Integration with Reduction Formula

[Matty R]

Hello

I was hoping someone could help me with this. I'm going round in circles and don't understand how to solve it.

Homework Statement

The integral $$I_m$$ is defined as:

$$I_m = \int^{\frac{\pi}{2}} _{0} (cos^m(x))dx$$

where m is a positive integer.

By representing $$cos^m(x)$$ by $$cos(x)cos^{m-1}(x)$$, show that:

$$I_m = \left( \frac{m-1}{m} \right) I_{m-2}$$

,for $$m \geq 2$$

Homework Equations

Integration by parts

$$I = \int^{b}_{a} \left( u \frac{dv}{dx} \right)dx = \left[uv \right] ^{b}_{a} - \int ^{b}_{a} \left( v \frac{du}{dx} \right)dx$$

The Attempt at a Solution

I've tried using each of the two functions as u and dv/dx, but can't get anywhere.

I would very much appreciate any help with this.

Thanks.

 

[LeonhardEuler]

Can you show what you got when you tried dv/dx=cos(x) and u=cos^(m-1)(x) and explain where you're stuck?

 

[Matty R]

Thanks for the reply.

Part of my problem is that I don't understand how to incorporate the limits for something like $$sin^{m-1}(x)$$.

$$I_{m} = \int^{\frac{\pi}{2}}_{0} \left(cos(x)cos^{m-1}(x) \right)dx$$

Integration by parts

$$I_{m} = \left[sin(x)cos^{m-1}(x) \right]^{b}_{a}+\int^{b}_{a} \left(sin(x)sin^{m-1}(x) \right) dx$$

$$= \left[sin(x)cos^{m-1}(x) \right]^{b}_{a}+J$$

Integration by parts

$$J = - \left[cos(x)sin^{m-1}(x) \right]^{b}_{a} + I_{m}$$

$$I_{m} = \left[sin(x)cos^{m-1}(x) \right]^{b}_{a} - \left[cos(x)sin^{m-1}(x) \right]^{b}_{a} + I_{m}$$

Thats what I get to. I'm not even confident of how I've dealt with u and dv/dx, but I haven't been able to find any information about that. I also don't understand how to get to $$I_{m-2}$$

I can follow the example from my tutorials easily enough ((ln|x|)^n), but I'm completely lost with this one.

EDIT

I haven't differentiated properly. I didn't use the chain rule. I'll try again.

 

[LeonhardEuler]

Yeah, when you do use the chain rule, it should work out.

 

[inline]

Integration by parts

$$I_{m} = \left[sin(x)cos^{m-1}(x) \right]^{b}_{a}+\int^{b}_{a} \left(sin(x)sin^{m-1}(x) \right) dx$$

$$= \left[sin(x)cos^{m-1}(x) \right]^{b}_{a}+J$$

Integration by parts

$$J = - \left[cos(x)sin^{m-1}(x) \right]^{b}_{a} + I_{m}$$

Why $$J = - \left[cos(x)sin^{m-1}(x) \right]^{b}_{a} + I_{m}$$? Where did you found $$sin^{m - 1}\left(x\right)$$?

 

[Matty R]

Yeah, when you do use the chain rule, it should work out.

Yup. I've got it now. I still had a hiccup, with (m - 1) multiplied by a couple of terms, I tried to fully expand it. I think I understand it now.

I found an example applying sin^m(x) in my lecture notes, so I'll use that for applying this integral.

Thanks for your help.

Why $$J = - \left[cos(x)sin^{m-1}(x) \right]^{b}_{a} + I_{m}$$? Where did you found $$sin^{m - 1}\left(x\right)$$?

I think I got it from integrating cos^{m - 1}(x) incorrectly.