Question

Integrate:

$\int \frac{1}{1+\tan x}dx$

Answer 1

Consider the given integral.

$I=\int \frac{1}{1+\tan x}dx$

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

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

Now,

$u=\sin x+\cos x$

$\frac{d}{dx}(\sin x+\cos x)=\cos x-\sin x$

Again, let

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

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

Therefore,

$A+B=1$

$A-B=0$

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

Therefore,

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

$I=\frac{1}{2}\int \frac{1}{u}du+\frac{1}{2}\int dx$

$I=\frac{1}{2}\ln u+\frac{x}{2}+C$

$I=\frac{1}{2}\ln |\sin x+\cos x|+\frac{x}{2}+C$

Hence, this is the required value of the integral.