How to solve the integral $\int \frac{y}{\cos^2(y)}~dy$.

Question

I would like to compute the integral of

$$\int \frac{y}{\cos^2(y)}~dy.$$

Here is what I tried:

$$u=\cos(y),\quad du=-\sin(y)~dy,$$

$$dv=y~dy,\quad v=y^2/2,$$

$$uv-\int v~du=\frac{y^2}{2}\cos(y)+\int \frac{y^2}{2}\sin(y)~dy.$$

The expected answer is $$y(\tan y)+\ln(\cos(y)).$$

How can I get to this? Thanks guys.

Answer 1

Since you are taking "u= cos(y) dv= ydy"m=,you appear to be using integration by parts to integrate $\int cos(y) ydy$ rather than $\int \frac{y}{cos^2(y)} dy$.

Answer 2

Use integration by parts using $$u=y \qquad du=dy$$ $$dv=\frac {dy}{\cos^2(y)}\implies v=\tan(y)$$ as a result $$y\tan(y)-\int \tan(y) \,dy=y\tan(y)-\int \frac {\sin(y)}{\cos(y)} \,dy$$ which looks simple.