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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}\;dx}$.We know that $1 = (\sin^2 x + \cos^2 x)^2 = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x$,so the denominator $1 - 2\sin^2 x \cos^2 x = \sin^4 x + \cos^4 x$.Thus,$I = \int {\frac{{(\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)}}{{\sin^4 x + \cos^4 x}}\;dx}$.$I = \int {(\sin^4 x - \cos^4 x)\,dx}$.Using the identity $a^2 - b^2 = (a-b)(a+b)$,we have $\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)$.Since $\sin^2 x + \cos^2 x = 1$,the integral becomes $I = \int {(\sin^2 x - \cos^2 x)\,dx}$.Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we have $\sin^2 x - \cos^2 x = -\cos 2x$.Therefore,$I = \int {-\cos 2x\,dx} = -\frac{\sin 2x}{2} + c$.

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

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