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    So the rule is, ∫f'g dx=f.g - ∫fg' dx
    And this was an example question written on the board: ∫sinxcosx dx
    =-cosxcosx-∫-cosx(-sinx) dx
    =-cosxcosx-∫cosxsinx dx
    Now, I'm confused about the next step:
    2∫sinxcosx dx = -cos^2 x + c
    Where did the 2 before the integral sign come from?
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    (Original post by Airess3)
    So the rule is, ∫f'g dx=f.g - ∫fg' dx
    And this was an example question written on the board: ∫sinxcosx dx
    =-cosxcosx-∫-cosx(-sinx) dx
    =-cosxcosx-∫cosxsinx dx
    Now, I'm confused about the next step:
    2∫sinxcosx dx = -cos^2 x + c
    Where did the 2 before the integral sign come from?
    You have:

    ∫sinxcosx dx = -cosxcosx-∫sinxcosx dx

    By adding ∫sinxcosx dx to both sides, you get the 2 before the integral sign.
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    (Original post by Airess3)
    And this was an example question written on the board: ∫sinxcosx dx
    This is a poor example.

    You would be better advised to spot \sin x \cos x = \frac{\sin 2x}{2} or make a substitution u=\sin x or u=\cos x.
 
 
 
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Updated: January 28, 2015
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