Help me with indefinite integral

Question

Can You help me with it

$$\int\ \frac{tg^2 x}{ \cos^2x}\ dx$$

I've already tried to decompose tg. But I don't know what to do after that.

Answer 1

Let $f(x)=\tan x$ so $f'(x)=\dfrac{1}{\cos^2 x}$ so $$\int\frac{\tan^2x}{\cos^2x} dx =\int f^2(x)f'(x)dx=\frac 1 3 f^3(x)+C$$

Answer 2

If your tg^2x is $\tan^2 x$, then the following is what you want.

$$\int\frac{\tan^2x}{\cos^2x} dx =\frac 13\tan^3x+C.$$

Notice that $$(\tan x)^\prime =\frac{1}{\cos^2x}.$$