Trigonometric anti derivative [closed]
Solution 1:
Using integration by parts gives $$ \int \dfrac{x}{(1 + \cos x)^2}dx = xv - \int v dx + C $$ where $v$ is an anti-derivative of $\displaystyle\int \dfrac{1}{(1 + \cos x)^2}dx$ and $C$ is a constant
Notice that, $1 + \cos x = 2\cos^2\left(\dfrac{x}{2}\right)$. Hence $$ \int \dfrac{1}{(1 + \cos x)^2}dx = \int \dfrac{dx}{4\cos^4 \frac{x}{2}} = \dfrac{1}{4}\int \left[1 + \tan^2 \left(\dfrac{x}{2}\right)\right]^2 dx $$ From here, let $t = \tan\left(\dfrac{x}{2}\right)$ and you'll be able to find $v$