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Evaluate: int(sin2xcos2x)/(sqrt(9-cos^4 ...

Evaluate: `int(sin2xcos2x)/(sqrt(9-cos^4 2x))\ dx`

Text Solution

Verified by Experts

Let `cos^2x=t`
then`dx=dt/(-4cos2xsin2x)`
So equation becomes
`int(sin2xcos2x)/(sqrt(9-cos^4 2x))\ dx`=`int(sin2xcos2x)/(sqrt(9-t^2))\ dx`
=`int(sin2xcos2x)/(sqrt(9-t^2)) xx dt/(-4cos2xsin2x)`
=-1/4`int(1)/(sqrt(3^2-t^2)) dt`
=`1/4(sin^(-1)(t/3)+c)`
=`1/4(sin^(-1)(cos^2 x/3)+c)`
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