indefinite integration of trigonometric function involving $\cos$

Question

integration of $\displaystyle \frac{1}{1+\cos^3 x}$ with respect to $x$

after using $a^3+b^3 = (a+b)(a^2-ab+b^2)$

$\displaystyle \int\frac{1}{(1+\cos x)(1+\cos^2 x-\cos x)}dx = $

I can't go further, could some help me with this?

Answer 1

For the integration of $\displaystyle \frac{1}{1+\cos^3 x}$ with respect to $x$ substitute $u = \tan (x/2) $ and then $$ du=\frac{1}{2} dx \sec (x/2)^2. $$ Then transform the integrated using the substitution $\sin x = \frac {2u}{u^2+1} $, $ \cos x= \frac {1-u^2}{1+u^2} $ and $ dx=\frac {2 du}{1+u^2} $. Then the integrand can be easily solved. Hope it helps.