Integrals, oh yeah

1. Feb 26, 2005

ziddy83

integrals, oh yeah......

hey whats 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?.... :uhh: yeah...i need some help..thanks.

2. Feb 26, 2005

dextercioby

No,need for part integration,a simple,obvious substitution would do it.Pay attention to the change of limits (of integration).

Daniel.

3. Feb 26, 2005

ziddy83

wait maybe i dont see it....would i let u= sin pi t/12? and then get du......?

Last edited: Feb 26, 2005
4. Feb 26, 2005

Jameson

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

$$C*\int sin(u)du$$

5. Feb 26, 2005

BobG

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 $$\frac{-\cos kt}{k}$$

6. Feb 26, 2005

Jameson

Or you could do it like that. Good point. :tongue2:

7. Feb 26, 2005

ziddy83

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.