#### Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 63 Maths Textbook Solution.

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

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

Solution: $\int_{0}^{2 \pi} \sqrt{1+\sin \frac{x}{2}} d x=\int_{0}^{2 \pi} \sqrt{\cos ^{2} \frac{x}{4}+\sin ^{2} \frac{x}{4}+2 \sin \frac{x}{4} \cos \frac{x}{4}} d x$

${\left[1=\cos ^{2} \theta+\sin ^{2} \theta, \sin 2 \theta=2 \sin \theta \cos \theta\right]}$

$=\int_{0}^{2 \pi} \sqrt{\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right)^{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[a^{2}+b^{2}+2 a b=(a+b)^{2}\right] \\$

$=\int_{0}^{2 \pi}\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right) d x$                                                                $\left[\begin{array}{l} \int \sin a x d x=\frac{-\cos a x}{a} \\ \int \cos a x d x=\frac{\sin a x}{a} \end{array}\right]$

$=\int_{0}^{2_{\pi}} \cos \frac{x}{4} d x+\int_{0}^{2 \pi} \sin \frac{x}{4} d x$

$=\left[\frac{\sin \frac{x}{4}}{\frac{1}{4}}\right]_{0}^{2 \pi}-\left[\frac{\cos \frac{x}{4}}{\frac{1}{4}}\right]_{0}^{2 \pi}$

$=4\left[\sin \frac{2 \pi}{4}-\sin 0\right]-4\left[\cos \frac{2 \pi}{4}-\cos 0\right]$                                             $\left[\begin{array}{l} \sin \frac{\pi}{2}=\cos 0=1 \\ \sin 0=\cos \frac{\pi}{2}=0 \end{array}\right]$

$=4\left[\sin \frac{\pi}{2}-\sin 0\right]-4\left[\cos \frac{\pi}{2}-\cos 0\right]$

$=4[1-0]-4[0-1]$

$=4+4$

$=8$