Question

Solve :$\int \frac{1-\cot x}{1+\cot x}dx$

Answer 1

Given,$\int \frac{1-\cot x}{1+\cot x}dx=\int \frac{(1-\frac{\cos x}{\sin x})}{(1+\frac{\cos x}{\sin x})}dx=\int \frac{\frac{\sin x-\cos x}{\sin x}}{\frac{\sin x+\cos x}{\sin x}}dx=\int \frac{\sin x-\cos x}{\sin x+\cos x}dxLetusassume,\sin x+\cos x=u.differentiatingbothsidesweget,(\cos x-\sin x)dx=du\Rightarrow -(-\cos x+\sin x)dx=du\Rightarrow (-\cos x+\sin x)dx=-du\Rightarrow (\sin x-\cos x)dx=-dunowsubstituingthesevaluesweget,\int \frac{\sin x-\cos x}{\sin x+\cos x}dx=\int \left(\frac{-du}{u}\right)=\int (-\frac{1}{u})du=-\log u=-\log (\sin x+\cos x).$