int_0^{pi / 2} sin ^5left(frac{x}{2}right) cd | vedclass

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(C) Let $I = \int_0^{\pi / 2} \sin ^5 \left(\frac{x}{2}\right) \sin x \, dx$. Using the identity $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get

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$\frac{1}{14 \sqrt{2}}$

Vedclass · Answer

(C) Let $I = \int_0^{\pi / 2} \sin ^5 \left(\frac{x}{2}\right) \sin x \, dx$. Using the identity $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get: $I = \int_0^{\pi / 2} \sin ^5 \left(\frac{x}{2}\right) \cdot 2 \sin \frac{x}{2} \cos \frac{x}{2} \, dx$ $I = 2 \int_0^{\pi / 2} \sin ^6 \left(\frac{x}{2}\right) \cos \frac{x}{2} \, dx$ Let $t = \sin \frac{x}{2}$. Then $dt = \frac{1}{2} \cos \frac{x}{2} \, dx$,which implies $2 \, dt = \cos \frac{x}{2} \, dx$. When $x = 0$,$t = \sin(0) = 0$. When $x = \pi / 2$,$t = \sin(\pi / 4) = \frac{1}{\sqrt{2}}$. Substituting these into the integral: $I = 2 \int_0^{1/\sqrt{2}} t^6 (2 \, dt) = 4 \int_0^{1/\sqrt{2}} t^6 \, dt$ $I = 4 \left[ \frac{t^7}{7} \right]_0^{1/\sqrt{2}} = \frac{4}{7} \left( \frac{1}{\sqrt{2}} \right)^7$ Since $(\sqrt{2})^7 = 2^3 \cdot \sqrt{2} = 8 \sqrt{2}$,we have: $I = \frac{4}{7 \cdot 8 \sqrt{2}} = \frac{1}{7 \cdot 2 \sqrt{2}} = \frac{1}{14 \sqrt{2}}$.

$\int_0^{\pi / 2} \sin ^5\left(\frac{x}{2}\right) \cdot \sin x \, dx =$

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