int_{ - pi /2}^{pi /2} {sqrt {frac{1}{2}(1 - | vedclass

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Question

(B) We know that $1 - \cos 2x = 2 \sin^2 x$. Substituting this into the integral,we get: $\int_{ - \pi /2}^{\pi /2} \sqrt{\frac{1}{2}(2 \sin^2 x)} \,d

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) We know that $1 - \cos 2x = 2 \sin^2 x$. Substituting this into the integral,we get: $\int_{ - \pi /2}^{\pi /2} \sqrt{\frac{1}{2}(2 \sin^2 x)} \,dx = \int_{ - \pi /2}^{\pi /2} \sqrt{\sin^2 x} \,dx = \int_{ - \pi /2}^{\pi /2} |\sin x| \,dx$. Since $|\sin x|$ is an even function,we can write: $2 \int_0^{\pi /2} \sin x \,dx$. Evaluating the integral: $2 [-\cos x]_0^{\pi /2} = 2 [-\cos(\pi/2) - (-\cos 0)] = 2 [0 + 1] = 2$.

$\int_{ - \pi /2}^{\pi /2} {\sqrt {\frac{1}{2}(1 - \cos 2x)} } \,dx = $

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