Integration using partial metod

Question

Need help on solving integrals using partial integration. As I have only solved ones with Newton-Leibniz, I don't know how to solve this types: $$\int_0^{\pi/4} \frac{x\sin(x)}{\cos^2(x)}dx$$

Answer 1

Hint:

$$\int\frac{x\sin x}{\cos^2x}\,dx=\int x\cdot\frac{\sin{x}}{\cos{x}}\cdot\frac{1}{\cos{x}}\,dx=\int x\tan{x}\sec{x}\,dx=x\sec{x}-\int{\sec{x}\,dx}$$

See Ways to evaluate $\int \sec \theta \, \mathrm d \theta$ on how to integrate $\sec{x}$.

Answer 2

First, use integration by parts:

$$\int uv'\,dx=uv-\int u'v\,dx\;,\;\;\text{with}\;\;u=x\;,\;\;v'=\frac{\sin x}{\cos^2x}$$

Observe also that

$$\int\frac{f'}{f^2}=-\frac1f$$

and also that

$$\int\frac{dx}{\cos x}=\int\frac{\cos x}{\cos^2x}dx=\int\frac{\cos x}{1-\sin^2x}dx=\frac12\int\left(\frac{\cos x}{1-\sin x}+\frac{\cos x}{1+\sin x}\right)dx$$