intlimits_0^{frac{pi }{4}} (cos 2x)^{3/2} cos | vedclass

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(C) Let $I = \int\limits_0^{\frac{\pi}{4}} (\cos 2x)^{3/2} \cos x \,dx$.Using the identity $\cos 2x = 1 - 2\sin^2 x$,we have:$I = \int\limits_0^{\frac

Vedclass · Accepted Answer

$\frac{3\pi}{16\sqrt{2}}$

Vedclass · Answer

(C) Let $I = \int\limits_0^{\frac{\pi}{4}} (\cos 2x)^{3/2} \cos x \,dx$.Using the identity $\cos 2x = 1 - 2\sin^2 x$,we have:$I = \int\limits_0^{\frac{\pi}{4}} (1 - 2\sin^2 x)^{3/2} \cos x \,dx$.Let $\sqrt{2} \sin x = \sin 	heta$. Then $\sqrt{2} \cos x \,dx = \cos 	heta \,d	heta$,which implies $\cos x \,dx = \frac{1}{\sqrt{2}} \cos 	heta \,d	heta$.When $x = 0$,$\sin 	heta = 0 \implies 	heta = 0$. When $x = \frac{\pi}{4}$,$\sin 	heta = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \implies 	heta = \frac{\pi}{2}$.Substituting these into the integral:$I = \int\limits_0^{\frac{\pi}{2}} (1 - \sin^2 	heta)^{3/2} \cdot \frac{1}{\sqrt{2}} \cos 	heta \,d	heta = \frac{1}{\sqrt{2}} \int\limits_0^{\frac{\pi}{2}} (\cos^2 	heta)^{3/2} \cos 	heta \,d	heta$.$I = \frac{1}{\sqrt{2}} \int\limits_0^{\frac{\pi}{2}} \cos^3 	heta \cdot \cos 	heta \,d	heta = \frac{1}{\sqrt{2}} \int\limits_0^{\frac{\pi}{2}} \cos^4 	heta \,d	heta$.Using Wallis' formula $\int\limits_0^{\frac{\pi}{2}} \cos^n 	heta \,d	heta = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2}$ for even $n$:$I = \frac{1}{\sqrt{2}} \left( \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} ight) = \frac{1}{\sqrt{2}} \cdot \frac{3\pi}{16} = \frac{3\pi}{16\sqrt{2}}$.

$\int\limits_0^{\frac{\pi }{4}} (\cos 2x)^{3/2} \cos x \,dx =$

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