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Math Help - Somemore integration help on trigo needed

  1. #1
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    Somemore integration help on trigo needed

    1.
    \int \frac {x\arcsin (2x)}{\sqrt{1-4x^2}} dx
    i tried using substitution by letting u= \arcsin (2x) but got stucked.

    2.
    Evaluate \int \sqrt{x} \arctan\sqrt{x}   dx
    totally no idea on how to start this.

    thanks in advance.
    Last edited by tempq1; December 17th 2012 at 01:46 AM.
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  2. #2
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    Re: Somemore integration help on trigo needed

    Hey tempq1.

    Try using a substitution u = 2x instead of arcsin(2x) and then see what the form of the integral is (and after that consider integrating by parts to get a cancellation).
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  3. #3
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    Re: Somemore integration help on trigo needed

    Quote Originally Posted by chiro View Post
    Hey tempq1.

    Try using a substitution u = 2x instead of arcsin(2x) and then see what the form of the integral is (and after that consider integrating by parts to get a cancellation).
    sry, but i don't get you. How would letting u=2x solve the equation?
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  4. #4
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    Re: Somemore integration help on trigo needed

    It certainly wouldn't hurt to substitute u=2x (I'll use w=2x):

    \int \frac {x\arcsin (2x)}{\sqrt{1-4x^2}}\ dx = \frac{1}{4} \int \frac {w\arcsin (w)}{\sqrt{1-w^2}}\ dw

    Now try integrating by parts using u = \arcsin(w) and dv = \frac {w}{\sqrt{1-w^2}}\ dw.

    - Hollywood
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  5. #5
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    Re: Somemore integration help on trigo needed

    Quote Originally Posted by hollywood View Post
    It certainly wouldn't hurt to substitute u=2x (I'll use w=2x):

    \int \frac {x\arcsin (2x)}{\sqrt{1-4x^2}}\ dx = \frac{1}{4} \int \frac {w\arcsin (w)}{\sqrt{1-w^2}}\ dw

    Now try integrating by parts using u = \arcsin(w) and dv = \frac {w}{\sqrt{1-w^2}}\ dw.

    - Hollywood
    thanks. Anyone knows Q2?
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  6. #6
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    Re: Somemore integration help on trigo needed

    Substitute w=sqrt(x) to get rid of the radicals, then integrate by parts with u=arctan(w) and dv=w^2\ dw.

    - Hollywood
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  7. #7
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    Re: Somemore integration help on trigo needed

    Quote Originally Posted by tempq1 View Post
    thanks. Anyone knows Q2?
    We know that:

    \int u \, dv=uv-\int v \, du.\!

    So for \int \sqrt{x} \tan^{-1}(\sqrt{x})\;\;dx

    \text{Let } u = \tan^{-1}(\sqrt{x}) \therefore du = \frac{1}{2\sqrt{x}(1+x)}\;\; dx

    \text{Let } dv = \sqrt{x} \therefore v = \frac{2x^{\frac{3}{2}}}{3}

    So now:

    \begin{align*}\int \sqrt{x} \tan^{-1}(\sqrt{x})\;\;dx =& \frac{2x^{\frac{3}{2}}}{3}\tan^{-1}(\sqrt{x}) - \int \left(\frac{2x^{\frac{3}{2}}}{3}\right)\left(\frac  {1}{2\sqrt{x}(1+x)}\right) dx\\ =&  \frac{2x^{\frac{3}{2}}}{3} \tan^{-1}(\sqrt{x}) -\int \frac{x}{3(1+x)} dx \end{align*}

    For:
    \int \frac{x}{3(1+x)} dx

    Let u = 1 + x \text{ so } x = u - 1 \text{ and } du = dx \text{ and } x = u -1


    \begin{align*}-\int \frac{x}{3(1+x)} dx =& -\int \frac{(u-1)}{3u} du\\ =& -\int (\frac{1}{3} -\frac{1}{3u}) du\\ =& -\frac{u}{3} + \frac{\ln{(u)}}{3} + C \\ =& -\frac{1+x}{3} +\frac{\ln{(1+x)}}{3} + C\end{align*}


    \therefore \int (\sqrt{x} \tan^{-1}(\sqrt{x}))\;\; dx = \frac{2x^{\frac{3}{2}}}{3} \tan^{-1}(\sqrt{x}) + \frac{\ln{(1+x)}}{3} - \frac{1+x}{3}  + C


    Check:

    http://www.wolframalpha.com/input/?i=differentiation+2*x^(3%2F2)*tan^(-1)(x^(1%2F2))%2F3+%2B+ln(1%2Bx)%2F3-(1%2Bx)%2F3+%2B+C
    Last edited by x3bnm; December 21st 2012 at 04:34 PM.
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