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. $\int \frac{x^{4}-3 x^{3}+4 x^{2}+6 x-8}{x^{2}} d x$.
2. $\int\left(\sqrt[3]{x^{2}}+\frac{1}{\sqrt[3]{x}}\right) d x$.
3. $\int \cos ^{2} \frac{x}{2} d x$.
4. $\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2} d x$.
5. $\int \frac{3-4 \cos ^{3} x}{\cos ^{2} x} d x$.
6. $\int(a x-b)^{3} d x$.
7. $\int \frac{\left(1-x^{n}\right)^{2}}{\sqrt{x}} d x$.
8. $\int e^{x}\left(1+\frac{e^{-x}}{x^{3}}\right) d x$.
9. $\int\left(\frac{4}{1+x^{2}}-\frac{5}{\sqrt{1-x^{2}}}\right) d x$.
10. $\int \frac{x^{2}}{1+x^{2}} d x$.

Answers

$$
\begin{aligned}
& \text { 1. } \frac{x^{3}}{3}-\frac{3}{2} x^{2}+4 x+6 \ln x+\frac{8}{x}+\text { C. 2. } \frac{3}{5} x \sqrt[3]{x^{2}}+\frac{3}{2} \sqrt[3]{x^{2}}+\text { C. 3. } \frac{1}{2} x+ \\
& +\frac{1}{2} \sin x+C . \quad 4 . x-\cos x+C . \quad 5.3 \operatorname{tg} x-4 \sin x+C . \quad 6 . \frac{a^{3}}{4} x^{4}-a^{2} b x^{3}+ \\
& +\frac{3}{2} a b^{2} x^{2}-b^{3} x+C . \quad \text { 7. } \sqrt{x}\left(2-\frac{4 x^{n}}{2 n+1}+\frac{2 x^{2 n}}{4 n+1}\right)+C . \quad \text { 8. } e^{x}-\frac{1}{2 x^{2}}+C
\end{aligned}
$$

## § 4.2. Independence of the form of the indefinite integral from the choice of the function argument

If

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

Then

$$
F(u)=\int f(u) d u
$$

where $u=\varphi(x)$ is any differentiable function of $x$. Formula (4.17) is obtained from formula (4.16) by formally replacing $x$ with $u$, which significantly expands the table of the simplest integrals. Based on this, we obtain:

$$
\begin{aligned}
& \int u^{\alpha} d u=\frac{u^{\alpha+1}}{\alpha+1}+C(\alpha \neq-1) \\
& \int \frac{d u}{u}=\ln |u|+C \\
& \int e^{u} d u=e^{u}+C \\
& \int a^{u} d u=\frac{a^{u}}{\ln a}+C \\
& \int \cos u d u=\sin u+C \\
& \int \sin u d u=-\cos u+C
\end{aligned}
$$

$$
\begin{aligned}
& \int \frac{d u}{\cos ^{2} u}=\operatorname{tg} u+C \\
& \int \frac{d u}{\sin ^{2} u}=-\operatorname{ctg} u+C \\
& \int \frac{d u}{\sqrt{1-u^{2}}}=\arcsin u+C=-\arccos u+C_{1} \\
& \int \frac{d u}{1+u^{2}}=\operatorname{arctg} u+C=-\operatorname{arcctg} u+C_{1} \\
& \int \frac{d u}{u^{2}-1}=\frac{1}{2} \ln \left|\frac{u-1}{u+1}\right|+C
\end{aligned}
$$

When using formulas ( $9.5^{\prime}$ ) $-\left(9.15^{\prime}\right)$, it is necessary to keep in mind the simplest transformations of the differential:

1. $d x=d(x+b)$, where $b-$ is a constant.
2. $d x=\frac{1}{a} d(a x)$, where the constant $a \neq 0$.
3. $d x=\frac{1}{a} d(a x+b)$, where the constant $a \neq 0$.
4. $x d x=\frac{1}{2} d\left(x^{2}\right)$.
5. $x d x=\frac{1}{2} d\left(x^{2}+b\right)$.
6. $\sin x d x=-d(\cos x)$.
7. $\cos x d x=d \sin x$.

In general

$$
\varphi^{\prime}(x) d x=d \varphi(x)
$$