Begin by rewriting ∫tan^{4}xdx as ∫tan^{2}xtan^{2}xdx.
Now we can apply the Pythagorean Identity, tan^{2}x+1=sec^{2}x, or tan^{2}x=sec^{2}x−1
∫tan^{2}x tan^{2}x dx = ∫(sec^{2}x−1)tan^{2}xdx
Distributing the tan^{2}x:
= ∫sec^{2}xtan^{2}x − tan^{2}xdx
Applying the sum rule:
= ∫sec^{2}xtan^{2}xdx − ∫tan^{2}xdx
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 ∫tan^{2}xdx is by just looking at it, try applying the tan^{2}x = sec^{2}x−1
identity again:
∫tan^{2}xdx = ∫(sec^{2}x−1)dx
Using the sum rule, the integral boils down to:
∫sec^{2}xdx − ∫1dx
The first of these, ∫sec^{2}xdx, is just tanx + C.
The second one, the so-called "perfect integral", is simply x+C.
Putting it all together, we can say:
∫tan^{2}xdx = tanx + C − x + C
And because C+C is just another arbitrary constant, we can combine it into a general constant C:
∫tan^{2}xdx = tanx − x + C
Combining the two results, we have:
∫tan^{4}xdx=∫sec^{2}xtan^{2}xdx−∫tan^{2}xdx
=(tan^{3}x/3 + C) − (tanx − x + C)
=tan^{3}x/3 − tanx + x + C
Again, because C+C is a constant, we can join them into one C.