# Trig Integrals

1. Nov 8, 2006

### americanforest

Quick question: What is the integral of

$$\int^{\frac{1}{2}}_{0}cos^2((\pi)x)$$?

Is it

$$\frac{1}{3}(cos^3((\pi)x))\frac{1}{\pi}sin((\pi)x)$$

and then plug in or is there something wrong with that?

Last edited: Nov 8, 2006
2. Nov 8, 2006

### acm

Cos^2(pix) = 1/2 (1 + Cos(2pix))
This is simply integrated.

3. Nov 8, 2006

### americanforest

Do you know where I can get a proof of that?

4. Nov 8, 2006

### quasar987

$$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$

Square that and you have your result.

5. Nov 8, 2006

### quasar987

Or square the power series of cosx and rearange but that's more difficult.

Last edited: Nov 8, 2006
6. Nov 9, 2006

### dextercioby

Or simply use that

$$\cos 2x=\cos^{2} x -\sin^{2} x$$

and

$$1=\cos^{2} x +\sin^{2} x$$

Daniel.

7. Nov 9, 2006

### VietDao29

Or basically, we can use the Power Reduction Fomulae:
$$\cos ^ 2 x = \frac{1 + \cos (2x)}{2}$$
$$\sin ^ 2 x = \frac{1 - \cos (2x)}{2}$$
-------------
Now, back to your problem:
$$\int_{0} ^ {\frac{\pi}{2}} \cos ^ 2 (\pi x) dx = \int_{0} ^ {\frac{\pi}{2}} \left( \frac{1 + \cos (2x)}{2} \right) dx$$
Now, all you need to do is to use a u-substitution to solve it.
Can you go from here? :)

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