Q233 / 270 · NDA 2015 (Phase 2) · Mathematics

$\int_0^{2 \pi} \sin ^5\left(\frac{x}{4}\right) d x$ is equal to

  • A
    8/15
  • B
    16/15
  • Correct answer
    C
    32/15
  • D
    0

Solution

$\int_0^{2 \pi} \sin ^5\left(\frac{x}{4}\right) d x=\int_0^{2 \pi}\left(1-\cos ^2 \frac{x}{4}\right)\left(1-\cos ^2 \frac{x}{4}\right) \sin \frac{x}{4} \cdot d x$ Put $\cos \left(\frac{x}{4}\right)=t$ $\begin{aligned} & \Rightarrow-\sin \left(\frac{x}{4}\right) \cdot \frac{d x}{4}=d t \\ & \Rightarrow \sin \left(\frac{x}{4}\right) d x=-4 d t \\ & \Rightarrow \int_0^{2 \pi} \sin ^5\left(\frac{x}{4}\right) d x=-4 \int\left(1-t^2\right)\left(1-t^2\right) d t \\ & =-4\left[\left(1+t^4-2 t^2\right) d t\right. \\ & =-4\left[t+\frac{t^5}{5}-\frac{2 t^3}{3}\right] \\ & =-4\left[\cos \left(\frac{x}{4}\right)+\frac{\cos ^5\left(\frac{x}{4}\right)}{5}-\frac{2 \cos ^3\left(\frac{x}{4}\right)}{3}\right]_0^{2 \pi} \end{aligned}$ $=-4\left[(0+0-0)-\left(1+\frac{1}{5}-\frac{2}{3}\right)\right]=\frac{32}{15}$

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

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