#### Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 25 maths.

Answer:$\frac{\pi}{2}$

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: $\int_{0}^{\frac{\pi}{4}}\left ( \sqrt{\tan x} + \sqrt{\cot x} \right )dx$

Solution: $I=\int_{0}^{\frac{\pi}{4}}\left ( \sqrt{\tan x} + \sqrt{\cot x} \right )dx$

\begin{aligned} &=\int_{0}^{\frac{\pi}{4}}\left(\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\operatorname{cox} x}{\sin x}}\right) d x \\ &=\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{\sin x} \sqrt{\cos x}} d x \\ &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{2 \sin x \cdot \cos x}} d x \end{aligned}

\begin{aligned} &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{1-1+2 \sin x \cdot \cos x}} d x \\ &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{1-\left(\sin ^{2} x+\cos ^{2} x\right)+2 \sin x \cdot \cos x}} d x \end{aligned}

\begin{aligned} &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{1-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cdot \cos x\right)}} d x \\ &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^{2}}} d x \end{aligned}

Put $\sin x-\cos x=t$

$\Rightarrow \left [ \cos x -(-\sin x) \right ]dx=dt$

$\Rightarrow \left ( \cos x + \sin x \right )dx=dt$

When x=0  then t=-1  and when $x=\frac{\pi}{4}$  then t=0

\begin{aligned} &I=\sqrt{2} \int_{-1}^{0} \frac{1}{\sqrt{1-t^{2}}} d t \\ &=\sqrt{2}\left[\sin ^{-1} t\right]_{-1}^{0} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}\right] \end{aligned}

\begin{aligned} &=\sqrt{2}\left[\sin ^{-1} 0-\sin ^{-1}(-1)\right] \\ &=\sqrt{2}\left[0-\sin ^{-1}(-1)\right]\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\sin ^{-1} 0=0\right] \\ &=\sqrt{2}\left[-\left(-\frac{\pi}{2}\right)\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\sin ^{-1}(-1)=-\frac{\pi}{2}\right] \\ &=\sqrt{2}\left[\frac{\pi}{2}\right] \\ &=\frac{\pi}{\sqrt{2}} \end{aligned}