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

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Question

(D) We use the Wallis formula for the definite integral $\int_0^{\pi / 2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}

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

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(D) We use the Wallis formula for the definite integral $\int_0^{\pi / 2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$.Here, $m = 8$ and $n = 2$.Substituting these values, we get $\int_0^{\pi / 2} \sin^8 x \cos^2 x \, dx = \frac{\Gamma(\frac{9}{2}) \Gamma(\frac{3}{2})}{2 \Gamma(\frac{12}{2})} = \frac{\Gamma(\frac{9}{2}) \Gamma(\frac{3}{2})}{2 \Gamma(6)}$.Using $\Gamma(n+1) = n \Gamma(n)$, we have $\Gamma(\frac{9}{2}) = \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi}$ and $\Gamma(\frac{3}{2}) = \frac{1}{2} \sqrt{\pi}$.Also, $\Gamma(6) = 5! = 120$.Thus, the integral is $\frac{(\frac{7 \cdot 5 \cdot 3 \cdot 1}{16} \sqrt{\pi}) \cdot (\frac{1}{2} \sqrt{\pi})}{2 \cdot 120} = \frac{\frac{105}{32} \pi}{240} = \frac{105 \pi}{7680} = \frac{7 \pi}{512}$.

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

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