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Math Help - Integration by Substitution

  1. #1
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    Integration by Substitution

    <br />
\int 5\cos(5x) dx<br />

    <br />
g'(x) = 5, f(g(x)) = cos5x<br />

    <br />
\int (cos5x)(5) dx<br />

    <br />
(1/5) \sin5x + C<br />

    Those are my steps above, but the book says the answer is sin5x. It seems like it got (1/5)*sin5x*5 after integrating and then canceled the 5 and 1/5, but I thought that \int f(g(x))g'(x) = F(g(x)) + C , which to me means that you drop the g'(x) and find the antiderivative of f(g(x)). My other thought is that I just find the antiderivative of f(u), which would be cosu in this case, which yields sinu + C. Am I correct?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by tom ato View Post
    <br />
\int 5\cos(5x) dx<br />

    <br />
g'(x) = 5, f(g(x)) = cos5x<br />

    <br />
\int (cos5x)(5) dx<br />

    <br />
(1/5) \sin5x + C<br />

    Those are my steps above, but the book says the answer is sin5x. It seems like it got (1/5)*sin5x*5 after integrating and then canceled the 5 and 1/5, but I thought that \int f(g(x))g'(x) = F(g(x)) + C , which to me means that you drop the g'(x) and find the antiderivative of f(g(x)). My other thought is that I just find the antiderivative of f(u), which would be cosu in this case, which yields sinu + C. Am I correct?
    \int (cos5x)(5) dx

    = (1/5) \sin5x(5) + C

    = sin 5x+C
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  3. #3
    Super Member Deadstar's Avatar
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    You can take constants outside the integration sign.

    5 \int \cos(5x) dx = 5 \bigg{(}\sin(5x) \cdot \frac{1}{5} \bigg{)} + C

    (note that you don't have to multiply the constant by 5 because the result will still be a constant.)

    Then cancel the 5 and 1/5
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  4. #4
    Super Member Deadstar's Avatar
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    Just noticed the mistake in your answer, you only integrate f(x) which is cos(x)

    Then you 'sub' in x -> g(x). So you get... sin(x) -> sin(5x)

    I think the proper way to write it is...

    \int f(g(t))g'(t) dt = \int f(x) dx = F(x) + c

    where x = g(t)
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