evaluating the integral

jojo · **Joined:** Wed, 16 Jul 2008 18:09:23 UTC **Posts:** 10

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.

dmoran · **Posted:** Mon, 18 Aug 2008 18:21:33 UTC

Where are you stuck? Try a substitution.

Dave

jojo · **Joined:** Wed, 16 Jul 2008 18:09:23 UTC **Posts:** 10

well , it needs to be evaluated

jojo · **Joined:** Wed, 16 Jul 2008 18:09:23 UTC **Posts:** 10

i need to substitute in the values of t, separate substitutions and then subtract the first from the second. ???

peccavi_2006 · **Posted:** Wed, 27 Aug 2008 15:13:31 UTC

$\int \cos 2t dt = \frac{1}{2} \sin 2t$

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 ^^

S.O.S. Mathematics CyberBoard

evaluating the integral

Who is online