# Integration homework

1. Mar 14, 2013

### VertexOperator

1. The problem statement, all variables and given/known data

Find an expression for

$$\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta)$$

Hence prove that
$$\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2}$$

2. Relevant equations

$$\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta)$$
and
$$\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2}$$

3. The attempt at a solution

I found an expression for $$\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta)$$ which was $$\sum_{k=1}^{n}1-2sin^{2}k\theta$$ but couldn't continue because it doesn't look like the appropriate expression.

Last edited: Mar 14, 2013
2. Mar 14, 2013

### SammyS

Staff Emeritus
What is 1 - sin2(u) ?

3. Mar 14, 2013

### VertexOperator

oh it should have been 1-2sin^2u
edited*

4. Mar 14, 2013

### SammyS

Staff Emeritus
cos(2u) = ?

5. Mar 15, 2013

### VertexOperator

Yes, that is how I go 1-2sin^2(u) in the first place (because the integral had sin) but I don't see how it can help me evaluate the integral :(

6. Mar 15, 2013

### raopeng

Do you know the identity that $cos θ= \frac{e^{iθ} + e^{-iθ}}{2}$?

7. Mar 15, 2013

### raopeng

It seems helpful to use this identity to reduce the question to geometric progression, then what it takes is some computation.

8. Mar 15, 2013

### vela

Staff Emeritus
Along the same lines as raopeng's hint, but a little bit less tedious:

$\cos 2x + \cos 4x + \cdots + \cos (2nx)$ is equal to the real part of $e^{2ix} + e^{4ix} + \cdots + e^{2nix}$

9. Mar 16, 2013

### VertexOperator

How does that help me do the integral though?

10. Mar 16, 2013

### vela

Staff Emeritus
If you calculate the sum correctly, it'll become clear.

11. Mar 16, 2013

### VertexOperator

I have never learnt how to express cis in terms of e so can you please give me a hint on how to do it?
Thank you!

12. Mar 16, 2013

### vela

Staff Emeritus
Not really sure what kind of hint you're looking for.

13. Mar 16, 2013

### VertexOperator

How do I find the real part of e^2ix...

14. Mar 16, 2013

### vela

Staff Emeritus
Use Euler's formula, keeping in mind that $x$ is real.

15. Mar 17, 2013

### VertexOperator

I get
e^2ix=cos2x+ison2x
e^4ix=cos4x+ison4x
.
.
.
e^nix=cosnx+isonnx

I still don't see how this can help with the integral :(

16. Mar 17, 2013

### vela

Staff Emeritus
Hint: $e^{2ix} + e^{4ix} + \cdots + e^{2nix} = (e^{2ix})^1 + (e^{2ix})^2 + \cdots + (e^{2ix})^n$. You have a geometric progression on your hands, as raopeng noted earlier.

17. Mar 18, 2013

### VertexOperator

Ok, so I get $$\frac{e^{ix}+e^{-ix}}{2} =cosx$$
$$\frac{e^{ix}-e^{-ix}}{2} =sinx$$

18. Mar 18, 2013

### VertexOperator

How can I make use of these identities now?

19. Mar 18, 2013

### vela

Staff Emeritus
We're not here to spoon-feed you the solution. You need to show some initiative at solving the problem yourself.

20. Mar 18, 2013

### VertexOperator

I know this is dumb but this is what I did :(
$$\int_{0}^{\frac{\pi }{2}}\frac{e^{i(2n+1)x}-e^{-i(2n+1)x}}{e^{ix}-e^{-ix}}dx$$