The value of the integral int_0^{pi / 2} sin^ | vedclass

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(C) To evaluate the integral $I = \int_0^{\pi / 2} \sin^5 x \, dx$,we use the Wallis formula for $\int_0^{\pi / 2} \sin^n x \, dx$,which is $\frac{(n-

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$\frac{8}{15}$

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(C) To evaluate the integral $I = \int_0^{\pi / 2} \sin^5 x \, dx$,we use the Wallis formula for $\int_0^{\pi / 2} \sin^n x \, dx$,which is $\frac{(n-1)!!}{n!!} 	imes \frac{\pi}{2}$ if $n$ is even,and $\frac{(n-1)!!}{n!!}$ if $n$ is odd.For $n = 5$ (which is odd),the formula gives:$I = \frac{(5-1) 	imes (5-3)}{5 	imes 3 	imes 1} = \frac{4 	imes 2}{5 	imes 3 	imes 1} = \frac{8}{15}$.Alternatively,using substitution:$I = \int_0^{\pi / 2} \sin^4 x \sin x \, dx = \int_0^{\pi / 2} (1 - \cos^2 x)^2 \sin x \, dx$.Let $u = \cos x$,then $du = -\sin x \, dx$. When $x = 0, u = 1$; when $x = \pi / 2, u = 0$.$I = -\int_1^0 (1 - u^2)^2 \, du = \int_0^1 (1 - 2u^2 + u^4) \, du$.$I = [u - \frac{2u^3}{3} + \frac{u^5}{5}]_0^1 = 1 - \frac{2}{3} + \frac{1}{5} = \frac{15 - 10 + 3}{15} = \frac{8}{15}$.

The value of the integral $\int_0^{\pi / 2} \sin^5 x \, dx$ is

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