#### Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 4 maths

4
Given:

$\int_{0}^{2 \pi} \sqrt{1+\sin {x} / 2} d x$

Hint:

Using $\int \sin x \; d x, \int \cos x \; d x$

Explanation:

Let

$I=\int_{0}^{2 \pi} \sqrt{1+\sin {x} / 2} d x$

\begin{aligned} &{\left[\sin 2 \theta=2 \sin \theta \cos \theta, \quad \sin ^{2} \theta+\cos ^{2} \theta=1\right]} \\\\ &=\int_{0}^{2 \pi} \sqrt{\left(\sin {x} / 4+\cos {x} / 4\right)^{2}} d x \\\\ &=\int_{0}^{2 \pi}\left(\sin x / 4+\cos {x} / 4\right) d x \end{aligned}

\begin{aligned} &=\int_{0}^{2 \pi} \sin x / 4 d x+\int_{0}^{2 \pi} \cos ^{x} / 4 d x \\\\ &=\left[-4 \cos \frac{x}{4}\right]_{0}^{2 \pi}+\left[4 \sin \frac{x}{4}\right]_{0}^{2 \pi} \end{aligned}

\begin{aligned} &=-4 \cos [x / 4]_{0}^{2 \pi}+\left[4 \sin \frac{x}{4}\right]_{0}^{2 \pi} \\\\ &=\left[-4 \cos \frac{\pi}{2}+4 \cos 0\right]+\left[4 \sin \frac{\pi}{2}-4 \sin 0\right] \end{aligned}

\begin{aligned} &=0+4+0 \\\\ &=4 \end{aligned}