int_0^pi sin^2 x , dx is equal to | vedclass

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(B) Let $I = \int_0^\pi \sin^2 x \, dx$. Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a - x) = f(x)$: Here,$f(x) = \sin^

Vedclass · Accepted Answer

$\frac{\pi}{2}$

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(B) Let $I = \int_0^\pi \sin^2 x \, dx$. Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a - x) = f(x)$: Here,$f(x) = \sin^2 x$ and $2a = \pi$,so $a = \frac{\pi}{2}$. Since $f(\pi - x) = \sin^2(\pi - x) = \sin^2 x = f(x)$,the property holds. Thus,$I = 2 \int_0^{\pi/2} \sin^2 x \, dx$. Using the Wallis formula or $\sin^2 x = \frac{1 - \cos 2x}{2}$: $I = 2 \int_0^{\pi/2} \frac{1 - \cos 2x}{2} \, dx = \int_0^{\pi/2} (1 - \cos 2x) \, dx$. $I = [x - \frac{\sin 2x}{2}]_0^{\pi/2} = (\frac{\pi}{2} - 0) - (0 - 0) = \frac{\pi}{2}$.

$\int_0^\pi \sin^2 x \, dx$ is equal to

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