Integrate the function: frac{sin ^{8} x-cos ^ | vedclass

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

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(D) Let $I = \int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} \, dx$.
We know that $\sin ^{8} x - \cos ^{8} x = (\sin ^{4} x - \cos ^{4} x)(\sin ^{4} x + \cos ^{4} x) = (\sin ^{2} x - \cos ^{2} x)(\sin ^{2} x + \cos ^{2} x)(\sin ^{4} x + \cos ^{4} x)$.
Since $\sin ^{2} x + \cos ^{2} x = 1$,we have $\sin ^{8} x - \cos ^{8} x = (\sin ^{2} x - \cos ^{2} x)(\sin ^{4} x + \cos ^{4} x)$.
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 these into the integrand:
$\frac{(\sin ^{2} x - \cos ^{2} x)(1 - 2 \sin ^{2} x \cos ^{2} x)}{1 - 2 \sin ^{2} x \cos ^{2} x} = \sin ^{2} x - \cos ^{2} x = -(\cos ^{2} x - \sin ^{2} x) = -\cos 2x$.
Therefore,$I = \int -\cos 2x \, dx = -\frac{\sin 2x}{2} + C$.

Integrate the function: $\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}$

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