Indefinite integral help

uman

Hi,

I need help evaluating the following integral by integration by parts: [tex]\int(a^2-x^2)^n\,dx[/tex]. Specifically I am supposed to prove the following formula: [tex]\int(a^2-x^2)^n\,dx=\frac{x(a^2-x^2)^n}{2n+1}+\frac{2a^2n}{2n+1}\int(a^2-x^2)^{n-1}\,dx+C[/tex] Any hints would be appreciated.

Also, does anyone have any general tips for learning how to do these things? I understand the principle behind integration by parts but I just can never seem to decide how to go about solving them. I'm learning from Tom Apostol's book and this is by far the hardest chapter so far for me as far as the exercises go even though other chapters were much more difficult to understand conceptually.

Vid

Start with the substitution asin(theta) = x. This gives you (a^2cos^2)^n*(acos). Rewrite this as (a^2cos^2)^(n+1/2). Use the formula for the integral of cos^k (I used a table of integrals here, but I'm guessing the method used to prove it uses integration by parts.) and back substitute to get the write answer. If you're not supposed to know how to do trig subs yet then I'm not really sure how to go about proving it using just integration by parts.

Vid

Ok, to the proof of the integral of cos^n(x)dx does use integration by parts.
u = cos(x)^(n-1)
dv = cos(x)dx.

cos(x)^n = -cos(x)^(n-1)sin(x) + int[ (n-1)cos(x)^(n-2)sin^2(x)dx
Now, using the fact that 1-cos^2 = sin^2 we get:
cos(x)^n = -cos(x)^(n-1)sin(x) + int[ (n-1)cos^(n-2)] + int[(n-1)cos(x)^n
Now, moving the right most integral of the left side and dividing by -n gives us:
cos(x)^n = cos(x)^(n-1)sin(x)/n + (n-1)/n*int[cos^(n-2)

Of course in your case it's not n, it's 2n+1.