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$\int_0^{\pi / 2} \sin ^{2 m} x d x=$
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Verified Answer
The correct answer is:
$\frac{(2 m)!}{\left(2^m \cdot m!\right)^2} \cdot \frac{\pi}{2}$
Here the power is even, so from formula
$\begin{aligned} & \int_0^{\pi / 2} \sin ^{2 m} x d x=\frac{(2 m-1)}{2 m} \cdot \frac{(2 m-3)}{(2 m-2)} \cdots \cdots \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} \\ & =\frac{2 m \cdot(2 m-1)(2 m-2) \ldots \cdot 3 \cdot 2 \cdot 1 \cdot \frac{\pi}{2}}{[2 m \cdot(2 m-2)(2 m-4) \ldots . .4 \cdot 2]^2}\end{aligned}$
Multiplying the numerator and the denominator by $2 m(2 m-2) \ldots . .4 .2$ $=\frac{(2 m)!}{\left[2^m \cdot m(m-1)(m-2) \ldots . .2 .1\right]^2} \frac{\pi}{2}$
$=\frac{(2 m)!}{\left(2^m \cdot m!\right)^2} \frac{\pi}{2}$
$\begin{aligned} & \int_0^{\pi / 2} \sin ^{2 m} x d x=\frac{(2 m-1)}{2 m} \cdot \frac{(2 m-3)}{(2 m-2)} \cdots \cdots \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} \\ & =\frac{2 m \cdot(2 m-1)(2 m-2) \ldots \cdot 3 \cdot 2 \cdot 1 \cdot \frac{\pi}{2}}{[2 m \cdot(2 m-2)(2 m-4) \ldots . .4 \cdot 2]^2}\end{aligned}$
Multiplying the numerator and the denominator by $2 m(2 m-2) \ldots . .4 .2$ $=\frac{(2 m)!}{\left[2^m \cdot m(m-1)(m-2) \ldots . .2 .1\right]^2} \frac{\pi}{2}$
$=\frac{(2 m)!}{\left(2^m \cdot m!\right)^2} \frac{\pi}{2}$
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