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Calculus 2 : Trigo Integration question

  1. Sep 26, 2008 #1
    Hi, this is my first post here.

    I managed to solve the question.
    integrate sin(3x)^3 cos(3x)^5 dx = -cos(3x)^6/2 + 3cos(3x)^8/8 + c

    That is the answer that I get when I differentiate somewhere in the equation, u = cos 3x,
    du/-3 = sin(3x) dx.

    My question is, why do I get 2 different answers when I do this?
    integrate sin(3x)^3 cos(3x)^5 dx =
    integrate cos(3x)^5 sin (3x) (1-cos x^2) dx (I get the answer with this method, see above)
    but when I
    integrate sin(3x)^3 (1-sin (3x)^2)^2 cos(3x) dx I get the wrong answer, u = sin (3x).

    P.S: sorry if my question seems abit hard to understand
  2. jcsd
  3. Sep 26, 2008 #2
    You decide u(x) as a substition, it is not an answer it merely expresses the integralin a form you may be more used to solving. Continue with you second part and perform the integration with both substitutions, u(x)=cos(3x), v(x)=sin(3x); you will get the same answer in both cases
  4. Sep 26, 2008 #3


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    Your first answer is a little off, are you sure you didn't accidentally multiply by -3 instead of dividing by -3?

    What do you get for your second answer?
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