The value of int frac{sin 2x}{sin^4 x + cos^4 | vedclass

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(B) Let $I = \int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx$.We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin

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

Vedclass · Answer

(B) Let $I = \int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx$.We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$.Using $\sin 2x = 2\sin x \cos x$,we have $2\sin^2 x \cos^2 x = \frac{1}{2}(2\sin x \cos x)^2 = \frac{1}{2}\sin^2 2x$.So,$I = \int \frac{\sin 2x}{1 - \frac{1}{2}\sin^2 2x} dx = \int \frac{2\sin 2x}{2 - \sin^2 2x} dx$.Using $\sin^2 2x = 1 - \cos^2 2x$,we get $I = \int \frac{2\sin 2x}{2 - (1 - \cos^2 2x)} dx = \int \frac{2\sin 2x}{1 + \cos^2 2x} dx$.Let $t = \cos 2x$,then $dt = -2\sin 2x dx$,so $2\sin 2x dx = -dt$.$I = \int \frac{-dt}{1 + t^2} = -	an^{-1}(t) + C$.Substituting back $t = \cos 2x$,we get $I = -	an^{-1}(\cos 2x) + C$.

The value of $\int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx$ is

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