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$\int_{0}^{2 \pi} \sin ^{5}\left(\frac{\mathrm{x}}{4}\right) \mathrm{d} \mathrm{x}$ is equal to
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Verified Answer
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$\frac{32}{15}$
$\int_{0}^{2 \pi} \sin ^{5}\left(\frac{x}{4}\right) d x=\int_{0}^{2 \pi}\left(1-\cos ^{2} \frac{x}{4}\right)\left(1-\cos ^{2} \frac{x}{4}\right) \sin \frac{x}{4} d x$
Put $\cos \left(\frac{x}{4}\right)=t$
$\Rightarrow-\sin \left(\frac{x}{4}\right) \cdot \frac{d x}{4}=d t$
$\Rightarrow \sin \left(\frac{x}{4}\right) d x=-4 d t$
$\Rightarrow \int_{0}^{2 \pi} \sin ^{5}\left(\frac{x}{4}\right) d x=-4 \int\left(1-t^{2}\right)\left(1-t^{2}\right) \mathrm{dt}$
$=-4 \int\left(1+t^{4}-2 t^{2}\right) \mathrm{dt}$
$=-4\left[t+\frac{t^{5}}{5}-\frac{2 t^{3}}{3}\right]$
$=-4\left[\cos \left(\frac{x}{4}\right)+\frac{\cos ^{5}\left(\frac{x}{4}\right)}{5}-\frac{ \left.2 \cos ^{3}\left(\frac{x}{4}\right)\right]_{0}^{2 \pi}}{3}\right]$
$=-4\left[(0+0-0)-\left(1+\frac{1}{5}-\frac{2}{3}\right)\right]=\frac{32}{15}$
Put $\cos \left(\frac{x}{4}\right)=t$
$\Rightarrow-\sin \left(\frac{x}{4}\right) \cdot \frac{d x}{4}=d t$
$\Rightarrow \sin \left(\frac{x}{4}\right) d x=-4 d t$
$\Rightarrow \int_{0}^{2 \pi} \sin ^{5}\left(\frac{x}{4}\right) d x=-4 \int\left(1-t^{2}\right)\left(1-t^{2}\right) \mathrm{dt}$
$=-4 \int\left(1+t^{4}-2 t^{2}\right) \mathrm{dt}$
$=-4\left[t+\frac{t^{5}}{5}-\frac{2 t^{3}}{3}\right]$
$=-4\left[\cos \left(\frac{x}{4}\right)+\frac{\cos ^{5}\left(\frac{x}{4}\right)}{5}-\frac{ \left.2 \cos ^{3}\left(\frac{x}{4}\right)\right]_{0}^{2 \pi}}{3}\right]$
$=-4\left[(0+0-0)-\left(1+\frac{1}{5}-\frac{2}{3}\right)\right]=\frac{32}{15}$
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