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If $\mathrm{m} \in \mathrm{Z}^{+}, \mathrm{n}=2 \mathrm{~m}$ and $\int_0^{\frac{\pi}{2}} \sin ^{\mathrm{m}} \mathrm{x} \cos ^{\mathrm{n}} \mathrm{x} d \mathrm{x}=\mathrm{K}(\mathrm{m})$ $\int_0^{\frac{\pi}{2}} \sin ^m x d x$, then $\frac{2^{m-1}(m-1) !}{(2 m-1) !} K(m)=$
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Verified Answer
The correct answer is:
$\frac{1}{m+2} \frac{1}{m+4} \frac{1}{m+r} \ldots \frac{1}{3 m}$
From $\mathrm{eq}^{\mathrm{n}}$ (i)
$\mathrm{k}(\mathrm{m})=\frac{2 m-1}{3 m} \cdot \frac{2 m-3}{3 m-2} \cdot \frac{2 m-5}{3 m-4} \ldots \frac{3}{m+4} \cdot \frac{1}{m+2}$
Now, $\frac{2^{m-1}(\mathrm{~m}-1) !}{(2 \mathrm{~m}-1) !} k(\mathrm{~m})$
$\begin{aligned} & =\frac{2^{m-1}(m-1) !}{(2 m-1) !} \cdot \frac{(2 m-1)(2 m-3)(2 m-5) \ldots 3.1}{(3 m)(3 m-2)(3 m-4) \ldots(m+4)(m+2)} \\ & =\frac{1}{m+2} \cdot \frac{1}{m+4} \ldots \frac{1}{m+r} \ldots \frac{1}{3 m}\end{aligned}$
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