int_0^{pi / 2} sin ^8 x cos ^2 x , dx is equa | vedclass

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(D) We use the Wallis formula for the definite integral $\int_0^{\pi / 2} \sin^m x \cos^n x \, dx = \frac{[(m-1)(m-3)\dots] 	imes [(n-1)(n-3)\dots]}{

Vedclass · Accepted Answer

$\frac{7 \pi}{512}$

Vedclass · Answer

(D) We use the Wallis formula for the definite integral $\int_0^{\pi / 2} \sin^m x \cos^n x \, dx = \frac{[(m-1)(m-3)\dots] 	imes [(n-1)(n-3)\dots]}{(m+n)(m+n-2)\dots} 	imes \frac{\pi}{2}$ (if both $m, n$ are even).Here $m=8$ and $n=2$. Both are even.$\int_0^{\pi / 2} \sin^8 x \cos^2 x \, dx = \frac{(7 	imes 5 	imes 3 	imes 1) 	imes (1)}{(10 	imes 8 	imes 6 	imes 4 	imes 2)} 	imes \frac{\pi}{2}$$= \frac{105}{3840} 	imes \frac{\pi}{2}$$= \frac{105 \pi}{7680}$Dividing numerator and denominator by $15$,we get $\frac{7 \pi}{512}$.

$\int_0^{\pi / 2} \sin ^8 x \cos ^2 x \, dx$ is equal to

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