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Math Help - Im struggling with integrals

  1. #1
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    Im struggling with integrals

    Integral 2x^2 cos (1/2 x^3) dx

    I dont understand!! can any body help?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mdo112 View Post
    Integral 2x^2 cos (1/2 x^3) dx

    I dont understand!! can any body help?
    \int{2x^2\cos\left(\frac{1}{2}x^3\right)dx}

    We will use a method called u-substitution

    If we let u=x^3

    we would then have that

    du=3x^2dx\Rightarrow\frac{du}{3}=x^2dx
    So now lets replace each

    \int{2x^2\cos\left(\frac{1}{2}x^3\right)}dx\iff{2\  int\cos\left(\frac{1}{2}u\right)}\frac{du}{3}=\fra  c{2}{3}\int\cos\left(\frac{1}{2}u\right)du
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by mdo112 View Post
    Integral 2x^2 cos (1/2 x^3) dx

    I dont understand!! can any body help?
    Let u=\frac{1}{2}x^3 \implies du =\frac{3}{2}x^2dx \iff \frac{2}{3}du=x^2dx

    So we get

    2\int x^2\cos\left( \frac{1}{2}x^3\right)dx =2 \int \cos(u)\left( \frac{2}{3}du\right) =\frac{4}{3}\int \cos(u)du

    Now we can take the antiderivative to get

    \frac{4}{3}\sin(u )+C We can now sub back in what u is in terms of x to get

    \frac{4}{3}\sin\left( \frac{1}{2}x^3\right)+C
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Let u=\frac{1}{2}x^3 \implies du =\frac{3}{2}x^2dx \iff \frac{2}{3}du=x^2dx

    So we get

    2\int x^2\cos\left( \frac{1}{2}x^3\right)dx =2 \int \cos(u)\left( \frac{2}{3}du\right) =\frac{4}{3}\int \cos(u)du

    Now we can take the antiderivative to get

    \frac{4}{3}\sin(u )+C We can now sub back in what u is in terms of x to get

    \frac{4}{3}\sin\left( \frac{1}{2}x^3\right)+C
    I really love seeing integrals done in two different ways..its neat everyone has their own style
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  5. #5
    Senior Member polymerase's Avatar
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    Let u=\frac{1}{2}x^3\to du=\frac{3}{2}x^2\;dx

    \frac{4}{3} \int \cos u \;du= \frac{4}{3} \sin u+C

    Therefore, 2x^2\cos\left(\frac{1}{2}x^3\right)=\frac{4}{3}\si  n\left(\frac{1}{2}x^3\right)+C

    Darn...to slow
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