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  4. ∫ ( sin 8 x- cos 8 x/1-2 sin 2 x cos 2 x) d x is equal to

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Q. $\int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x$ is equal to

Integrals

Solution:

$I =\int \frac{\sin ^{8} x -\cos ^{8} x }{1-2 \sin ^{2} x \cos ^{2} x } dx$
$=\int \frac{\left(\sin ^{4} x -\cos ^{4} x \right)\left(\sin ^{4} x +\cos ^{4} x \right)}{1-2 \sin ^{2} x \cos ^{2} x } dx$
$=\int \frac{\left(\sin ^{2} x -\cos ^{2} x \right)\left(\sin ^{2} x +\cos ^{2} x \right)\left(\sin ^{4} x +\cos ^{4} x \right)}{1-2 \sin ^{2} x \cos ^{2} x } dx$
$1 .\left(\sin ^{2} x -\cos ^{2} x \right)\left[\left(\sin ^{2} x +\cos ^{2} x \right)^{2}\right.$
$=\int \frac{ \left.-2 \sin ^{2} x \cos ^{2} x\right]}{1-2 \sin ^{2} x \cos ^{2} x}$
$=\int \frac{\left(\sin ^{2} x-\cos ^{2} x\right)\left(1-2 \sin ^{2} x \cos ^{2} x\right)}{1-2 \sin ^{2} x \cos ^{2} x} d x$
$=-\int \cos 2 x d x=-\frac{1}{2} \sin 2 x+ C$