Question

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

Answer 1

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`