int_0^{pi /2} sin^2 x cos^3 x , dx = | vedclass

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Question

(B) 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

Vedclass · Accepted Answer

$\frac{2}{15}$

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(B) 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,$m = 2$ and $n = 3$.Substituting these values,we get:$\int_0^{\pi/2} \sin^2 x \cos^3 x \, dx = \frac{\Gamma(\frac{2+1}{2}) \Gamma(\frac{3+1}{2})}{2 \Gamma(\frac{2+3+2}{2})} = \frac{\Gamma(\frac{3}{2}) \Gamma(2)}{2 \Gamma(\frac{7}{2})}$.We know that $\Gamma(\frac{3}{2}) = \frac{1}{2} \sqrt{\pi}$,$\Gamma(2) = 1! = 1$,and $\Gamma(\frac{7}{2}) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi} = \frac{15}{8} \sqrt{\pi}$.Thus,the integral is $\frac{(\frac{1}{2} \sqrt{\pi}) \cdot 1}{2 \cdot (\frac{15}{8} \sqrt{\pi})} = \frac{\frac{1}{2} \sqrt{\pi}}{\frac{15}{4} \sqrt{\pi}} = \frac{1}{2} \cdot \frac{4}{15} = \frac{2}{15}$.

$\int_0^{\pi /2} \sin^2 x \cos^3 x \, dx = $

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