Question

${\int }_0^{\frac{\pi }{2}}\left(\frac{1}{1+\cot x}\right)dx$

Answer 1

Consider the given integral.

$I={\int }_0^{\frac{\pi }{2}}\left(\frac{1}{1+\cot x}\right)dx$

$I={\int }_0^{\frac{\pi }{2}}\left(\frac{1}{1+\frac{\cos x}{\sin x}}\right)dx$

$I={\int }_0^{\frac{\pi }{2}}\left(\frac{\sin x}{\sin x+\cos x}\right)dx$ …….. (1)

We know that

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Therefore,

$I={\int }_0^{\frac{\pi }{2}}\left(\frac{\sin (\frac{\pi }{2}-x)}{\sin (\frac{\pi }{2}-x)+\cos (\frac{\pi }{2}-x)}\right)dx$

$I={\int }_0^{\frac{\pi }{2}}\left(\frac{\cos x}{\cos x+\sin x}\right)dx$ …........(2)

On adding equation (1) and (2), we get

$2I={\int }_0^{\frac{\pi }{2}}\left(\frac{\cos x+\sin x}{\cos x+\sin x}\right)dx$

$2I={\int }_0^{\frac{\pi }{2}}1dx$

$2I={\left(x\right)}_0^{\frac{\pi }{2}}$

$2I=\frac{\pi }{2}-0$

$I=\frac{\pi }{4}$

Hence, the value of integral is $\frac{\pi }{4}$.