How do you integrate #int tan^5(x/4)#?
1 Answer
Given
Begin with a substitution for
#u=x/4 => du=1/4dx => 4du=dx#
We now have:
#4inttan^5(u)du#
Break up
#4inttan^2(u)*tan^3(u)du#
Use the trigonometric identity
#4int(sec^2(u)-1)tan^3(u)du#
Distribute
#4intsec^2(u)tan^3(u)-tan^3(u)du#
Split the integral:
#4intsec^2(u)tan^3(u)du-4inttan^3(u)du#
For the LH integral, we can perform a substitution:
#4intsec^2(u)tan^3(u)du=>4intz^3dz#
This is basic integral. We will now move on to the RHS.
#4inttan^3(u)du#
Break up
#4inttan^2(u)*tan(u)du#
Apply trigonometric identity
#4int(sec^2(u)-1)tan(u)du#
Distribute
#4intsec^2(u)tan(u)-tan(u)du#
Split the integral:
#4intsec^2(u)tan(u)du-4inttan(u)du#
For the LH integral, a substitution:
#r=tan(u) => dr=sec^2(u)du#
#4intsec^2(u)tan(u)du=4intrdr#
This is a basic integral, as is
We have:
#4intz^3dz-[4intrdr-4inttan(u)du]#
We integrate, then substitute back in for all of the variables.
#4(1/4z^4)-[4(1/2r^2)-4ln|sec(u)|]+C#