Question

Integrate $\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

Answer 1

$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

$\Rightarrow \frac{dy}{dx}=\frac{2{\sin }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}}$

$\Rightarrow \frac{dy}{dx}={\tan }^2\frac{x}{2}$

$dy=\left({\tan }^2\frac{x}{2}\right)dx$

Integrating both sides

$\int dy=\int {\tan }^2\frac{x}{2}dx$

$y=\int ({\sec }^2\frac{x}{2}-1)dx$

$y=2\tan \frac{x}{2}-x+c$.