Technical question on integrals.

[MathematicalPhysicist]

i have to show that:
1)(-1)^n\int_{-1}^{1}(x^2-1)^ndx=2^{2n+1}(n!)^2/(2n+1)!
2) \binom{n}{k}=[(n+1)\int_{0}^{1}x^k(1-x)^{n-k}dx]^{-1}

for the first part i thought to use Newton's binomial, i.e:
(1-x^2)^n=\sum_{k=0}^{n}\binom{n}{k}(-x^2)^k
but it didn't get me far.
for the second part i don't have a clue, i don't think you can integrate the integral by parts or substitution can you?

 

[StatusX]

Where'd you get stuck after using the binomial theorem? Just integrate term by term.

 

[MathematicalPhysicist]

this is what i got:
2-2n/3+2n(n-1)/10+...+2(-1)^n/(2n+1)
i don't know how to procceed from here.

 

[tim_lou]

yeah, i think number 2 can be worked out using integral by parts... the tabular method.
all except the last term of the resulting series is zero when the bounds are substituted in.