# 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.