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

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Question

(A) Let $I = \int \frac{\sin ^8 x - \cos ^8 x}{1 - 2 \sin ^2 x \cos ^2 x} \, dx$. Using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$,we

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{\sin ^8 x - \cos ^8 x}{1 - 2 \sin ^2 x \cos ^2 x} \, dx$. Using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$,we can write the numerator as: $\sin ^8 x - \cos ^8 x = (\sin ^4 x - \cos ^4 x)(\sin ^4 x + \cos ^4 x) = (\sin ^2 x - \cos ^2 x)(\sin ^2 x + \cos ^2 x)(\sin ^4 x + \cos ^4 x)$. Since $\sin ^2 x + \cos ^2 x = 1$,the numerator becomes $(\sin ^2 x - \cos ^2 x)(\sin ^4 x + \cos ^4 x)$. Also,note that $\sin ^4 x + \cos ^4 x = (\sin ^2 x + \cos ^2 x)^2 - 2 \sin ^2 x \cos ^2 x = 1 - 2 \sin ^2 x \cos ^2 x$. Substituting this into the integral: $I = \int \frac{(\sin ^2 x - \cos ^2 x)(1 - 2 \sin ^2 x \cos ^2 x)}{1 - 2 \sin ^2 x \cos ^2 x} \, dx$. $I = \int (\sin ^2 x - \cos ^2 x) \, dx = \int -(\cos ^2 x - \sin ^2 x) \, dx$. Using the identity $\cos 2x = \cos ^2 x - \sin ^2 x$,we get: $I = -\int \cos 2x \, dx = -\frac{\sin 2x}{2} + C$.

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

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