Question

What is the integral of `int tan^3(x) dx`?

Answer 1

`tan^2(x)/2+ln(abscos(x))+C`

Explanation:

Split up `tan^3(x)` into `tan^2(x)tan(x)` then rewrite `tan^2(x)` using the identity `tan^2(theta)+1=sec^2(theta)=>tan^2(theta)=sec^2(theta)-1`.

`inttan^3(x)dx=inttan^2(x)tan(x)dx=int(sec^2(x)-1)tan(x)dx`

Distribute:

`=intsec^2(x)tan(x)dx-inttan(x)dx`

For the first integral, apply the substitution `u=tan(x)=>du=sec^2(x)dx`, both of which are already in the integral.

`=intucolor(white).du-inttan(x)dx`

`=u^2/2-inttan(x)dx`

`=tan^2(x)/2-inttan(x)dx`

Now rewrite `tan(x)` as `sin(x)/cos(x)` and apply the substitution `v=cos(x)=>dv=-sin(x)dx`.

`=tan^2(x)/2-intsin(x)/cos(x)dx`

`=tan^2(x)/2+int(-sin(x))/cos(x)dx`

`=tan^2(x)/2+int(dv)/v`

This is a common integral:

`=tan^2(x)/2+ln(absv)+C`

`=tan^2(x)/2+ln(abscos(x))+C`