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cosecxtanxdx=

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a
2cosx+C
b
−2sinx+C
c
2sinx+C
d
2cosx+C

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detailed solution

Correct option is B

I=∫cos⁡ecxtan⁡xdx=∫cos⁡x(sin⁡x)32dx Let sin⁡x=tcos⁡xdx=dt I=∫t−32dt=t1−321−32+C=−2t+C=−2sin⁡x+C Level of question: Moderate


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