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Thread: Does wolfram work?

  1. #1
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    Does wolfram work?

    Hi guys,

    In the attachment you can see me trying to find the indefinite integral of a standard function:

    My log book gives:

    $\int \dfrac{dx}{\sqrt(a^2 - x^2)} = sin^{-1} \frac{x}{a} $

    I guess I am doing something wrong in wolfram. Can anyone explain this?
    Attached Thumbnails Attached Thumbnails Does wolfram work?-forum_standard_int.jpeg  
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  2. #2
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    Re: Does wolfram work?

    Didn't you consider the possibility that they are both correct? And these are just different ways of writing the same thing?

    If $\displaystyle \theta= sin^{-1}\left(\frac{x}{a}\right)$ then $\displaystyle sin(\theta)= \frac{x}{a}$. I presume you know that $\displaystyle sin^2(\theta)+ cos^2(\theta)= 1$ so that $\displaystyle cos(\theta)= \sqrt{1- sin^2(\theta)}= \sqrt{1- \frac{x^2}{a^2}}= \frac{\sqrt{a^2- x^2}}{a}$.

    Then $\displaystyle tan(\theta)= \frac{sin(\theta)}{cos(\theta)}= \frac{x}{\sqrt{a^2- x^2}}$.
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  3. #3
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    Re: Does wolfram work?

    You will also find with integration that answers can look very different, but in reality differ by only a constant. I do not have an example off the top of my head, but I remember running into examples twenty years ago when I studied integration theory.
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  4. #4
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    Re: Does wolfram work?

    The simplest example that I know of is
    \begin{align*}
    \int 2+2x \, \mathrm dx &= \int 2u \, \mathrm du &\text{where $u=1+x$} \\
    2x + x^2 + c_x &= u^2 + c_u \\
    &= (1+x)^2 + c_u
    \end{align*}
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  5. #5
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    Re: Does wolfram work?

    Here's a harder one to see:

    $\displaystyle \int \dfrac{1}{\sqrt{1+x^2}}dx=\text{sinh}^{-1} x+c_1 = \ln\left| x+\sqrt{1+x^2}\right|+c_2 $

    These actually have the same constant of integration. Nm.
    Last edited by SlipEternal; May 23rd 2018 at 07:01 PM.
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  6. #6
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    Re: Does wolfram work?

    The "standard" trigonometric example is $$\int 4\sin{(x)}\cos{(x)} \, \mathrm dx$$
    which we can solve by trigonometric manipulation:
    \begin{align*}
    \int 4\sin{(x)}\cos{(x)} \, \mathrm dx &= \int 2\sin{(2x)} \, \mathrm dx \\
    &= -\cos{(2x)} + c_1
    \end{align*}
    Or by the substitution $u=\sin{(x)}$:
    \begin{align*}
    \int 4\sin{(x)}\cos{(x)} \, \mathrm dx &= \int 4u \, \mathrm du \\
    &= 2u^2 + c_2 \\
    &= 2\sin^2{(x)} + c_2
    \end{align*}
    Or by the substitution $u=\cos{(x)}$:
    \begin{align*}
    \int 4\sin{(x)}\cos{(x)} \, \mathrm dx &= \int -4u \, \mathrm du \\
    &= -2u^2 + c_3 \\
    &= -2\cos^2{(x)} + c_3
    \end{align*}
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  7. #7
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    Re: Does wolfram work?

    Thanks for all the examples guys. I get the picture now!
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