Question
$\int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x$ is equal to(A) $\frac{1}{2} \sin 2 x+c$(B) $-\frac{1}{2} \sin 2 x+c$(C) $-\frac{1}{2} \sin x+c$(D) $\quad-\sin ^{2} x+c$
Step 1
We can write $\sin^{8}x - \cos^{8}x$ as $(\sin^{4}x)^{2} - (\cos^{4}x)^{2}$. This is a difference of squares, which can be factored as $(\sin^{4}x - \cos^{4}x)(\sin^{4}x + \cos^{4}x)$. Show more…
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