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Question

Evaluate
dx23sin2x4cos2x

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Solution

We have,

dx23sin2x4cos2xdx

dx23sin2x3cos2xcos2xdx

dx23(sin2x+cos2x)cos2xdx

dx23cos2x

dx1cos2x

1dx1+cos2x

Now, using formula,

11+x2dx=logx+1+x2

Then,

dx1+cos2x=logcosx+1+cos2x+C

Hence, this is the answer.

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