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

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(B) Using Wallis's Formula for the integral of $\sin^{n} x$ from $0$ to $\pi/2$ where $n = 2m$ is even:$\int_0^{\pi /2} \sin^{2m} x \, dx = \frac{2m-1

Vedclass · Accepted Answer

$\frac{(2m)!}{(2^m \cdot m!)^2} \cdot \frac{\pi}{2}$

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(B) Using Wallis's Formula for the integral of $\sin^{n} x$ from $0$ to $\pi/2$ where $n = 2m$ is even:$\int_0^{\pi /2} \sin^{2m} x \, dx = \frac{2m-1}{2m} \cdot \frac{2m-3}{2m-2} \cdot \dots \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}$To simplify this,multiply the numerator and denominator by the product of even numbers $2m \cdot (2m-2) \cdot \dots \cdot 2$:Numerator: $(2m-1)(2m-3)\dots(1) \cdot [2m(2m-2)\dots(2)] = (2m)!$Denominator: $[2m(2m-2)\dots(2)]^2 = [2^m \cdot m(m-1)\dots(1)]^2 = (2^m \cdot m!)^2$Thus,the integral is $\frac{(2m)!}{(2^m \cdot m!)^2} \cdot \frac{\pi}{2}$.

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

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