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 Post subject: evaluating the integral
PostPosted: Mon, 18 Aug 2008 17:51:56 UTC 
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Joined: Wed, 16 Jul 2008 18:09:23 UTC
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can someone please help me.

pi/3
~ 7 cos 2tdt
0


the line in middle in the integral sign lol, dont know how you do it.
thanks.


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 Post subject:
PostPosted: Mon, 18 Aug 2008 18:21:33 UTC 
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Where are you stuck? Try a substitution.

Dave

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 Post subject: integral
PostPosted: Fri, 22 Aug 2008 17:33:24 UTC 
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well , it needs to be evaluated


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 Post subject: working out
PostPosted: Fri, 22 Aug 2008 17:35:57 UTC 
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i need to substitute in the values of t, separate substitutions and then subtract the first from the second. ???


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 Post subject:
PostPosted: Wed, 27 Aug 2008 15:13:31 UTC 
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\int \cos 2t dt = \frac{1}{2} \sin 2t

Therefore,

\\
\int_0^\frac{\pi}{3} 7 \cos 2t dt = [\frac{7}{2} \sin 2t]_0^\frac{\pi}{3}\\
\\
= \frac{7}{2} \sin \frac{2 \pi}{3} - \frac{7}{2} \sin (0)

\sin \frac{\pi}{3} = \sin \frac{2 \pi}{3} = \frac{\sqrt 3}{2}

So

\\
\frac{7}{2} \sin \frac{2 \pi}{3} - \frac{7}{2} \sin (0) = \frac{7 \sqrt 3}{4}

as \sin(0) = 0

Hmm...I think that's about right...unless I screwed up somewhere (summer holidays have a habit of killing my brain cells)

That's also assuming you meant \cos 2t - if you meant \cos^{2}t then you would have to use the formula:

\\
\cos 2t = 2 \cos^{2}t - 1

Unless i'm missing something obvious, which makes this integral far more complex than I realised, I think this thread belongs in 'Calculus' rather than Advanced Maths ^^

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