# What is antiderivative of y=`sqrt(tgx)` /(cosx)^2?

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### 1 Answer

antiderivative of y=`int(sqrt(tgx)) /cos^2x dx`

`=int sqrttgx*sec^2x dx`

Let `u=tgx` so, `du=sec^2xdx`

and the integral becomes

`int sqrtu du=intu^(1/2)du`

`=u^(3/2)/(3/2)+c`

`=2/3u^(3/2)+c`

Substituting the value of u we get

`=2/3*tg^(3/2)x+c`

where, `tgx=tanx`