Thread: 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)



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!

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

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

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)

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

Dude, this problem is ridiculous.