Evaluate the integral: int frac{sin ^8 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 can fac

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

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 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 $\sin^2 x + \cos^2 x = 1$,this simplifies to $(\sin^2 x - \cos^2 x)(\sin^4 x + \cos^4 x)$. We know 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 back into the integral: $I = \int \frac{(\sin^2 x - \cos^2 x)(1 - 2 \sin^2 x \cos^2 x)}{1 - 2 \sin^2 x \cos^2 x} dx$. $I = \int (\sin^2 x - \cos^2 x) dx$. Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we have $\sin^2 x - \cos^2 x = -\cos 2x$. $I = \int -\cos 2x dx = -\frac{1}{2} \sin 2x + C$.

Evaluate the integral: $\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} dx$ (where $C$ is the constant of integration).

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