#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 25 Maths Textbook Solution.

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{0}^{\frac{\pi }{2}}\sqrt{1+\cos xdx}$

Solution:

$\int_{0}^{\frac{\pi }{2}}\sqrt{1+\cos xdx}$

$=\int_{0}^{\frac{\pi }{2}}\sqrt{2\cos ^{2}\frac{x}{2}dx}$                                                                                  $\left [ 1+\cos 2\theta =2\cos ^{2}\theta \right ]$

$=\int_{0}^{\frac{\pi }{2}}\sqrt{2}\cos \frac{x}{2}dx$                                                                                                    $\left [ \int \cos \: axdx=\frac{\sin \: ax}{a} \right ]$

\begin{aligned} &=\sqrt{2} \int_{0}^{\frac{\pi}{2}} \cos \frac{x}{2} d x \\ &=\sqrt{2}\left[\frac{\sin \frac{x}{2}}{\frac{1}{2}}\right]_{0}^{\frac{\pi}{2}} \\ &=\sqrt{2} .2\left[\sin \frac{x}{2}\right]_{0}^{\frac{\pi}{2}} \\ &=2 \sqrt{2}\left[\sin \frac{\pi}{2 \times 2}-\sin \frac{0}{2}\right] \end{aligned}                                                                    $\left[\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 0=0\right]$

\begin{aligned} &=2 \sqrt{2}\left[\sin \frac{\pi}{4}-0\right] \\ &=2 \sqrt{2}\left[\frac{1}{\sqrt{2}}-0\right] \\ &=2 \sqrt{2} \frac{1}{\sqrt{2}}=2 \end{aligned}