If m in Z^{+},n=2m and int_0^{frac{pi}{2}} si | vedclass

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(A) Using Wallis' formula,we have $\int_0^{\frac{\pi}{2}} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{

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$\frac{1}{m+2} \cdot \frac{1}{m+4} \cdot \ldots \cdot \frac{1}{3m}$

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(A) Using Wallis' formula,we have $\int_0^{\frac{\pi}{2}} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$.Given $n=2m$,the integral becomes $\int_0^{\frac{\pi}{2}} \sin^m x \cos^{2m} x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{2m+1}{2})}{2 \Gamma(\frac{3m+2}{2})}$.Also,$\int_0^{\frac{\pi}{2}} \sin^m x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{1}{2})}{2 \Gamma(\frac{m+2}{2})}$.Thus,$K(m) = \frac{\int_0^{\frac{\pi}{2}} \sin^m x \cos^{2m} x \, dx}{\int_0^{\frac{\pi}{2}} \sin^m x \, dx} = \frac{\Gamma(\frac{2m+1}{2}) \Gamma(\frac{m+2}{2})}{\Gamma(\frac{3m+2}{2}) \Gamma(\frac{1}{2})}$.Using the duplication formula $\Gamma(z) \Gamma(z + \frac{1}{2}) = 2^{1-2z} \sqrt{\pi} \Gamma(2z)$,we simplify $K(m)$ to obtain the product form:$K(m) = \frac{(2m-1)!! (m-1)!}{2^{m-1} (3m-1)!!} \cdot \dots$ (simplified via Wallis' product).Substituting into the expression $\frac{2^{m-1}(m-1)!}{(2m-1)!} K(m)$,we get $\frac{1}{m+2} \cdot \frac{1}{m+4} \cdot \dots \cdot \frac{1}{3m}$.

If $m \in Z^{+}$,$n=2m$ and $\int_0^{\frac{\pi}{2}} \sin ^{m} x \cos ^{n} x \, dx = K(m) \int_0^{\frac{\pi}{2}} \sin ^m x \, dx$,then $\frac{2^{m-1}(m-1)!}{(2m-1)!} K(m) =$

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