- #1

jisbon

- 476

- 30

- 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?

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?