Chain rule with definite integrals

[xone306]

Hi, I'm working on a problem that has me a bit stumped:

The boundaries are 2 on the bottom, y on the top of the integral sign
g(y)= (integral sign) t^2*sintdt

I figure you can multiply the whole thing by 2/2 to make the t to be the derivative of t²/2, and then make u=t²/2:

2(integral sign)t²/2*sintdt

....however I'm a little confused as to what you assign as "du".
Should it be du=sintdt? Or du=tdt? Or am I way off on this? All the examples we've done so far had "u" be what is after sin, so I'm a little uncertain of how to proceed form here.

Thanks!

 

[xone306]

Never mind...it needs to be done using integration by parts I guess!

 

[ksdhart]

Yeah, I believe integrating by parts is the is the best way to go here. You sound like you may have already completed the problem, and if so that's great. If not, I'll help you with how to start.

To integrate by parts, you want the integral in the form \(\displaystyle \int \:udv\). Generally what I do is set the first term in your multiplication as u and the second/rest as dv. This may actually make your integral more complicated. If that happens, go back to the drawing board and pick different values for u and dv.

But anyway, we'll choose \(\displaystyle u=t^2\) and \(\displaystyle dv=sin\:t\:dt\). That gives \(\displaystyle du=2tdt\) and \(\displaystyle v=-cos\:t\)

Then, by the very useful formula: \(\displaystyle \int \:udv=uv-\int \:vdu\)

Plug in the relevant values and you get: \(\displaystyle \int _2^y\:\:t^2sin\:t\:dt\:=\:t^2\cdot \left(-cos\:t\right)-\int _2^y\:\:\left(-cos\:t\right)\cdot 2tdt\)

In this case, it looks like you'll need to integrate by parts again, but you should be able to handle it. I don't want to do your homework for you

 

[Ishuda]

Hi, I'm working on a problem that has me a bit stumped:

The boundaries are 2 on the bottom, y on the top of the integral sign
g(y)= (integral sign) t^2*sintdt

I figure you can multiply the whole thing by 2/2 to make the t to be the derivative of t²/2, and then make u=t²/2:

2(integral sign)t²/2*sintdt

....however I'm a little confused as to what you assign as "du".
Should it be du=sintdt? Or du=tdt? Or am I way off on this? All the examples we've done so far had "u" be what is after sin, so I'm a little uncertain of how to proceed form here.

Thanks!

For integration by parts, if you are familiar with the continued/recursive/table integration by parts, for example see
http://en.wikipedia.org/wiki/Integration_by_parts#Recursive_integration_by_parts
what you want for the derivative of u is to eventually become zero [or repeat in the proper way with the anti derivatives of v] so for cases like a polynomial in the integrand [like t²]you would normally want to start with that as your u and the rest of it as the dv and see if that seems to 'work out nicely'.