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Question

dx4+5sin2xdx is equal to.

A

16tan1(3cotx2)+C
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B

16tan1(3tanx2)+C
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C

16cot1(3cotx2)+C
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D

16cot1(3tanx2)+C
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Solution

The correct option is B
16tan1(3tanx2)+C
These types of integrals of the form dxa+bsin2x or dxa+bcos2x
are solved by multiplying sec2x in both numerator and denominator of the integrand and then substituting t=tanx.
So, for our integral multiplying sec2x in both numerator and denominator we get,
I=sec2x(4+5sin2x) sec2xdx
I=sec2x dx4sec2x+5sin2x sec2x
I=sec2x dx4(1+tan2x)+5tan2xI=sec2x dx4+9tan2x
Now, if we assume t=tanx
we get dt=sec2xdx
So, our integral becomes I=dt4+9t2I=19dt(23)2+t2I=19×32tan1(3t2)+CI=16tan1(3 tanx2)+C

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