Question

What is the Integral of `tan^3(4x)`?

Answer 1

We have that `int(tan^3(4x)dx`

Let `u=4x`
so, `du/4=dx`

Substitute in the integral

`1/4int(tan^3(u))du`
`1/4int(tan^2u*tanu)du`

Use the trigonometric identity: `tan^2u=sec^2x-1`

Hence

`1/4int(sec^2(u)-1)(tan(u))du`
`1/4int(sec^2(u)tan(u)-tan(u))du`
`1/4(int(sec^2(u)tan(u)du-int(tan(u)du))`
For the first integral `(int(sec^2(u)tan(u))du)`:
Let `p=tan(u)`
`dp=sec^2(u)du`
so,
`int(p)dp =p^2/2+c`

For the second integral
#(int(tan(u))du) =-ln|cos(u)|+c #
put it all back in
`1/4[(p^2/2)+(ln|cos(u)|))+c`
substitute back:
`1/4(tan^2u/2+ln|cos(u)|+c`
substitute back for u:
`tan^2(4x)/8+ln|cos(4x)|/4+c`

finally

`int(tan^3(4x))dx=tan^2(4x)/8+ln|cos(4x)|/4+c`