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ObsessiveMathsFreak
#7
Nov7-06, 04:54 AM
P: 406
Quote Quote by mathman
sin^2 + cos^2 =1. The rest is trivial.
No. That was the trivial part. The rest is hard.

Using the double angle cossine formula;

[tex]\cos{2\theta} = cos^2{\theta} - sin^2{\theta} = 1 - 2\sin^2{\theta}[/tex]
Therefore;
[tex]2\sin^2{\theta}= 1-\cos{2\theta}[/tex]
[tex]\Rightarrow \sin^2{\theta}= \frac{1}{2} \left(1-\cos{2\theta} \right) [/tex]

So that means the integral can be evaluated as follows.

[tex]\int_0^{\pi} \sin^2{\theta} d\theta = \int_0^{\pi} \frac{1}{2} \left(1-\cos{2\theta} \right) d\theta[/tex]
[tex]= \frac{1}{2} \int_0^{\pi}1 d\theta -\frac{1}{2} \int_0^{\pi} \cos{2\theta} d\theta[/tex]
[tex] = \frac{1}{2} \left[ \theta |_0^{\pi}\right] + \frac{1}{4}\left[ \sin{2\theta}|_0^{\pi}\right][/tex]
[tex] = \frac{1}{2} ( \pi - 0) \frac{1}{4} ( 0 - 0) = \frac{\pi}{2}[/tex]