Question

Integral of `1/sqrt(tanx) dx=`?

Answer 1

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

Explanation:

We begin with a u-substitution with `u=sqrt(tanx)`

The derivative of `u` is:
`(du)/dx=(sec^2(x))/(2sqrt(tanx))`

so we divide by that to integrate with respect to `u` (and remember, dividing by a fraction is the same as multiplying by its reciprocal):

`int\ 1/sqrt(tanx)\ dx=int\ 1/sqrt(tanx)*(2sqrt(tanx))/sec^2x\ du=`

`=int\ 2/sec^2x\ du`

Since we can't integrate `x`'s with respect to `u`, we use the following identity:

`sec^2theta=tan^2theta+1`

This gives:

`int\ 2/(tan^2x+1)\ du=int\ 2/(1+u^4)\ du=2int\ 1/(1+u^4)\ du`

This remaining integral uses a rather tedious partial fraction decomposition, so I will not do it here. Have a look at this answer if you're interested in how it is worked out:

https://socratic.org/questions/how-do-you-evaluate-the-integral-int-dx-x-4-1

`2int\ 1/(1+u^4)\ du=2(1/(2sqrt2)tan^-1((u^2-1)/(sqrt2u))-1/(4sqrt2)ln|(u^2-sqrt2u+1)/(u^2-sqrt2u+1)|)+C=`

`=1/(sqrt2)tan^-1((u^2-1)/(sqrt2u))-1/(2sqrt2)ln|(u^2-sqrt2u+1)/(u^2-sqrt2u+1)|+C`

Resubstituting for `u=sqrt(tanx)`, we get:

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

Answer 2

`=1/sqrt(2)tan^-1((tanx-1)/(sqrt(2tanx)))-1/(2sqrt(2))ln|(tanx+1-sqrt(2tanx))/(tanx+1+sqrt(2tanx))|+c`

Explanation:

`I=int1/sqrt(tanx)dx`
Let, `sqrt(tanx)=t=>tanx=t^2=>sec^2xdx=2tdt`
`=>(1+tan^2x)dx=2tdt=>dx=(2tdt)/(1+(t^2)^2`
`:.I=int1/cancelt*(2*cancelt*dt)/(1+t^4)=int2/(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)/((t-1/t)^2+2)dt-int(1-1/t^2)/((t+1/t)^2-2)dt`
Take,`(t-1/t)=uand(t+1/t)=v` `=>(1+1/t^2)dt=duand(1-1/t^2)dt=dv` `=>I=int1/(u^2+(sqrt(2))^2)du-int1/(v^2-(sqrt(2))^2)dv=1/sqrt(2)tan^-1(u/sqrt(2))-1/(2sqrt(2))ln|(v-sqrt2)/(v+sqrt2)|+c=1/sqrt(2)tan^-1((t-1/t)/sqrt(2))-1/(2sqrt(2))ln|((t+1/t)-sqrt2)/((t+1/t)+sqrt2)|+c` `=1/sqrt(2)tan^-1((t^2-1)/(sqrt(2)t))-1/(2sqrt(2))ln|((t^2+1-sqrt(2)t))/((t^2+1+sqrt(2)t))|+c`
`=1/sqrt(2)tan^-1((tanx-1)/(sqrt(2tanx)))-1/(2sqrt(2))ln|(tanx+1-sqrt(2tanx))/(tanx+1+sqrt(2tanx))|+c`