Integral of #1/sqrt(tanx) dx=#?
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#1/(sqrt2)tan^-1((tanx-1)/(sqrt(2tanx)))-1/(2sqrt2)ln|(tanx-sqrt(2tanx)+1)/(tanx-sqrt(2tanx)+1)|+C#
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#
#=1/sqrt(2)tan^-1((tanx-1)/(sqrt(2tanx)))-1/(2sqrt(2))ln|(tanx+1-sqrt(2tanx))/(tanx+1+sqrt(2tanx))|+c#
#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#
