∫sin2xcos2x(9−cos42x)dx

put t=cos22x dt=−2cos2xsin2x(2)dx =−4sin2xcos2xdxI=−∫dt4(32−t2)=−14sin−1(t3)+c=I=−14sin−1(cos22x3)+c

∫sin2xcos2x(9−cos42x)dx

put t=cos22x dt=−2cos2xsin2x(2)dx =−4sin2xcos2xdxI=−∫dt4(32−t2)=−14sin−1(t3)+c=I=−14sin−1(cos22x3)+c

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

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$\frac{1}{4} \sin^{- 1} (\frac{\cos^{2} 2 x}{3}) + c$

$- \frac{1}{4} \sin^{- 1} (\frac{\cos^{2} 2 x}{3}) + c$

$4 \cos^{- 1} \cos^{4} 2 x + c$

$\frac{1}{3} \cos^{- 1} (\frac{\cos^{2} 2 x}{3}) + c$

answer is B.

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Detailed Solution

put $t = \cos^{2} 2 x$ $d t = - 2 \cos 2 x \sin 2 x (2) d x$

$= - 4 \sin 2 x \cos 2 x d x$

$I = - \int \frac{d t}{4 \sqrt{(3^{2} - t^{2})}} = - \frac{1}{4} \sin^{- 1} (\frac{t}{3}) + c$

= $I = - \frac{1}{4} \sin^{- 1} (\frac{\cos^{2} 2 x}{3}) + c$

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∫sin2xcos2x(9−cos42x)dx

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