Evaluate: `intsqrt("tan\ theta\ d\ theta")`

Explanation

Let `I=int sqrt(tan theta) d theta.`
Let `tan theta=x^(2). " Then, " d(tan theta)=d(x^(2)) " or " sec^(2) theta d theta =2xdx`
or ` d theta=(2x*dx)/(sec^(2) theta)=(2xdx)/(1+tan^(2) theta)=(2xdx)/(1+x^(4))`
`I=int sqrt(x^(2))xx(2xdx)/(1+x^(4))`
`=int(2x^(2))/(x^(4)+1)dx`
`=int(2)/(x^(2)+1//x^(2))dx`
`=int(1+1//x^(2)+1-1//x^(2))/(x^(2)+1//x^(2))dx`
`=int(1+1//x^(2))/(x^(2)+1//x^(2))dx+int(1-1//x^(2))/(x^(2)+1//x^(2))dx`
`=int(1+1//x^(2))/((x-1//x)^(2)+2)dx+int(1-1//x^(2))/((x+1//x)^(2)-2)dx`
Putting `x-(1)/(x)=u` in first integral and `x+(1)/(x)=v` in second integral,
we get `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)((x-1//x)/(sqrt(2)))+(1)/(2sqrt(2))log|(x+1//x-sqrt(2))/(x+1//x+sqrt(2))|+C`
`=(1)/(sqrt(2))tan^(-1)((x^(2)-1)/(x sqrt(2)))+(1)/(2sqrt(2))log|(x^(2)-x sqrt(2)+1)/(x^(2)+x sqrt(2)+1)|+C`
`=(1)/(sqrt(2))tan^(-1)((tan theta-1)/(sqrt(2tan theta)))+(1)/(2sqrt(2))log|(tan theta-sqrt(2tan theta)+1)/(tan theta+ sqrt(2tan theta)+1)|+C`