A trigonometric substitution problem

  • Thread starter celeramo
  • Start date
  • #1
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Homework Statement



[tex]\int[/tex]sin3cos2xdx



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 :)
 

Answers and Replies

  • #2
11
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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))
 
  • #3
9
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Thanks very much. I don't know what I was thinking
 

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