int frac{sin 2x}{sin^4 x + cos^4 x} dx का मान | vedclass

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Question

(B) माना $I = \int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx$.हम जानते हैं कि $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2

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$-	an^{-1}(\cos 2x) + C$

Vedclass · Answer

(B) माना $I = \int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx$.हम जानते हैं कि $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$.$\sin 2x = 2\sin x \cos x$ का उपयोग करने पर,$2\sin^2 x \cos^2 x = \frac{1}{2}(2\sin x \cos x)^2 = \frac{1}{2}\sin^2 2x$.अतः,$I = \int \frac{\sin 2x}{1 - \frac{1}{2}\sin^2 2x} dx = \int \frac{2\sin 2x}{2 - \sin^2 2x} dx$.$\sin^2 2x = 1 - \cos^2 2x$ का उपयोग करने पर,$I = \int \frac{2\sin 2x}{2 - (1 - \cos^2 2x)} dx = \int \frac{2\sin 2x}{1 + \cos^2 2x} dx$.माना $t = \cos 2x$,तो $dt = -2\sin 2x dx$,इसलिए $2\sin 2x dx = -dt$.$I = \int \frac{-dt}{1 + t^2} = -	an^{-1}(t) + C$.$t = \cos 2x$ वापस रखने पर,हमें $I = -	an^{-1}(\cos 2x) + C$ प्राप्त होता है।

$\int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx$ का मान ज्ञात कीजिए।

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