How do you integrate #int tan^5(x/4)#?

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Answer 1 · 2017-03-07T00:00:12

Given #inttan^5(x/4)dx#

Begin with a substitution for #x/4#:

#u=x/4 => du=1/4dx => 4du=dx#

We now have:

#4inttan^5(u)du#

Break up #tan^5(u)#:

#4inttan^2(u)*tan^3(u)du#

Use the trigonometric identity #tan^2(theta)=sec^2(theta)-1#:

#4int(sec^2(u)-1)tan^3(u)du#

Distribute #tan^3#:

#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:

#z=tan(u) => dz=sec^2(u)du#

#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 #tan^3(u)#:

#4inttan^2(u)*tan(u)du#

Apply trigonometric identity #tan^2(theta)=sec^2(theta)-1#:

#4int(sec^2(u)-1)tan(u)du#

Distribute #tan(u)#:

#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 #4inttan(u)du#.

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#

#=>tan^4(u)-2tan^(u)+4ln|sec(u)|+C#

#=>tan^4(x/4)-2tan^2(x/4)+4ln|sec(x/4)|+C#