Example 7. Find $\int \sin ^{5} x d x$.
Solution. The factor needed to be placed under the differential sign is obtained by splitting the power:

$$
\begin{aligned}
& \int \sin ^{5} x d x=\int \sin ^{4} x \cdot \sin x d x= \\
& \begin{array}{l}
=\left\{\begin{array}{l}
\text { obtaining the factor } \sin x=-(\cos x)^{\prime}, \\
\text { it is necessary to express } \sin ^{4} x \text { in terms of } \cos x
\end{array}\right\}= \\
=-\int\left(1-\cos ^{2} x\right)^{2} d(\cos x)=\{u=\cos x\}=-\int\left(1-2 u^{2}+u^{4}\right) d u= \\
=-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x+C .
\end{array}
\end{aligned}
$$

Here, the notation became more concise due to the transition to the new variable $u=\cos x$.

## 2.2. Method of Substitution

The method of substitution, or change of variable in an indefinite integral, is essentially a generalization of the method of bringing under the differential sign. The general method of substitution in the integral $\int f(x) d x$ involves replacing part of the integrand function $u=u(x)$ so that the new integral with respect to the variable $u$ has the form $\int g(u) d u$ and becomes simpler than the original. Sometimes the argument $x$ is replaced by some function: $x=\varphi(t)$, if the integral with respect to the new variable becomes simpler than the original.

## Examples with Solutions