int_0^{pi /2} sin^5 x , dx = | vedclass

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Question

(A) To evaluate the integral $I = \int_0^{\pi /2} \sin^5 x \, dx$,we use Wallis' Formula.Wallis' Formula states that for $n > 1$:$\int_0^{\pi /2} \sin

Vedclass · Accepted Answer

$\frac{8}{15}$

Vedclass · Answer

(A) To evaluate the integral $I = \int_0^{\pi /2} \sin^5 x \, dx$,we use Wallis' Formula.Wallis' Formula states that for $n > 1$:$\int_0^{\pi /2} \sin^n x \, dx = \frac{(n-1)(n-3)\dots(1)}{n(n-2)\dots(2)} 	imes \frac{\pi}{2}$ (if $n$ is even)$\int_0^{\pi /2} \sin^n x \, dx = \frac{(n-1)(n-3)\dots(2)}{n(n-2)\dots(3)}$ (if $n$ is odd)Here,$n = 5$,which is an odd number.Applying the formula:$I = \frac{(5-1)(5-3)}{5(5-2)(5-4)} = \frac{4 	imes 2}{5 	imes 3 	imes 1} = \frac{8}{15}$.Thus,the correct option is $A$.

$\int_0^{\pi /2} \sin^5 x \, dx = $

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