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Old 2011-04-10, 19:56   #4
Mathew's Avatar
Nov 2009

2·52·7 Posts

\int_{b}^aA\sin\left(\frac{\pi(t-6)}{12}\right)+B dt what I understood from Baz's post

\int_{a}^bA\sin\left(\frac{t\pi-6}{12}\right)+B dt what Christenson wrote

Let us look at Christenson's equation without the points


Break this into two integrals


Let us do the harder integral first


A is a constant and can be placed outside of the integral


Next is an u-substitution

let u=\frac{t\pi-6}{12}

then du=\frac{\pi}{12}dt

Therefore dt=\frac{12}{\pi}du

Substitute the values


Again since \frac{12}{\pi} is a constant move it out of the integral

Now you have


Which is


replace the u for what was substituted


Now do the second integral

\int_{}Bdt =Bt

The whole equation is


Evaluating from a to b i.e. \int_{a}^b it would be the following

\left(-A\frac{12}{\pi}\cos\left(\frac{b\pi-6}{12}\right)+B*b \right)-\left(-A\frac{12}{\pi}\cos\left(\frac{a\pi-6}{12}\right)+B*a\right)

Entering in all the values =6.25624
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