Q226 / 270 · NDA 2017 (Phase 2) · Mathematics

What is $\int_0^{2 \pi} \sqrt{1+\sin \frac{x}{2}} d x$ equal to?

  • Correct answer
    A
    $8$
  • B
    $4$
  • C
    $2$
  • D
    $0$

Solution

$\begin{aligned} & \int_0^{2 \pi} \sqrt{1+\sin \frac{x}{2}} \cdot d x \\ & \sin \frac{x}{2}=\sin 2\left(\frac{x}{4}\right)=2 \sin \frac{x}{4} \cos \frac{x}{4}\end{aligned}$ $\begin{aligned} & \therefore \int_0^{2 \pi} \sqrt{\sin ^2 \frac{x}{4}+\cos ^2 \frac{x}{4}+2 \sin \frac{x}{4} \cos \frac{x}{4}} \\ & =\int_0^{2 \pi} \sqrt{\left(\sin \frac{x}{4}+\cos \frac{x}{4}\right)^2} \cdot d x\end{aligned}$ $\begin{aligned} & =\int_0^{2 \pi}\left|\sin \frac{x}{4}+\cos \frac{x}{4}\right| \cdot d x=4\left[-\cos \frac{x}{4}+\sin \frac{x}{4}\right]_0^{2 \pi} \\ & =4\left[-\left(\cos \frac{2 \pi}{4}-\cos 0\right)+\sin \left(\frac{2 \pi}{4}-\sin 0\right)\right]\end{aligned}$ $\begin{aligned} & =4[-(-1)+(1)] \\ & =4 \times 2=8\end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions