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 $\displaystyle I_n = \frac{\pi ^n}{2^n} - n (n - 1) I_{n-2}, \ n \geq 2$.
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).
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.
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
$\displaystyle =[-x^n \sin x]_0^y + n \int_0^y x^{n-1} \sin x dx$
$\displaystyle =y^n+n \underbrace{\left( [x^{n-1} \cos x]_0^y \right)}_{=0} + n(n-1) \underbrace{\int_0^y x^{n-2} \cos x dx}_{I_{n-2}}$
$\displaystyle =\frac{\pi^n}{2^n}+n(n-1)I_{n-2}$
Is it your reduction formula or are you looking for the induction thing ?