Question

$\int \frac{\cos x}{1-\cos x}dx$

Answer 1

$\int \frac{\cos x}{1-\cos x}dx$

As we know that,

$\cos 2x=1-2{\sin }^{2}x\Rightarrow 1-\cos 2x=2{\sin }^{2}x$

Therefore,

$\int \frac{\cos x}{1-\cos x}dx$

$=\int \frac{1-2{\sin }^2\frac{x}{2}}{2{\sin }^2\frac{x}{2}}dx$

$=\int \frac{1}{2{\sin }^2\frac{x}{2}}dx-\int \frac{2{\sin }^2\frac{x}{2}}{2{\sin }^2\frac{x}{2}}dx$

$=\frac{1}{2}\int {\csc }^2\left(\frac{x}{2}\right)dx-\int dx$

$=\frac{1}{2}(-2\cot \left(\frac{x}{2}\right))-x+C$

$=-\cot \left(\frac{x}{2}\right)-x+C$

Hence $\int \frac{\cos x}{1-\cos x}dx=-\cot \left(\frac{x}{2}\right)-x+C$.