I am trying to find the integral of $\int \tan x \sec^3 x dx$
$\int \tan x(1+\tan^2 x)\sec x\, dx$
This gets me nowhere since I get a $\sec^2 x$ derivative with tan substitution so I try something else.
$\int \frac {\sin x}{\cos x} \frac{1}{\cos^3x} dx$
$\int \frac {\sin x}{\cos^4 x} dx$
$u = \cos x$ $du = -\sin x$
$\int \frac {-1}{u^4} du$
$-1\int {u^{-4}} du$
$-1 \frac{u^{-3}}{3}$ $\frac{-1}{3\cos^3 x}$
This for some reason is wrong.