Question

$\int \frac{\cos x}{1-\cos x}dx\mathrm{or}\int \frac{\cot x}{\mathrm{cosec}x-\cot x}dx$

Answer 1

$$\begin{gathered} \int \frac{\cot x}{\mathrm{cosec}x-\cot x}dx \\ =\int \frac{{\displaystyle \frac{\cos x}{\sin x}}}{{\displaystyle \frac{1}{\sin x}}-{\displaystyle \frac{\cos x}{\sin x}}}dx \\ =\int \left(\frac{\cos x}{1-\cos x}\right)\times \frac{\left(1+\cos x\right)}{\left(1+\cos x\right)}dx \\ =\int \left(\frac{\cos x+{\cos }^{2}x}{1-{\cos }^{2}x}\right)dx \\ =\int \left(\frac{\cos x+{\cos }^{2}x}{{\sin }^{2}x}\right)dx \\ =\int \left(\frac{\cos x}{\sin x}\times \frac{1}{\sin x}+\frac{{\cos }^{2}x}{{\sin }^{2}x}\right)dx \\ =\int \left[\left(\cot x\mathrm{cosec}x\right)+{\cot }^{2}x\right]dx \\ =\int \left[\mathrm{cosec}x\cot x+{\mathrm{cosec}}^{2}x-1\right]dx \\ =-\mathrm{cosec}x-\cot x-x+\mathrm{C} \end{gathered}$$