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

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(D) We have the integral $I = \int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$. Using the identity $a^2 - b^2 = (a-b)(a+b)$,we can factor

Vedclass · Accepted Answer

$\frac{-1}{2} \sin 2 x+c$

Vedclass · Answer

(D) We have the integral $I = \int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$. Using the identity $a^2 - b^2 = (a-b)(a+b)$,we can factor the numerator: $\sin ^8 x - \cos ^8 x = (\sin ^4 x - \cos ^4 x)(\sin ^4 x + \cos ^4 x) = (\sin ^2 x - \cos ^2 x)(\sin ^2 x + \cos ^2 x)(\sin ^4 x + \cos ^4 x)$. Since $\sin ^2 x + \cos ^2 x = 1$,the numerator becomes $(\sin ^2 x - \cos ^2 x)(\sin ^4 x + \cos ^4 x)$. Now,consider the denominator: $1 - 2 \sin ^2 x \cos ^2 x$. We know that $1 = (\sin ^2 x + \cos ^2 x)^2 = \sin ^4 x + \cos ^4 x + 2 \sin ^2 x \cos ^2 x$. Therefore,$1 - 2 \sin ^2 x \cos ^2 x = \sin ^4 x + \cos ^4 x$. Substituting these into the integral: $I = \int \frac{(\sin ^2 x - \cos ^2 x)(\sin ^4 x + \cos ^4 x)}{\sin ^4 x + \cos ^4 x} d x = \int (\sin ^2 x - \cos ^2 x) d x$. Using the identity $\cos 2x = \cos ^2 x - \sin ^2 x$,we get $\sin ^2 x - \cos ^2 x = -\cos 2x$. Thus,$I = \int -\cos 2x d x = -\frac{1}{2} \sin 2x + C$.

$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=$

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