Question

Evaluate $\int \left(\frac{1-\cos x}{1+\cos x}\right)dx$.

Answer 1

$\int \left(\frac{1-\cos x}{1+\cos x}\right)dx$

Sol.:

$\cos x=\left(2{\cos }^2\right(\frac{x}{2}\left)\right)-1=1-\left(2{\sin }^2\right(\frac{x}{2}\left)\right)$

$\Rightarrow 1-\cos x=2{\sin }^2\left(\frac{x}{2}\right)\&1+\cos x=2{\cos }^2\left(\frac{x}{2}\right)$

$\therefore I=\int \left(\frac{2{\sin }^2\left(\frac{x}{2}\right)}{2{\cos }^2\left(\frac{x}{2}\right)}\right)dx$

$I=\int \left({\tan }^2\right(\frac{x}{2}\left)\right)dxI=\int \left({\sec }^2\right(\frac{x}{2})-1)dxI=\int \left({\sec }^2\right(\frac{x}{2}\left)\right)dx-\int dxI=2\tan \left(\frac{x}{2}\right)-x+C$

Hence, correct answer is $2\tan \left(\frac{x}{2}\right)-x+C.$