#### Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 Question 3 maths textbook solution.

Answer : $\frac{\pi}{12}$

Given : $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}} d x$

Hint : Use the formula of $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

Solution : $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}} d x \quad-----(1)$

\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}{\tan \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)+\sqrt{\cot \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}} d x \\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan \left(\frac{\pi}{2}-x\right)}}{\tan \left(\frac{\pi}{2}-x\right)}+\sqrt{\cot \left(\frac{\pi}{2}-x\right)} d x \\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} d x \quad------(2) \end{aligned}

\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}} d x+\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{\tan x}+\sqrt{\cot x}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 d x \end{aligned}

\begin{aligned} &2 I=(x)_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\ &2 I=\frac{\pi}{6} \\ &I=\frac{\pi}{12} \end{aligned}