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

$\frac{\pi}{4}$

Given:

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

Hint:

using $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$

Explanation:

Let

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

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \ldots(i) \\\\ &I=\int_{0}^{\frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}-2\right)}{\cos \left(\frac{\pi}{2}-2\right)+\sin \left(\frac{\pi}{2}-2\right)} d x \end{aligned}

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

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