Question

Integrate with respect to x: $\frac{1}{1+\cot x}$

Answer 1

Consider the given integral.

$I=\int \frac{1}{1-\cot x}dx$

$I=\int \frac{1}{1-\frac{\cos x}{\sin x}}dx$

$I=\int \frac{\sin x}{\sin x-\cos x}dx$

Let,

$\frac{\sin x}{\sin x-\cos x}=\frac{A(\sin x+\cos x)+B(\sin x-\cos x)}{\sin x-\cos x}$

$\frac{\sin x}{\sin x-\cos x}=\frac{(A+B)\sin x+(A-B)\cos x}{\sin x-\cos x}$

Comparing both the sides, we have

$A=\frac{1}{2},B=\frac{1}{2}$

Therefore,

$I=\frac{1}{2}\int \frac{\sin x+\cos x}{\sin x-\cos x}dx+\frac{1}{2}\int \frac{\sin x-\cos x}{\sin x-\cos x}dx$

$I=\frac{1}{2}\int \frac{\sin x+\cos x}{\sin x-\cos x}dx+\frac{1}{2}\int 1dx$

$I=\frac{1}{2}\ln (\sin x-\cos x)+\frac{x}{2}+C$

Hence, this is the required value of the integral.