#### Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 15 textbook solution.

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

Given : $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}} d x$

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

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

$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x----(1)\left(\tan x=\frac{\sin x}{\cos x}\right)$

We know that $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \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 \\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x-------(2) \end{aligned}

\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 . d x \\ &2 I=(x)_{\frac{\pi}{6}}^{\frac{\pi}{3}} \end{aligned}

\begin{aligned} &2 I=\frac{2 \pi-\pi}{6} \\ &I=\frac{\pi}{12} \end{aligned}

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

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

Given : $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}} d x$

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

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

$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x----(1)\left(\tan x=\frac{\sin x}{\cos x}\right)$

We know that $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \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 \\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x-------(2) \end{aligned}

\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 . d x \\ &2 I=(x)_{\frac{\pi}{6}}^{\frac{\pi}{3}} \end{aligned}

\begin{aligned} &2 I=\frac{2 \pi-\pi}{6} \\ &I=\frac{\pi}{12} \end{aligned}