Example. Find the indefinite integral

$$
\int \operatorname{ctg} x \ln \sin x d x
$$
## Solution.

1. Represent the integrand as a product of two functions $F(x) g(x)$, where $g(x)$ has an obvious antiderivative $G(x)$, and $F(x)$ is a function of this antiderivative, i.e., $F(x)=u(G(x))$.

In this case,

$$
F(x)=\frac{\ln \sin x}{\sin x}, \quad g(x)=\cos x, \quad G(x)=\sin x, \quad F(x)=\frac{\ln G}{G}=u(G)
$$

2. Then

$$
\int \frac{\ln \sin x}{\sin x} \cos x d x=\int \frac{\ln \sin x}{\sin x} d \sin x=\int \frac{\ln G}{G} d G
$$

where $G=\sin x$.

3. The last integral is not a standard one, but the method of substitution under the differential sign can be applied again:

$$
\int \frac{\ln G}{G} d G=\int \ln G \frac{1}{G} d G=\int \ln G d \ln G=\frac{\ln ^{2} G}{2}+C=\frac{\ln ^{2} \sin x}{2}+C
$$

Answer. $\int \operatorname{ctg} x \ln \sin x d x=\frac{\ln ^{2} \sin x}{2}+C$.

Conditions of the Problems. Find the indefinite integrals.

1. $\int \frac{3 \operatorname{arctg}^{2} x}{x^{2}+1} d x$.
2. $\int \frac{\sin 2 x-\cos x}{\left(\cos ^{2} x+\sin x\right)^{2}} d x$.
3. $\int \frac{x^{2}+1}{\left(x^{3}+3 x+1\right)^{4}} d x$.
4. $\int \frac{3 \operatorname{tg}^{2} x}{\cos ^{2} x} d x$.
5. $\int \frac{1}{x \sqrt{\ln x}} d x$.
6. $\int \frac{x}{\sqrt{1-x^{4}}} d x$.
7. $\int \frac{x \cos x+\sin x}{(x \sin x)^{3}} d x$.
8. $\int \frac{2 \operatorname{arctg}(x+2)}{x^{2}+4 x+5} d x$.
9. $\int \frac{3 \sqrt{x}+1}{2 x \sqrt{x}+x} d x$.
10. $\int \frac{2 \arcsin x+x}{\sqrt{1-x^{2}}} d x$.

Answers.

1. $\operatorname{arctg}^{3} x+C$.
2. $\frac{1}{\cos ^{2} x+\sin x}+C$.
3. $-\frac{1}{9\left(x^{3}+3 x+1\right)^{3}}+C$.
4. $\operatorname{tg}^{3} x+C$.
5. $2 \sqrt{\ln x}+C$.
6. $\frac{1}{2} \arcsin x^{2}+C$.
7. $-\frac{1}{2(x \sin x)^{2}}+C$.
8. $\operatorname{arctg}^{2}(x+2)+C$.
9. $\ln (2 x \sqrt{x}+x)+C$.
10. $\arcsin ^{2} x-\sqrt{1-x^{2}}+C$.

## 7.2. Integration by Parts

Problem Statement. Find the indefinite integral

$$
\int F(x) g(x) d x
$$

Plan of Solution. Suppose $g(x)$ has an obvious antiderivative $G(x)$, and $F(x)$ is a differentiable function, with its derivative $f(x)=F^{\prime}(x)$ being a simpler function than $F(x)$. Then apply the integration by parts formula

$$
\int F(x) g(x) d x=F(x) G(x)-\int f(x) G(x) d x
$$

If the method is chosen correctly, the integral on the right side of this equation turns out to be a standard one or can be reduced to a standard one by repeated integration by parts.