Integration by parts with trig function insanity

[cuttlefish]

Homework Statement

Let I_n=the integral from 0 to pi/2 of sinⁿxdx
show that (I_2n+1)/(I_2n)=(2n+1)/(2n+2)

Homework Equations

integral from 0 to pi/2 of sin²ⁿ=(2n-1)pi/4n

The Attempt at a Solution

I can't write down all that I've done because it's just too ridiculous. I've tried lots of forms of just trying to manipulate my equation above, which didn't work. I've tried integration by parts for sin²ⁿ⁺¹*sinx but I've ended up with the integral of cos²xsin²ⁿx and I can't find a suitable substitution for that. I've been working on this problem forever and it's killing me. If anyone has any suggestions I would be very appreciative.

 

[JG89]

Have you seen the derivation of Wallis' infinite product for pi/2?

 

[cuttlefish]

unfortunately, I have to use this problem to prove that next.

 

[lanedance]

so you're trying to do the even integral first? think you might have been on the right track how about looking at some induction as a next step...

I_2n = $$\int$$dt sint²ⁿ = $$\int$$dt sint^2n-1.sint

integrating by parts

I_2n = (2n-1)$$\int$$dt sint^2(n-1)cost²

I_2n = (2n-1)$$\int$$dt sint^2(n-1) - (2n-1)$$\int$$dt sint²ⁿ

So
2n. I_2n = (2n-1).I_2(n-1)

 

[lanedance]

so you're trying to do the even integral? think you might have been on the right track how about looking at some induction as a next step...

so from I_2n = $$\int$$dt sint²ⁿ = $$\int$$dt sint^2n-1.sint

and integrating by parts like you did, I get to

2n. I_2n = (2n-1).I_2(n-1)

Also is that formula for I_2n in the post correct? i would of thought it should tend zero as n heads towrds infinity