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

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\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

\int_{}A\sin\left(\frac{t\pi-6}{12}\right)+Bdt

Break this into two integrals

\int_{}A\sin\left(\frac{t\pi-6}{12}\right)dt+\int_{}Bdt

Let us do the harder integral first

\int_{}A\sin\left(\frac{t\pi-6}{12}\right)dt

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

A\int_{}\sin\left(\frac{t\pi-6}{12}\right)dt

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

A\int_{}\frac{12}{\pi}\sin\left(u\right)du

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

Now you have

A\frac{12}{\pi}\int_{}\sin\left(u\right)du

Which is

-A\frac{12}{\pi}\cos\left(u\right)

replace the u for what was substituted

-A\frac{12}{\pi}\cos\left(\frac{t\pi-6}{12}\right)

Now do the second integral

\int_{}Bdt =Bt

The whole equation is

-A\frac{12}{\pi}\cos\left(\frac{t\pi-6}{12}\right)+Bt

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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