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

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(C) माना $I = \int\limits_0^{\frac{\pi}{4}} (\cos 2x)^{3/2} \cos x \,dx$.सर्वसमिका $\cos 2x = 1 - 2\sin^2 x$ का उपयोग करने पर:$I = \int\limits_0^{\fra

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

Vedclass · Answer

(C) माना $I = \int\limits_0^{\frac{\pi}{4}} (\cos 2x)^{3/2} \cos x \,dx$.सर्वसमिका $\cos 2x = 1 - 2\sin^2 x$ का उपयोग करने पर:$I = \int\limits_0^{\frac{\pi}{4}} (1 - 2\sin^2 x)^{3/2} \cos x \,dx$.माना $\sqrt{2} \sin x = \sin 	heta$. तब $\sqrt{2} \cos x \,dx = \cos 	heta \,d	heta$,जिसका अर्थ है $\cos x \,dx = \frac{1}{\sqrt{2}} \cos 	heta \,d	heta$.जब $x = 0$,तब $\sin 	heta = 0 \implies 	heta = 0$. जब $x = \frac{\pi}{4}$,तब $\sin 	heta = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \implies 	heta = \frac{\pi}{2}$.इन मानों को समाकलन में प्रतिस्थापित करने पर:$I = \int\limits_0^{\frac{\pi}{2}} (1 - \sin^2 	heta)^{3/2} \cdot \frac{1}{\sqrt{2}} \cos 	heta \,d	heta = \frac{1}{\sqrt{2}} \int\limits_0^{\frac{\pi}{2}} (\cos^2 	heta)^{3/2} \cos 	heta \,d	heta$.$I = \frac{1}{\sqrt{2}} \int\limits_0^{\frac{\pi}{2}} \cos^3 	heta \cdot \cos 	heta \,d	heta = \frac{1}{\sqrt{2}} \int\limits_0^{\frac{\pi}{2}} \cos^4 	heta \,d	heta$.वालिस के सूत्र $\int\limits_0^{\frac{\pi}{2}} \cos^n 	heta \,d	heta = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2}$ (जहाँ $n$ सम है) का उपयोग करने पर:$I = \frac{1}{\sqrt{2}} \left( \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} ight) = \frac{1}{\sqrt{2}} \cdot \frac{3\pi}{16} = \frac{3\pi}{16\sqrt{2}}$.

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

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