intsec^(2/3)xcosec^(4/3)xdx=...

`intsec^(2/3)xcosec^(4/3)xdx=`

A

(a) `3(tanx)^(1/3)+c`

B

(b) `3(cotx)^((-1)/3)+c`

C

(c) `-3(tanx)^((-1)/3)+c`

D

(d) `-3(cotx)^((-1)/3)+c`

Verified by Experts

The correct Answer is:

C

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