int frac{sin 2x cos 2x}{sqrt{4-cos^4 2x}} , d | vedclass

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Question

(B) Let $I = \int \frac{\sin 2x \cos 2x}{\sqrt{4-\cos^4 2x}} \, dx$. Substitute $u = \cos^2 2x$. Then $du = 2 \cos 2x (-\sin 2x) \cdot 2 \, dx = -4 \s

Vedclass · Accepted Answer

$-\frac{1}{4} \sin^{-1}\left(\frac{\cos^2 2x}{2}\right) + c$,where $c$ is the constant of integration.

Vedclass · Answer

(B) Let $I = \int \frac{\sin 2x \cos 2x}{\sqrt{4-\cos^4 2x}} \, dx$. Substitute $u = \cos^2 2x$. Then $du = 2 \cos 2x (-\sin 2x) \cdot 2 \, dx = -4 \sin 2x \cos 2x \, dx$. So,$\sin 2x \cos 2x \, dx = -\frac{1}{4} du$. The integral becomes $I = \int \frac{-\frac{1}{4} du}{\sqrt{4-u^2}} = -\frac{1}{4} \int \frac{du}{\sqrt{2^2-u^2}}$. Using the standard formula $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(\frac{x}{a}) + c$,we get: $I = -\frac{1}{4} \sin^{-1}\left(\frac{u}{2}\right) + c$. Substituting back $u = \cos^2 2x$,we obtain: $I = -\frac{1}{4} \sin^{-1}\left(\frac{\cos^2 2x}{2}\right) + c$.

$\int \frac{\sin 2x \cos 2x}{\sqrt{4-\cos^4 2x}} \, dx =$

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