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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Definite Integration

NDA 2017 · Mathematics · Definite Integration

What is ∫_0^ 2 π 1+ (x)/(2) d x equal to?

What is ∫02π1+sin⁡x2dx\int_0^{2 \pi} \sqrt{1+\sin \frac{x}{2}} d x equal to?

Options

  1. A88
  2. B44
  3. C22
  4. D00

Correct answer

A. 88

Step-by-step solution

∫02π1+sin⁡x2⋅dxsin⁡x2=sin⁡2(x4)=2sin⁡x4cos⁡x4\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}∴∫02πsin⁡2x4+cos⁡2x4+2sin⁡x4cos⁡x4=∫02π(sin⁡x4+cos⁡x4)2⋅dx\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}=∫02π∣sin⁡x4+cos⁡x4∣⋅dx=4[−cos⁡x4+sin⁡x4]02π=4[−(cos⁡2π4−cos⁡0)+sin⁡(2π4−sin⁡0)]\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}=4[−(−1)+(1)]=4×2=8\begin{aligned} & =4[-(-1)+(1)] \\ & =4 \times 2=8\end{aligned}$
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