\alpha, \beta$ - the new limits of integration.

3. Integration by parts in the definite integral:

$$
\int_{a}^{b} u(x) d v(x)=\left.u(x) v(x)\right|_{a} ^{b}-\int_{a}^{b} v(x) d u(x)
$$

or

$$
\int_{a}^{b} u(x) v^{\prime}(x) d x=u(b) v(b)-u(a) v(a)-\int_{a}^{b} v(x) u^{\prime}(x) d x
$$

4. Theorem of the mean value. If $f(x) \leq \Phi(x)$, for $a \leq x \leq b$, then

$$
\int_{a}^{b} f(x) d x \leq \int_{a}^{b} \Phi(x) d x
$$

If $f(x)$ and $\varphi(x)$ are continuous for $a \leq x \leq b$ and $\varphi(x) \geq 0$, then

$$
m \int_{a}^{b} \varphi(x) d x \leq \int_{a}^{b} f(x) \varphi(x) d x \leq M \int_{a}^{b} \varphi(x) d x
$$

where $m$ is the smallest and $M$ is the largest value of the function $f(x)$ on the interval $[a, b]$.

In particular, if $\varphi(x) \equiv 1$, then

$$
m(b-a) \leq \int_{a}^{b} f(x) d x \leq M(b-a)
$$

Inequalities (5.8) and (5.9) can be replaced by the equivalent equalities:

$$
\int_{a}^{b} f(x) \varphi(x) d x=f(c) \int_{a}^{b} \varphi(x) d x
$$

and

$$
\int_{a}^{b} f(x) d x=f(d)(b-a)
$$

where $c$ and $d$ are some numbers lying between $a$ and $b$.

The number

$$
\mu=\frac{1}{b-a} \int_{a}^{b} f(x) d x
$$

is called the mean value of the function $y=f(x)$ on the interval $a \leq x \leq b$.