int_0^{pi /2} sin^4 x cos^6 x , dx equals | vedclass

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(B) Using the Wallis formula for the integral $I = \int_0^{\pi /2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gam

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$\frac{3\pi}{512}$

Vedclass · Answer

(B) Using the Wallis formula for the integral $I = \int_0^{\pi /2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$.Here,$m = 4$ and $n = 6$.$I = \frac{\Gamma(\frac{4+1}{2}) \Gamma(\frac{6+1}{2})}{2 \Gamma(\frac{4+6+2}{2})} = \frac{\Gamma(5/2) \Gamma(7/2)}{2 \Gamma(6)}$.Using $\Gamma(n+1) = n\Gamma(n)$ and $\Gamma(1/2) = \sqrt{\pi}$:$\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3\sqrt{\pi}}{4}$.$\Gamma(7/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{15\sqrt{\pi}}{8}$.$\Gamma(6) = 5! = 120$.$I = \frac{(\frac{3\sqrt{\pi}}{4}) (\frac{15\sqrt{\pi}}{8})}{2 \cdot 120} = \frac{45\pi}{32 \cdot 240} = \frac{45\pi}{7680} = \frac{3\pi}{512}$.

$\int_0^{\pi /2} \sin^4 x \cos^6 x \, dx$ equals

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