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Math Help - Integration by Substitution

  1. #1
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    Integration by Substitution

    ∫5x 3√(1-x2)dx

    So I'm still a beginner with integrating by substituting.
    What I know is that you use 'u' from one of the thing in the equation to substitute.

    Though, I'm not entirely sure where 'u' should be coming from.
    So I did u = (1-x2)
    And since it has a cube root, I did this:
    1/3 du = 3√(1-x2)dx
    So I got: ∫1/3 u1/3 du
    4/3 u4/3 + C -> 4/3 (1-x2)3 + C
    But I'm not sure if this is entirely correct after:

    Someone else showed me how to do the steps, which was this:
    ∫5x 3√(1-x2)dx = -5/2∫(1-x2)1/3(-2x)dx = -5/2 * ((1-x2)4/3)/(4/3) + C = (-15/8) * (1-x2)4/3 + C
    This was the correct answer in the book.
    But I'm a bit confused on how the person got the -5/2, and the (-2x)
    Can someone explain to me that part?
    Thanks!
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  2. #2
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    Re: Integration by Substitution

    Integration by Substitution-integral-substi.png
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  3. #3
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    Re: Integration by Substitution

    Quote Originally Posted by Chaim View Post
    ∫5x 3√(1-x2)dx

    So I'm still a beginner with integrating by substituting.
    What I know is that you use 'u' from one of the thing in the equation to substitute.

    Though, I'm not entirely sure where 'u' should be coming from.
    So I did u = (1-x2)
    And since it has a cube root, I did this:
    1/3 du = 3√(1-x2)dx
    So I got: ∫1/3 u1/3 du
    4/3 u4/3 + C -> 4/3 (1-x2)3 + C
    But I'm not sure if this is entirely correct after:

    Someone else showed me how to do the steps, which was this:
    ∫5x 3√(1-x2)dx = -5/2∫(1-x2)1/3(-2x)dx = -5/2 * ((1-x2)4/3)/(4/3) + C = (-15/8) * (1-x2)4/3 + C
    This was the correct answer in the book.
    But I'm a bit confused on how the person got the -5/2, and the (-2x)
    Can someone explain to me that part?
    Thanks!
    The better substitution to use here is \displaystyle \begin{align*} x = \sin{(\theta)} \implies dx = \cos{(\theta)}\,d\theta \end{align*}. Then the integral becomes

    \displaystyle \begin{align*} \int{\frac{5x^3}{\sqrt{1 - x^2}}\,dx} &= \int{\frac{5\sin^3{(\theta)}}{\sqrt{1 - \sin^2{(\theta)}}}\, \cos{(\theta)}\,d\theta} \\ &= \int{\frac{5\sin^3{(\theta)}\cos{(\theta)}}{\sqrt{  \cos^2{(\theta)}}}\,d\theta} \\ &= \int{\frac{5\sin^3{(\theta)}\cos{(\theta)}}{\cos{(  \theta)}}\,d\theta} \\ &= \int{5\sin^3{(\theta)}\,d\theta} \\ &= \int{5\sin{(\theta)}\sin^2{(\theta)}\,d\theta} \\ &= -5\int{-\sin{(\theta)}\left[ 1 - \cos^2{(\theta)} \right] d\theta} \end{align*}

    Now make the substitution \displaystyle \begin{align*} u = \cos{(\theta)} \implies du = -\sin{(\theta)}\,d\theta \end{align*} and the integral becomes

    \displaystyle \begin{align*} -5\int{-\sin{(\theta)}\left[ 1 - \cos^2{(\theta)} \right] d\theta} &= -5\int{ 1 - u^2 \,du} \\ &= -5 \left( u - \frac{u^3}{3} \right) + C \\ &= -5 \left[ \cos{(\theta)} - \frac{\cos^3{(\theta)}}{3} \right] + C \end{align*}

    And remember that we originally made the substitution \displaystyle \begin{align*} x = \sin{(\theta)} \end{align*}, that means

    \displaystyle \begin{align*} x^2 &= \sin^2{(\theta)} \\ 1 - x^2 &= 1 - \sin^2{(\theta)} \\ 1 - x^2 &= \cos^2{(\theta)} \\ \sqrt{1 - x^2} &= \cos{(\theta)} \end{align*}

    which means

    \displaystyle \begin{align*} -5 \left[ \cos{(\theta)} - \frac{\cos^3{(\theta)}}{3} \right] +C &= -5 \left[ \sqrt{1 - x^2} - \frac{\left( \sqrt{1 -x^2} \right) ^3 }{3} \right] + C \end{align*}

    So finally we have \displaystyle \begin{align*} \int{ \frac{5x^3}{\sqrt{1 - x^2}}\,dx } = -5 \left[ \sqrt{1 - x^2} - \frac{\left( \sqrt{1 - x^2} \right) ^3}{3} \right] + C \end{align*}
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  4. #4
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    Re: Integration by Substitution

    Hello, Chaim!

    \int 5x\sqrt[3]{1-x^2}\,dx

    We have: . 5\int(1-x^2)^{\frac{1}{3}}(x\,dx)

    Let u \,=\,1-x^2 \quad\Rightarrow\quad du \,=\,\text{-}2x\,dx \quad\Rightarrow\quad x\,dx \,=\,\text{-}\tfrac{1}{2}du

    Substitute: . 5\int u^{\frac{1}{3}}(\text{-}\tfrac{1}{2}\,du) \;=\; \text{-}\tfrac{5}{2}\int u^{\frac{1}{3}}\,du \;=\; \text{-}\tfrac{15}{8}u^{\frac{4}{3}}+C

    Back-substitute: . \text{-}\tfrac{15}{8}(1-x^2)^{\frac{4}{3}} + C
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  5. #5
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    Re: Integration by Substitution

    Thanks for responding guys!
    I understand how u and du is coming from now.

    I'm a bit confused on one thing.
    The xdx

    Just wondering, how come xdx = -1/2?
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  6. #6
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    Re: Integration by Substitution

    It isn't and Soroban didn't say it was. From u= 1- x^2, we get \frac{du}{dx}= -2x so that du= -2x dx. Divide both sides by -2 to get -\frac{1}{2}du= xdx.
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