∫$$sin2xcos2x9$$−$$cos42xdx=$$

Correct option is 4−14sin−113cos22x+c

Questions

$\int \frac{\sin 2 x \cos 2 x}{\sqrt{9 - \cos^{4} (2 x)}} d x =$

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12cos−113cos22x+c

−14cos−113cos22x+c

−12sin−113cos22x+c

−14sin−113cos22x+c

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detailed solution

Correct option is D

Put cos2⁡2x=t⇒-2sin2xcos2x2dx=dt⇒sin2xcos2x2dx=-14dt -14∫19-t4dt=-14sin-1t3+c

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∫sin2xcos2x9−cos42xdx=

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$\int \frac{\sin 2 x \cos 2 x}{\sqrt{9 - \cos^{4} (2 x)}} d x =$