Intuitive explanation of proof of reduction formula?

Question

I've been asked to, using integration by parts, prove the reduction formula;

INT (tan^n (x)) dx = (tan(x)^n-1)/(n-1) - INT( tan(x)^(n-2) dx)

I can barely manage to get to the part with the recursion (if that's an appropriate way to describe it), in which I come up with;

I= (tan(x)^(n-1))/(n-1)

The next step is what throws me off. In the solutions manual, they go straight from that to the answer;

-------------------------------------------------
"Returning to the original integral, INT (tan^n (x)) dx = (tan(x)^n-1)/(n-1) - INT( tan(x)^(n-2) dx)"
-------------------------------------------------

Which, from my perspective, just seems to jump up out of nowhere, with no formal explanation. Can someone please explain the major steps of this proof? I'm particularly interested in that last step, but any other information would be greatly appreciated.

kb · Accepted Answer

Do you really need integration by parts to do this?

Rewrite the integral as follows:

∫ tan^n(x) dx.
  = ∫ tan^(n-2)(x) * tan^2(x) dx, by splitting the powers of tangent  
  = ∫ tan^(n-2)(x) * (sec^2(x) - 1) dx, via sec^2(x) = 1 + tan^2(x)
  = ∫ [tan^(n-2)(x) * sec^2(x) - tan^(n-2)(x)] dx  
  = ∫ tan^(n-2)(x) * sec^2(x) dx - ∫ tan^(n-2)(x) dx.

This first term can be evaluated by substitution.
u = tan x, du = sec^2(x) dx

So, ∫ tan^(n-2)(x) * sec^2(x) dx
= ∫ u^(n-2) du
= u^(n-1)/(n-1) + C, assuming n is not 1.
= tan^(n-1)(x)/(n - 1) + C

Thus, 
∫ tan^n(x) dx = tan^(n-1)(x)/(n - 1) - ∫ tan^(n-2)(x) dx.

I hope this helps!

Trending News

Intuitive explanation of proof of reduction formula?

1 Answer