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Math Help - integration

  1. #1
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    integration

     \int_0^{1/5} x\sin^{-1}(5 x) dx =

    i did the integration by parts first:

    u = arcsin(5x), dv = x,
    du = \frac{5}{\sqrt{1-25x^2}} dx, v = \frac{1}{2}x^2,

    uv-\int{vdu} = arcsin(5x)\frac{1}{2}x^2-\int_0^{1/5}\frac{1}{2}x^2(\frac{5}{\sqrt{1-25x^2}})dx

    need help from this point
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  2. #2
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    First, set \mu=5x, then the integration by parts process won't be so tedious.
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  3. #3
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    After this, a trig. substitution would work with the remaining integral.
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  4. #4
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    Re:

    RE:

    I haven't worked much with trig sub. At this time I would probably go for a table/manual...
    Attached Thumbnails Attached Thumbnails integration-31.jpg  
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  5. #5
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    Hello, viet!

     \int_0^{\frac{1}{5}} x\sin^{-1}(5x)\,dx

    i did the integration by parts first:

    \begin{array}{ccccccc}u &= & \arcsin(5x) & \quad & dv & = & x \\ <br />
du &= &\frac{5}{\sqrt{1-25x^2}} dx& \quad & v & = & \frac{1}{2}x^2<br />
\end{array}

    uv-\int{v\,du} \;= \;\arcsin(5x)\frac{1}{2}x^2 \,-\,\int_0^{\frac{1}{5}}\frac{1}{2}x^2\left(\frac{5}  {\sqrt{1-25x^2}}\right)dx . Good!

    We have: .  \frac{1}{2}x^2\arcsin(5x) - \frac{5}{2}\int\frac{x^2}{\sqrt{1 - 25x^2}}\,dx

    . . Let: 5x = \sin\theta\quad\Rightarrow\quad x = \frac{1}{5}\sin\theta\quad\Rightarrow\quad dx = \frac{1}{5}\cos\theta\,d\theta

    . . The integral becomes: . \int\frac{\frac{1}{25}\sin^2\!\theta}{\cos\theta}\  left(\frac{1}{5}\cos\theta\,d\theta\right) \;=\;\frac{1}{125}\int\sin^2\!\theta\,d\theta

    . . . . =\;\frac{1}{250}\int (1 - \cos2\theta)d\theta\;= \;\frac{1}{250}\left(\theta - \frac{1}{2}\sin2\theta\right) \;=\;\frac{1}{250}(\theta - \sin\theta\cos\theta)

    . . Back-substitute: . \frac{1}{250}\left[\arcsin(5x) - 5x\sqrt{1 - 25x^2}\right]


    We have: . \frac{1}{2}x^2\arcsin(5x) - \frac{5}{2}\left[\frac{1}{250}\left(\arcsin(5x) - 5x\sqrt{1-25x^2}\right)\right]

    . . = \;\frac{1}{2}x^2\arcsin(5x) - \frac{1}{250}\arcsin(5x) + \frac{1}{20}x\sqrt{1-25x^2}

    . . =\;\frac{1}{500}\left[250x^2\arcsin(5x) - 2\arcsin(5x) + 25x\sqrt{1-25x^2}\right]^{\frac{1}{5}}_0


    Now evaluate . . .
    . . \left(\text{I got: }\frac{\pi}{125}\right)

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  6. #6
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    \int {x\arcsin 5x~dx}

    Let's set \theta=5x\implies{d\theta=5~dx}

    The integral becomes to

    \frac{1}<br />
{25}\int {\theta \arcsin \theta~d\theta }

    Now let u=\arcsin\theta\implies{du=\frac1{\sqrt{1-\theta^2}}~d\theta} and dv=\theta~d\theta\implies{v=\frac12\theta^2}, this yields


    \int {x\arcsin 5x~dx} = \frac{1}<br />
{2}x^2 \arcsin 5x - \frac{1}<br />
{{100}}\arcsin 5x + \frac{1}<br />
{{20}}x\sqrt {1 - 25x^2 } + c

    That's my contribution
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