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sin2tsin4t+cos4tdt is equal to 

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a
tan−1tan2t+C
b
−tan−1tan2t+C
c
tan−1cot2t+C
d
cot−1tan2t+C

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detailed solution

Correct option is A

Let I=∫sin⁡2tsin4⁡t+cos4⁡tdt=∫2sin⁡tcos⁡tcos4⁡ttan4⁡t+1dtI=∫2sintcostsec2tdt(tant)4+1 Put tan2⁡t=x⇒2tan⁡tsec2⁡tdt=dxI=∫dxx2+1=tan−1⁡(x)+C=tan−1⁡tan2⁡t+C


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