Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Intergration by parts for sin(x)cos(x)

  1. May 30, 2014 #1
    I know its easier to use the substitution method, by I'm trying to see how it'll work for integration by parts. I follow the LIATE method for integration by parts.

    attachment.php?attachmentid=70182&stc=1&d=1401500731.jpg

    Now if I take u=cos(x) and dv = sin(x), the answer changes.
    attachment.php?attachmentid=70184&stc=1&d=1401501227.jpg


    Can you please explain this to me? Which is the 'right' answer?
     

    Attached Files:

  2. jcsd
  3. May 30, 2014 #2
    Rather than answer your question directly, let me provide another example of the same phenomenon that might make it easier to identify why both answers are correct.

    If you compute ##\int (x+1)\ dx## using the sum rule for antiderivatives and the "reverse power rule", you get $$\int x+1\ dx=\frac{1}{2}x^2+x+C.$$ If instead you use a ##u##-sub, with ##u=x+1##, you get $$\int (x+1)\ dx=\int u\ du=\frac{1}{2}u^2+C=\frac{1}{2}(x+1)^2+C.$$

    Now the exact same thing is happening in your example as is happening here; you have two different-looking answers that are both correct. How can that be?
     
  4. May 30, 2014 #3

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    And closer to home, let's do ##\int \sin(x) \cos(x) dx## a third way, using the trig substitution ##\sin(2x)=2\sin(x) \cos(x)##. Thus ##\int \sin(x) \cos(x) dx = \int \frac {\sin(2x)}2 dx = -\frac {\cos(2x)} 4 + C##.

    To quote gopher_p,
     
  5. May 30, 2014 #4
    dear friend , all these answers are different only by a constant, so they are the same answers.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Intergration by parts for sin(x)cos(x)
  1. X = (cos@)/(sin@) (Replies: 9)

  2. Cos ax/sin pi*x (Replies: 9)

Loading...