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  1. #1
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    integration

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

    i did the integration by parts first:

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

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

    need help from this point
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  2. #2
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    First, set $\displaystyle \mu=5x$, 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...
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  5. #5
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    Hello, viet!

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

    i did the integration by parts first:

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

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

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

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

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

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

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


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

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

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


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

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

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

    The integral becomes to

    $\displaystyle \frac{1}
    {25}\int {\theta \arcsin \theta~d\theta }$

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


    $\displaystyle \int {x\arcsin 5x~dx} = \frac{1}
    {2}x^2 \arcsin 5x - \frac{1}
    {{100}}\arcsin 5x + \frac{1}
    {{20}}x\sqrt {1 - 25x^2 } + c$

    That's my contribution
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