1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration by parts question

  1. Jan 3, 2015 #1

    mrg

    User Avatar

    The problem statement, all variables and given/known data
    $$ \int x^{3}cos(x^{2})dx$$

    The attempt at a solution
    OK, so I am aware that there is a way in which to do this problem where you do a substitution (let $$u=x^{2}$$ to do a substitution before you integrate by parts), and I was able to get the answer right using this method. The thing is, I tried it a different way first, and after triple checking it, I feel like I should have gotten it right. It doesn't matter what I choose to let u and dv equal, right? I should get the right answer no matter what I choose (assuming I did things correctly, of course).
    Here's my work:

    $$u=cos(x^{2})$$
    $$du=-2xsin(x^{2})$$
    $$v=\frac{x^{4}}{4}$$
    $$dv=x^{3}dx$$

    $$\frac{x^{4}cos(x^{2})}{4}+\frac{1}{2}\int x*sin(x^{2})dx$$

    $$u_2=sin(x^{2})$$
    $$du_2=2xcos(x)$$
    $$v_2=\frac{x^{2}}{2}$$
    $$dv_2=x$$

    $$\frac{x^{4}cos(x^{2})}{4}+\frac{1}{2}(\frac{x^{2}sin(x^{2})}{2}-\int x^{3}cos(x^{2})dx)$$

    Since this last integral is the same as the one we started with, we can now say that
    $$I=A+B-\frac{1}{2}I$$
    Hence
    $$\frac{3}{2}I=A+B$$
    Multiplying both sides by two over three...
    $$\frac{x^{4}cos(x^{2})}{6}+\frac{x^{2}sin(x^{2})}{6}$$


    Thanks for your time and any help you can offer me.
     
  2. jcsd
  3. Jan 3, 2015 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    So, are you asking why your result isn't correct?

    ##\displaystyle v\,du = (-1/2)x^5\sin(x^2)\,dx\ \text{ not } (-1/2)x\sin(x^2)\,dx\ .##
     
  4. Jan 3, 2015 #3

    mrg

    User Avatar

    I knew it was something stupid. No idea where I got x from. Thanks Sammy.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integration by parts question
Loading...