Integral of cos^2((π)x): Is it \frac{1}{3}?

[americanforest]

Quick question: What is the integral of

[tex]\int^{\frac{1}{2}}_{0}cos^2((\pi)x)[/tex]?

Is it

[tex]\frac{1}{3}(cos^3((\pi)x))\frac{1}{\pi}sin((\pi)x)[/tex]

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

 

[acm]

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

 

[americanforest]

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

 

[quasar987]

[tex]\cos x=\frac{e^{ix}+e^{-ix}}{2}[/tex]

Square that and you have your result.

 

[quasar987]

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

 

[dextercioby]

Or simply use that

[tex] \cos 2x=\cos^{2} x -\sin^{2} x [/tex]

and

[tex] 1=\cos^{2} x +\sin^{2} x [/tex]

Daniel.

 

[VietDao29]

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