Integrating 50 + 14sin(πt/12) from 0 to 12

[ziddy83]

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?... yeah...i need some help..thanks.

 

[dextercioby]

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

Daniel.

 

[ziddy83]

wait maybe i don't see it...would i let u= sin pi t/12? and then get du...?

 

[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

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

 

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

 

[Jameson]

Or you could do it like that. Good point.

 

[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.