Showing that integral is an induction

[jisbon]

Homework Statement:

Show that $\int _0^{\frac{\pi }{2}}\:sin^{2n+1}x\:dx = \frac{2*4*6*...2n}{3*5*7*...(2n+1)}$#

Relevant Equations:

$\int _0^{\frac{\pi }{2}}\:sin^{n}x\:dx\:=\:\frac{n-1}{n}\int _0^{\frac{\pi }{2}}\:sin^{n-2}x\:dx$

Hi all,

Having this equation derived:
$\int _0^{\frac{\pi }{2}}\:sin^{n}x\:dx\:=\:\frac{n-1}{n}\int _0^{\frac{\pi }{2}}\:sin^{n-2}x\:dx$
What I will do is simply substitue n with n+2, and I will get the following:
$\frac{2n}{2n+1}\int_{0}^{\pi /2}(sinx)^{2n-1}dx$
What should I do from here?

 

[PeroK]

Do you understand the concept of proof by induction?

 

[Delta2]

You can either do induction or use recursion.

I found it to be more interesting to follow the recursive way on this.

By applying the equation you give at homework equations where you substitute n with 2n-1 you get something and then you ll apply it again for n=2n-3 and then for n=2n-5 and so on and with this recursive way you ll get to prove the desired result given that at the very end you ll be left with $\int_0^{\frac{\pi}{2}}\sin x dx$ which is equal to 1.