Reduction formula for definate integrals

[the tree]

Given that [math]I_n = \int_{0}^{\frac{\pi}{2}}x^n \sin(x)\cdot dx[/math]

Show that, for [math]x \ge 2[/math]:

[math]I_n = \left (\frac{\pi}{2} \right )-n(n-1)I_{n-2}[/math]

Okay if that were an indefinite integral then I'd be happy to say that [math]I_n = -x^{n}\cos{x}-n I_{n-1}[/math] but processing just that as a definite integral, [math][-x^{n}\cos{x}-n I_{n-1}]_{0}^{\frac{\pi}{2}}[/math], causes the [math]I[/math] term to be canceled out and I am nowhere near what I am trying to prove.

So what do I do?

[Tom Mattson]

Okay if that were an indefinite integral then I'd be happy to say that [math]I_n = -x^{n}\cos{x}-n I_{n-1}[/math] but processing just that as a definite integral, [math][-x^{n}\cos{x}-n I_{n-1}]_{0}^{\frac{\pi}{2}}[/math], causes the [math]I[/math] term to be canceled out

 

Say what? The [math]I[/math] term does not cancel out. And you don't get [math]I_{n-1}[/math] after integrating by parts once. That's because you get a cosine (not a sine) in the integrand after doing it once. You have to integrate by parts twice to get back to the sine function.

[Dave]

Given that [math]I_n = \int_{0}^{\frac{\pi}{2}}x^n \sin(x)\cdot dx[/math]

Show that, for [math]x \ge 2[/math]:

[math]I_n = \left (\frac{\pi}{2} \right )-n(n-1)I_{n-2}[/math]

Okay if that were an indefinite integral then I'd be happy to say that [math]I_n = -x^{n}\cos{x}-n I_{n-1}[/math] but processing just that as a definite integral, [math][-x^{n}\cos{x}-n I_{n-1}]_{0}^{\frac{\pi}{2}}[/math], causes the [math]I[/math] term to be canceled out and I am nowhere near what I am trying to prove.

So what do I do?

 

It doesn't quite work like that:

 

[math]\int_a^b uv' \, dx = [uv]_a^b - \int_a^b u'v\, dx[/math]

[the tree]

you get a cosine (not a sine) in the integrand after doing it once. You have to integrate by parts twice to get back to the sine function.

Taking [math]\int \underbrace{x^{n}}_u \underbrace{\sin{x}}_{v'} \cdot dx[/math], we get [math]-x^{n} \cos{x} - \int nx^{n-1} \sin{x} \cdot dx[/math]

You would get a cosing in the integrand if you'd started the other way around though.

 

It doesn't quite work like that

Thanks, now I should be able to do this.

[MolotovCocktail]

`Deleted`. I had an example, but somebody beat me to it.

[Tom Mattson]

Taking [math]\int \underbrace{x^{n}}_u \underbrace{\sin{x}}_{v'} \cdot dx[/math], we get [math]-x^{n} \cos{x} - \int nx^{n-1} \sin{x} \cdot dx[/math]

 

Nope. [math]v^{\prime}[/math] does not appear in the right hand side of the formula for integration by parts.

[Dave]

Taking [math]\int \underbrace{x^{n}}_u \underbrace{\sin{x}}_{v'} \cdot dx[/math], we get [math]-x^{n} \cos{x} - \int nx^{n-1} \sin{x} \cdot dx[/math]

 

I'm pretty sure the second integral is wrong there. (It should be [math]u'v[/math] and you have [math]u'v'[/math]).

[the tree]

I'm pretty sure the second integral is wrong there. (It should be [math]u'v[/math] and you have [math]u'v'[/math]).

DAMMIT! Thanks for pointing that one out (both Dave and Tom).