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Thread: Integral. Substituting makes it more complex

  1. #1
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    Integral. Substituting makes it more complex

    $\displaystyle \int x \arccos{\sqrt{x}}dx$
    $\displaystyle \arccos{\sqrt{x}}=u$
    $\displaystyle -\frac{1}{2\sqrt{x-x^2}}dx=du$
    $\displaystyle \int udv=uv-\int vdu$
    then
    $\displaystyle \frac{\arccos{\sqrt{x}}x^2}{2}+\frac{1}{4}\int\fra c{{x^2}}{{\sqrt{x-x^2}}}dx$
    Last edited by totalnewbie; Jan 3rd 2007 at 12:48 PM.
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  2. #2
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    Quote Originally Posted by totalnewbie View Post
    $\displaystyle \int x \arccos{\sqrt{x}}dx$
    $\displaystyle \arccos{\sqrt{x}}=u$
    $\displaystyle -\frac{1}{2\sqrt{x-x^2}}dx=du$
    $\displaystyle \int udv=uv-\int vdu$
    then
    $\displaystyle \frac{\arccos{\sqrt{x}}x^2}{2}+\frac{1}{4}\int\fra c{{x^2}}{{\sqrt{x-x^2}}}dx$
    I am note sure what you are doing.
    Given the integral,
    $\displaystyle \int x\cos^{-1} \sqrt{x} dx$
    You can do it by parts,
    $\displaystyle u=x$ and $\displaystyle v'=\cos^{-1} \sqrt{x}$
    Thus,
    $\displaystyle u'=1$ and $\displaystyle v=\int \cos^{-1} \sqrt{x} dx$

    But the question is,
    What is,
    $\displaystyle \int \cos^{-1} \sqrt{x} dx$
    You can write,
    $\displaystyle 2\int \sqrt{x} \cos^{-1} \sqrt{x} \cdot \frac{1}{2\sqrt{x}} dx$
    Let,
    $\displaystyle u=\sqrt{x}$ then, $\displaystyle u'=\frac{1}{2\sqrt{x}}$
    Thus,
    $\displaystyle 2\int u\cos^{-1} u du$
    You can let,
    $\displaystyle a=u$ and $\displaystyle b'=\cos^{-1}u$
    Thus,
    $\displaystyle a'=1$ and $\displaystyle b=\int \cos^{-1} u du$
    Then, we need to find what,
    $\displaystyle \int \cos^{-1}udu$ is?
    To do that we can write,
    $\displaystyle \int (1)\cos^{-1} udu$
    Let,
    $\displaystyle c'=1$ and $\displaystyle d=\cos^{-1} u$
    Then,
    $\displaystyle c=u$ and $\displaystyle d'=\frac{1}{\sqrt{1-x^2}}$.
    Thus, by parts,
    $\displaystyle u\cos^{-1} u-\int \frac{u}{\sqrt{1-u^2}} du$
    But then what is?
    $\displaystyle \int \frac{u}{\sqrt{1-u^2}} du$
    You can write,
    $\displaystyle -\frac{1}{2}\int \frac{-2u}{\sqrt{1-u^2}} du$
    Let, $\displaystyle t=1-u^2$ then $\displaystyle t'=-2u$.
    Thus,
    $\displaystyle -\frac{1}{2} \int \frac{1}{\sqrt{t}}dt$
    This is certainly doable!

    The details, and the actual integral I leave to you do complete. (Maybe, there is an easier way, but this is one of the longest I seen).
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  3. #3
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    Hello, totalnewbie!

    Make a substitution first . . . then do it by parts.


    $\displaystyle \int x\cdot\arccos(\sqrt{x})\,dx$

    Let $\displaystyle \arccos(\sqrt{x}) \:=\:y\quad\Rightarrow\quad \sqrt{x} \:=\:\cos y\quad\Rightarrow\quad x \:=\:$ $\displaystyle \cos^2\!y\quad\Rightarrow\quad dx \:=\:-2\cos y\!\cdot\!\sin y\!\cdot\!dy$

    Substitute: .$\displaystyle \int\left(\cos^2\!y\right)(y)(-2\cos y\!\cdot\!\sin y\!\cdot\!dy) \;=\;-2\int y\cdot\cos^3\!y\!\cdot\!\sin y\!\cdot\!dy$

    By parts: .$\displaystyle \begin{array}{cc} u \\ du\end{array}\begin{array}{cc}= \\ = \end{array}\begin{array}{cc} y \\ dy\end{array}\quad \begin{array}{cc}dv\\ v\end{array} \begin{array}{cc}=\\=\end{array} \begin{array}{cc}\cos^3\!y\!\cdot\!\sin y\!\cdot\!dy \\ -\frac{1}{4}\cos^4\!y\end{array}$

    And we have: .$\displaystyle -\frac{1}{4}y\cos^4\!y \;+ \;\frac{1}{4}\!\int\cos^4\!y\,dy$

    Can you finish it now?

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