Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 31

$\frac{\pi}{4}$

Hint:

To solve this equation we should simplify cot x.

Given:

$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot x} d x$

Solution:

Let

$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot x} d x$

$I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\frac{\cos x}{\sin x}} d x, \quad\left[\therefore \cot x=\frac{\cos x}{\sin x}\right]$

$=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\cos x+\sin x} d x \ldots(i), \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$

$=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\cos \left(\frac{\pi}{2}-x\right)+\sin \left(\frac{\pi}{2}-x\right)} d x$

$=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \ldots(i i)$

\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x+\sin x}{\cos x+\sin x} d x \\\\ &2 I=\int_{0}^{\frac{\pi}{2}} 1 d x \end{aligned}

\begin{aligned} &I=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}} \\\\ &I=\frac{1}{2}\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{4} \end{aligned}