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 16tan−1(3tanx2)+CThese 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)+5tan2x⇒I=∫sec2x dx4+9tan2x Now, if we assume t=tanx we get dt=sec2xdx So, our integral becomes I=∫dt4+9t2⇒I=19∫dt(23)2+t2⇒I=19×32tan−1(3t2)+C⇒I=16tan−1(3 tanx2)+C

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