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Integrate the function
$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$
$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$
Solution:
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Verified Answer
Let $\sin ^8 x-\cos ^8 x=\left(\sin ^4 x+\cos ^4 x\right)\left(\sin ^4 x-\cos ^4 x\right)$
$=\left(1-2 \sin ^2 x \cos ^2 x\right)(1)(-\cos 2 x)$
$\therefore I=\int \frac{\left(1-2 \sin ^2 x \cos ^2 x\right)(-\cos 2 x)}{1-2 \sin ^2 x \cos ^2 x} d x$
$\Rightarrow I=-\int \cos 2 x d x=-\frac{1}{2} \sin 2 x+c$
$=\left(1-2 \sin ^2 x \cos ^2 x\right)(1)(-\cos 2 x)$
$\therefore I=\int \frac{\left(1-2 \sin ^2 x \cos ^2 x\right)(-\cos 2 x)}{1-2 \sin ^2 x \cos ^2 x} d x$
$\Rightarrow I=-\int \cos 2 x d x=-\frac{1}{2} \sin 2 x+c$
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