#### Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 6 maths textbook solution.

Answer:-  $\frac{\pi}{4}$

Hints:-  You must know the integration rules of trigonometric functions and its limits

Given:-  $\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x$

Solution :

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

$I=\int_{0}^{\pi / 2} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}+\sqrt{\sin \left(\frac{\pi}{2}-x\right)}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$

$I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

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

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