If int frac{sin ^8 x-cos ^8 x}{1-2 sin ^2 x c | vedclass

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Question

(A) Let $I = \int \frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx$. We know that $\sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(A) Let $I = \int \frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx$. We know that $\sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)$. Also,$1 - 2 \sin^2 x \cos^2 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = \sin^4 x + \cos^4 x$. Substituting these into the integral,we get: $I = \int \frac{(\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)}{\sin^4 x + \cos^4 x} dx$. $I = \int (\sin^4 x - \cos^4 x) dx$. Using the identity $\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) = -\cos 2x(1) = -\cos 2x$. $I = \int -\cos 2x dx = -\frac{\sin 2x}{2} + B$. Comparing this with $I = A \sin 2x + B$,we find $A = -\frac{1}{2}$.

If $\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=A \sin 2 x+B$,then $A$ is equal to

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