# Thread: twice integration by part?

1. ## twice integration by part?

This is a differential equation problem, but since I have problem with integration, I am posting here for help. Thank you very much.

I need to integrate the right side

I will list my steps.

1. perform integration by part

2. result

3. again, integration by part

4. result (edited)
$\frac{9}{5}\left [ (200+t)^{2}(t+sin(t))-\left [ (2(200+t))(\frac{1}{2}t^{2}-cos(t))-\int \frac{1}{2}t^{2}-cos(t)2dt \right ] \right ]$
$\frac{9}{5}\left [ (200+t)^{2}(t+sin(t))-\left [ (2(200+t))(\frac{1}{2}t^{2}-cos(t))- \left [ \frac{1}{3}t^{3}-2sin(t) \right ] \right ] \right ]$

The last result is not right. This is the result from the website (Paul's Note)
Pauls Online Notes : Differential Equations - Modeling with First Order DE's (first problem)

I don't even have (200+t)^3

Can someone help me on this? Thank you very much!

2. First of all, int[(1/2)t^2 dt] = (1/2)(1/3)t^3 = (1/6)t^3, not (1/3)t^3...

3. Hi Prove it, you are very right. I think I first distributed the 2, but I forgot about 2*2 cos(t).

4. It's been a while since I did Calc II, but I do believe that at the end you have Integral (1/2 t ^2 - cos(t) dt)

Someone already pointed out that the inegral of 1/2 t^2 is 1/6 t ^3, but on the second part you have

-2cos(t)*2dt

The integral should be 4sin(t)

5. Hi Guys. Thanks for the help.
I re-edited the post.
Again, I distributed the 2 dt (so the integrand reduces to t^2 - 2cos(t) dt)

Previously the integrand has a typo. @ naikn05, it was my fault for ahving a typo. It was suppose to be -cos(t) * 2dt

6. Dude, this problem is ridiculous.