int_0^{pi / 2} sin^8 x , dx = | vedclass

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(C) To evaluate the integral $I = \int_0^{\pi / 2} \sin^8 x \, dx$,we use Wallis' Formula: $\int_0^{\pi / 2} \sin^n x \, dx = \frac{(n-1)(n-3) \dots (

Vedclass · Accepted Answer

$\frac{35 \pi}{256}$

Vedclass · Answer

(C) To evaluate the integral $I = \int_0^{\pi / 2} \sin^8 x \, dx$,we use Wallis' Formula: $\int_0^{\pi / 2} \sin^n x \, dx = \frac{(n-1)(n-3) \dots (1)}{n(n-2) \dots (2)} 	imes \frac{\pi}{2}$ for even $n$. Here,$n = 8$. $I = \frac{7 	imes 5 	imes 3 	imes 1}{8 	imes 6 	imes 4 	imes 2} 	imes \frac{\pi}{2}$ $I = \frac{105}{384} 	imes \frac{\pi}{2}$ $I = \frac{105 \pi}{768}$ Dividing numerator and denominator by $3$: $I = \frac{35 \pi}{256}$.

$\int_0^{\pi / 2} \sin^8 x \, dx =$

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