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Math Help - Inverse Trig Func: Integration

  1. #1
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    Inverse Trig Func: Integration

    Hi, I’m a bit confused on how to solve a problem like this.
    My friend helped me a bit, but didn’t explain very clearly, so I was wondering if you guys could help me understand this.
    01/6 1/√(1-9x2)dx
    So I was told that u=3x, and du=3
    So it would become 01/6 1/√(1-(3x)2dx
    Now for u-substitution.
    1/√(1-u2)du

    I get [arcsin(3x)]1/60
    Though, one thing I missed is the 1/3 in front of the integration, and my friend showed me this:
    (1/3)01/6 (1/√(1-(3x)2) (3)dx

    So yeah… there’s a 3 outside…
    And there’s a 1/3 in front of the integration.
    Can someone explain to me why this was done?
    Also, can someone explained why u=3x, instead of something like u=9x?

    Thanks J
    Hi, I’m a bit confused on how to solve a problem like this.
    My friend helped me a bit, but didn’t explain very clearly, so I was wondering if you guys could help me understand this.
    01/6 1/√(1-9x2)dx
    So I was told that u=3x, and du=3
    So it would become 01/6 1/√(1-(3x)2dx
    Now for u-substitution.
    1/√(1-u2)du

    I get [arcsin(3x)]1/60
    Though, one thing I missed is the 1/3 in front of the integration, and my friend showed me this:
    (1/3)01/6 (1/√(1-(3x)2) (3)dx

    So yeah… there’s a 3 outside…
    And there’s a 1/3 in front of the integration.
    Can someone explain to me why this was done?
    Also, can someone explained why u=3x, instead of something like u=9x?

    Thanks J
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  2. #2
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    Re: Inverse Trig Func: Integration

    Hey Chaim.

    The answer for the 3 and 1/3 is that du = 3dx.

    This means that 1/3 * 3 dx = dx = 1/3 * du and hence if you are integrating with respect to u then you need a 1/3 term to balance everything out.
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  3. #3
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    Re: Inverse Trig Func: Integration

    You have \displaystyle \begin{align*} \int_0^{\frac{1}{6}}{\frac{1}{\sqrt{1 - 9x^2}}\,dx} = \int_0^{\frac{1}{6}}{\frac{1}{\sqrt{1 - \left( 3x \right) ^2 } } \,dx} \end{align*}. The standard integral is \displaystyle \begin{align*} \int{\frac{1}{\sqrt{1 - X^2}}\,dX} = \arcsin{(X)} + C \end{align*}, so the substitution \displaystyle \begin{align*} u = 3x \end{align*} here is appropriate to turn it into a multiple of this standard integral.
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  4. #4
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    Re: Inverse Trig Func: Integration

    Oh, now I see where it is coming from!

    Thanks you guys!
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