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Prove that, for even powers of sine,

$$ \int_0^{\frac{\pi}{2}} \sin^{2n} x dx = \frac{1 \cdot 3 \cdot 5 \cdots \cdots (2n - 1)}{2 \cdot 4 \cdot 6 \cdots \cdots 2n} \frac{\pi}{2} $$

Hence, $[(2 n+1)+1] \int_{0}^{\frac{\pi}{2}} \sin ^{2(n+1)} x d x=(2 n+1) \int_{0}^{\frac{\pi}{2}} \sin ^{2 n} x d x=(2 n+$

1) $\frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 n-1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 n} \cdot \frac{\pi}{2}=\frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 n-1) \cdot(2(n+1)-1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 n} \frac{\pi}{2}$ . And

then we have that $\int_{0}^{\frac{\pi}{2}} \sin ^{2(n+1)} x d x=\frac{1}{2(n+1)} \frac{1 \cdot 3 \cdot 5 \cdot \ldots(2 n-1) \cdot(2(n+1)-1)}{2 n} \frac{\pi}{2}=$

$\frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 n-1) \cdot(2(n+1)-1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 n \cdot 2(n+1)} \frac{\pi}{2}$

Thus the formula is true for $n+1$ and then we have that it's true for every $n$ integer with $n \geq 1 .$

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