intlimits_0^{frac{pi }{2}} {sqrt {sin 2theta | vedclass

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(B) Let $I = \int\limits_0^{\frac{\pi}{2}} \sqrt{\sin 2	heta} \sin 	heta \,d	heta$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we h

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$\frac{\pi}{4}$

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(B) Let $I = \int\limits_0^{\frac{\pi}{2}} \sqrt{\sin 2	heta} \sin 	heta \,d	heta$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int\limits_0^{\frac{\pi}{2}} \sqrt{\sin 2(\frac{\pi}{2}-	heta)} \sin(\frac{\pi}{2}-	heta) \,d	heta = \int\limits_0^{\frac{\pi}{2}} \sqrt{\sin 2	heta} \cos 	heta \,d	heta$. Adding the two expressions for $I$: $2I = \int\limits_0^{\frac{\pi}{2}} \sqrt{\sin 2	heta} (\sin 	heta + \cos 	heta) \,d	heta$. Since $\sin 2	heta = 1 - (\sin 	heta - \cos 	heta)^2$,let $t = \sin 	heta - \cos 	heta$. Then $dt = (\cos 	heta + \sin 	heta) \,d	heta$. When $	heta = 0, t = -1$. When $	heta = \frac{\pi}{2}, t = 1$. $2I = \int_{-1}^{1} \sqrt{1 - t^2} \,dt$. This is the area of a semicircle of radius $1$,which is $\frac{1}{2} \pi (1)^2 = \frac{\pi}{2}$. Therefore,$2I = \frac{\pi}{2} \implies I = \frac{\pi}{4}$.

$\int\limits_0^{\frac{\pi }{2}} {\sqrt {\sin 2\theta } } \sin \theta \,d\theta$ is equal to :

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