> 
> By creating a reduction formula for 
>  $$I\_n=\int\_0 ^\pi x^n \sin x \,dx$$ Show that $I\_4=\pi^4 -12\pi^2 + 48$
> 
> 
> 

So by using integration by parts I wrote $I\_n$ as $$I\_n=\int\_0 ^\pi x^n \sin x \,dx = \left[-x^n\cos x \right]\_0 ^\pi -\int\_0 ^\pi nx^{n-1}\times -\cos x\, dx$$ and after messing about with this I got $$I\_n = \pi^n +nI\_{n-1}$$ However, the answer shows that this is wrong and I don't get the correct answer for $I\_4$. The correct answer is $I\_n = \pi^n -n(n-1)I\_{n-2}$. Can someone help explain how you get this reduction formula as I can't seem to get it and why my one doesn't work.