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

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12sin2x+C

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Correct option is B

∫sin8⁡x−cos8⁡x1−2sin2⁡xcos2⁡xdx=∫sin2⁡x−cos2⁡xsin2⁡x+cos2⁡xsin4⁡x+cos4⁡x1−2sin2⁡xcos2⁡x =∫sin2⁡x−cos2⁡x1−2sin2⁡xcos2⁡x1−2sin2⁡xcos2⁡xdx=∫-cos 2x dx =-sin 2x +c

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∫sin8⁡x−cos8⁡x1−2sin2⁡xcos2⁡xdx is equal to

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