Please help with this question. Thanks in advance. :)

Q:

$\displaystyle I_n = \displaystyle\int^\frac{\pi}{2}_0 x^n \cos x \, \mathrm{d}x$

Prove that $\displaystyle I_n = \frac{\pi ^n}{2^n} - n (n - 1) I_{n-2}, \ n \geq 2$.

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- Apr 24th 2008, 01:14 PMSimplicityReduction Formula
Please help with this question. Thanks in advance. :)

Q:

$\displaystyle I_n = \displaystyle\int^\frac{\pi}{2}_0 x^n \cos x \, \mathrm{d}x$

Prove that $\displaystyle I_n = \frac{\pi ^n}{2^n} - n (n - 1) I_{n-2}, \ n \geq 2$. - Apr 24th 2008, 01:19 PMMathstud28
- Apr 24th 2008, 01:23 PMSimplicity
- Apr 24th 2008, 01:25 PMMathstud28
- Apr 24th 2008, 01:27 PMJhevon
- Apr 24th 2008, 01:33 PMDanshader
generally i will go for u = x^n and dv/dt = cos x. My reasoning is that if x^n is taken as dv/dt then the power of x will increase which brings you nowhere closer in solving the integral. However if you set u = x^n for each part the power of x decreases till x^0 (this gives an integration of a trigonometry which is simple to solve).

- Apr 26th 2008, 09:06 AMSimplicity
Thanks all.

I've done it and obtained the correct answer however my method was pretty long and slightly doubtful. Could someone also do this question so I can see if they do this 'Reduction Formula' method. I will be grateful if someone could help. Thanks in advance. - Apr 26th 2008, 09:19 AMMoo
Hello,

I don't think you could avoid it...

$\displaystyle I_n =\int^\frac{\pi}{2}_0 x^n \cos x \, \mathrm{d}x$

I'll name $\displaystyle \frac{\pi}{2}$ y because it's really awful to write :D

$\displaystyle =[-x^n \sin x]_0^y + n \int_0^y x^{n-1} \sin x dx$

$\displaystyle =y^n+n \underbrace{\left( [x^{n-1} \cos x]_0^y \right)}_{=0} + n(n-1) \underbrace{\int_0^y x^{n-2} \cos x dx}_{I_{n-2}}$

$\displaystyle =\frac{\pi^n}{2^n}+n(n-1)I_{n-2}$

Is it your reduction formula or are you looking for the induction thing ? - Apr 26th 2008, 09:33 AMSimplicity
- Apr 26th 2008, 09:35 AMMoo
I know I've missed some steps, but it should add one or two more lines for each line..

I hope this will really help (Bow)