Solve Integral: 50 + 14sin(πt/12) | 0 to 12

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The integral discussed is ∫ from 0 to 12 of (50 + 14sin(πt/12) dt). The consensus is that a simple substitution, u = πt/12, is sufficient for solving it, rather than using integration by parts. Participants emphasize the importance of adjusting the limits of integration accordingly. The anti-derivative of sin(kt) is highlighted as -cos(kt)/k, which simplifies the process. Ultimately, the original poster successfully solved the integral using the suggested substitution and constants.
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integrals, oh yeah...

hey what's up,
Ok so...i am having a little bit of a problem on solving the following integral...

\int_{0}^{12} 50 + 14 sin\frac{\pi t}{12} dt

would i use...trig substitution or by parts?... :rolleyes: yeah...i need some help..thanks.
 
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No,need for part integration,a simple,obvious substitution would do it.Pay attention to the change of limits (of integration).

Daniel.
 
wait maybe i don't see it...would i let u= sin pi t/12? and then get du...?
 
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The \int (a + b) = \int a + \int b

So the only thing that seems tricky is the second part.

Let u = \frac{\pi{t}}{12} , du = \frac{\pi}{12}

Set \frac{\pi}{12} = 14dx and put the new integral in the form of

C*\int sin(u)du
 
You don't need any substitution. You have the following:

The second part is:

14 \int \sin (kt) dt

where k=\frac{\pi}{12}

The anti-derivative of sin kt is \frac{-\cos kt}{k}
 
Or you could do it like that. Good point. :rolleyes: :-p
 
great...thanks a lot guys...i got the right answer for the problem. i just used the substitution of u = \frac{\pi{t}}{12} and brought out all of the constants. Thanks again.
 
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