`intsqrt[tanx].dx`

Answer

N/a

Explanation

Let ` I = intsqrt(tanx)dx`
Put `tanx = t^(2) rArr sec^(2) x dx = 2t dt`
`:. I = intt. (2t)/(sec^(2)x) dt = 2int(t^(2))/(1+t^(4))dt`
`= int((t^(2)+1)+(t^(2)-1))/((1+t^(4))) dt`
`= int(t^(2)+1)/(1+t^(4))dt + int(t^(2)-1)/(1+t^(4))dt`
`= int(1+1/(t^(2)))/(t^(2)+(1)/(t^(2))) dt + int (1-(1)/(t^(2)))/(t^(2)+(1)/(t^(2))) dt`
` = int(1-(-1/(t^(2))dt))/((t-(1)/(t))^(2)+2) + int(1+(-(1)/(t^(2))))/((t+(1)/(t))^(2)-2) dt`
Put `u=t-1/t rArr du = (1+1/(t^(2)))dt`
and `v = t + (1)/(t) rArr dv = (1-1/(t^(2)))dt`
`:. I = int(du)/(u^(2)+(sqrt(2))^(2)) + int(dv)/(v^(2)-(sqrt(2))^(2))`
`= 1/(sqrt(2))tan^(-1)'(u)/(sqrt(2)) + 1/(2sqrt(2)) "log"|(v-sqrt(2))/(v+sqrt(2))|+C`
`= 1/(sqrt(2))tan^(-1)((tanx-1)/(sqrt(2)tanx))+(1)/(2sqrt(2))"log" |(tanx-sqrt(2tanx)+1)/(tanx+sqrt(2tanx)+1)|+C`