Question

How do you integrate `∫ (tan2x + tan4x) dx`?  

Answer 1

Re-write it and use substitution.

`int(tan2x+tan4x)dx=int((sin2x)/(cos2x)+(sin4x)/(cos4x))dx`.

Now do the integrals seperately:

`int(sin2x)/(cos2x)dx`.

Let `u=cos2x`. this makes `du=-2sin2x dx`, so `sin2x dx = -1/2du`.

`int(sin2x)/(cos2x)dx=-1/2int1/(cos2x)sin2x dx=-1/2 int 1/udu`

So, `int(sin2x)/(cos2x)dx=-1/2 ln abs (cos 2x)`.

In a similar way the second integral is found to be

`int(sin4x)/(cos4x)dx=-1/4 ln abs (cos 4x)`.

So,
`int(tan2x+tan4x)dx=-1/2 ln abs (cos 2x)-1/4 ln abs (cos 4x)+C`.

Of course, there are many way of rewriting this expression..