Begin by rewriting ∫tan4xdx as ∫tan2xtan2xdx.
Now we can apply the Pythagorean Identity, tan2x+1=sec2x, or tan2x=sec2x−1
∫tan2x tan2x dx = ∫(sec2x−1)tan2xdx
Distributing the tan2x:
= ∫sec2xtan2x − tan2xdx
Applying the sum rule:
= ∫sec2xtan2xdx − ∫tan2xdx
We'll evaluate these integrals one by one.
First Integral
This one is solved using a
Let u = tanx
Applying the substitution,
Because u = tanx,
Second Integral
Since we don't really know what ∫tan2xdx is by just looking at it, try applying the tan2x = sec2x−1
identity again:
∫tan2xdx = ∫(sec2x−1)dx
Using the sum rule, the integral boils down to:
∫sec2xdx − ∫1dx
The first of these, ∫sec2xdx, is just tanx + C.
The second one, the so-called "perfect integral", is simply x+C.
Putting it all together, we can say:
∫tan2xdx = tanx + C − x + C
And because C+C is just another arbitrary constant, we can combine it into a general constant C:
∫tan2xdx = tanx − x + C
Combining the two results, we have:
∫tan4xdx=∫sec2xtan2xdx−∫tan2xdx
=(tan3x/3 + C) − (tanx − x + C)
=tan3x/3 − tanx + x + C
Again, because C+C is a constant, we can join them into one C.