Question

Evaluate :

`int(sqrt(cotx)+sqrt(tanx))dx`

Answer 1

`I=int(sqrt(cotx)+sqrt(tanx))dx`

`=int(sqrt(tanx)(1+cotx))dx`

`Let tanx=t^2`

Differentiating both sides w.r.t. x, we get

`sec^2 x dx=2t dt`

`=> dx=(2tdt)/(1+t^4)`

`therefore I=intt(1+1/t^2)xx(2t)/(1+t^4)dt`

`=2int(t^2+1)/(t^4+1)dt`

`=2int(1+1/t^2)/(t^2+1/t^2)dt`

`=2int(1+1/t^2)/((t-1/t)^2+2)dt`

`Let (t−1)/t=y`

`=>(1+1/t^2)dt=dy`

`therefore I=2int 1/(y^2+(sqrt2)^2) dy`

`=2xx1/sqrt2 tan^-1(y/sqrt2)+C`

`=sqrt2 tan^-1 (t-1/t)/sqrt2+C`

`=sqrt2 tan^-1 ((t^2-1)/(sqrt2t))+C`

`=sqrt2 tan^-1((tanx-1)/sqrt(2tanx))+C`