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$Antiderivative of \frac{\sqrt[3]{\tan x}}{\cos^{2} x} d x is$

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43tanx34+K

43tanx43+K

34tanx34+K

34tanx43+K

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Correct option is D

Let I=tan⁡x3cos2⁡xdx=∫tan⁡x3sec2⁡xdx Put tan⁡x=t⇒sec2⁡xdx=dtI=∫t13dt=34(t)43+K=34(tan⁡x)43+K

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Antiderivative of tan⁡x3cos2⁡xdx is

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$Antiderivative of \frac{\sqrt[3]{\tan x}}{\cos^{2} x} d x is$