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!

A trigonometric substitution problem

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex]sin3cos2xdx



    3. The attempt at a solution

    I've successfully solved this problem by factoring out 1 sinx and changing the sin2x to (1-cos2x then assigning u=cosx and du=-sinx and so on.

    What I'm wondering is why does letting u=sin3x in the original integral not work. Then du=3cos2xdx and there is a cos2x in the original integral. Why does this fail? Please and thank you :)
     
  2. jcsd
  3. Nov 22, 2008 #2
    Well, the reason that doesn't work is that that is not the correct derivative. the derivative is 3sin(x)^2*cos(x) :) remember taking the derivative of v^3 is 3v^2*dv (where I let v=sin(x))
     
  4. Nov 24, 2008 #3
    Thanks very much. I don't know what I was thinking
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A trigonometric substitution problem
Loading...