int_{-pi/2}^{pi/2} sin^2 x , dx = | vedclass

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(B) Let $I = \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx$. Since $f(x) = \sin^2 x$ is an even function,we have $I = 2 \int_0^{\pi/2} \sin^2 x \, dx$. Using t

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$\frac{\pi}{2}$

Vedclass · Answer

(B) Let $I = \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx$. Since $f(x) = \sin^2 x$ is an even function,we have $I = 2 \int_0^{\pi/2} \sin^2 x \, dx$. Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we get: $I = 2 \int_0^{\pi/2} \frac{1 - \cos 2x}{2} \, dx$ $I = \int_0^{\pi/2} (1 - \cos 2x) \, dx$ $I = [x - \frac{\sin 2x}{2}]_0^{\pi/2}$ $I = (\frac{\pi}{2} - \frac{\sin \pi}{2}) - (0 - \frac{\sin 0}{2})$ $I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

$\int_{-\pi/2}^{\pi/2} \sin^2 x \, dx = $

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