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

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(B) Let $I = \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$Using the identity $a^2 - b^2 = (a - b)(a + b)$,we have:$

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

Vedclass · Answer

(B) Let $I = \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$Using the identity $a^2 - b^2 = (a - b)(a + b)$,we have:$I = \int {\frac{{({{\sin }^4}x - {{\cos }^4}x)({{\sin }^4}x + {{\cos }^4}x)}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$Since ${{\sin }^4}x + {{\cos }^4}x = ({{\sin }^2}x + {{\cos }^2}x)^2 - 2{{\sin }^2}x{{\cos }^2}x = 1 - 2{{\sin }^2}x{{\cos }^2}x$,the expression simplifies to:$I = \int {({{\sin }^4}x - {{\cos }^4}x)} dx$Further,${{\sin }^4}x - {{\cos }^4}x = ({{\sin }^2}x - {{\cos }^2}x)({{\sin }^2}x + {{\cos }^2}x) = -\cos 2x \cdot 1 = -\cos 2x$Thus,$I = \int {-\cos 2x} dx = -\frac{{\sin 2x}}{2} + c$

$\int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$ is equal to

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