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Thread: Two easy integrals

  1. #1
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    Two easy integrals

    Integrate

    $\displaystyle f(x)=\sqrt{1-x^2}$ & $\displaystyle h(x)=\sqrt{x^2-1}$
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  2. #2
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    Quote Originally Posted by liyi View Post
    Integrate

    $\displaystyle f(x)=\sqrt{1-x^2}$ & $\displaystyle h(x)=\sqrt{x^2-1}$
    Do the substitutions: x=sint for the first one and x=sect for the second.
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  3. #3
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    I'm not sure what the user above means but I'd;

    Rearrange
    $\displaystyle f'(x)=\sqrt{1-x^2}$
    $\displaystyle f'(x)=\sqrt{1} - \sqrt{x^2}$
    $\displaystyle f'(x)= 1^{1/2} - x$
    Intergrate; add one to power, divide by new power and add c:
    $\displaystyle f(x)= 2/3^{3/2} - 1 + c $
    I don't think that can be simplified
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by markcantdomaths View Post
    I'm not sure what the user above means but I'd;

    Rearrange
    $\displaystyle f'(x)=\sqrt{1-x^2}$
    $\displaystyle f'(x)=\sqrt{1} - \sqrt{x^2}$
    In general $\displaystyle \sqrt{a^2 + b^2} \neq a + b$

    curvature's substitutions should work nicely.

    -Dan
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  5. #5
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    Quote Originally Posted by markcantdomaths View Post
    Rearrange
    $\displaystyle f'(x)=\sqrt{1-x^2}$
    $\displaystyle f'(x)=\sqrt{1} - \sqrt{x^2}$
    No the second equality is wrong.
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  6. #6
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    Quote Originally Posted by liyi View Post
    Integrate

    $\displaystyle f(x)=\sqrt{1-x^2}$
    Define the function $\displaystyle u(x) = x\sqrt {1 - x^2 } \implies u'(x) = \sqrt {1 - x^2 } - \frac{{x^2 }}
    {{\sqrt {1 - x^2 } }} = 2\sqrt {1 - x^2 } - \frac{1}
    {{\sqrt {1 - x^2 } }}.$

    Make-up: $\displaystyle \sqrt {1 - x^2 } = \frac{1}
    {2}\left[ {u'(x) + \frac{1}
    {{\sqrt {1 - x^2 } }}} \right].$

    Integrate $\displaystyle \int {\sqrt {1 - x^2 } \,dx} = \frac{1}
    {2}\left( {x\sqrt {1 - x^2 } + \arcsin x} \right) + k.$

    Quote Originally Posted by liyi View Post
    Integrate

    $\displaystyle h(x)=\sqrt{x^2-1}$
    Consider the function $\displaystyle \varphi (x) = x\sqrt {x^2 - 1} \implies \varphi '(x) = \sqrt {x^2 - 1} + \frac{{x^2 }}
    {{\sqrt {x^2 - 1} }} = 2\sqrt {x^2 - 1} + \frac{1}
    {{\sqrt {x^2 - 1} }}.$

    Make-up & integrate

    $\displaystyle \int {\sqrt {x^2 - 1} \,dx} = \frac{1}
    {2}\left( {x\sqrt {x^2 - 1} - \int {\frac{1}
    {{\sqrt {x^2 - 1} }}\,dx} } \right)+k_1.$

    For the remaining integral, substitute $\displaystyle u=x+\sqrt{x^2-1},$

    $\displaystyle \int {\frac{1}
    {{\sqrt {x^2 - 1} }}\,dx} = \int {\frac{1}
    {u}\,du} = \ln \left| {x + \sqrt {x^2 - 1} } \right| + k_2.$

    And we happily get $\displaystyle \int {\sqrt {x^2 - 1} \,dx} = \frac{1}
    {2}\left( {x\sqrt {x^2 - 1} - \ln \left| {x + \sqrt {x^2 - 1} } \right|} \right) + k.$
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  7. #7
    Super Member angel.white's Avatar
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    Quote Originally Posted by Krizalid View Post
    Define the function $\displaystyle u(x) = x\sqrt {1 - x^2 } \implies u'(x) = \sqrt {1 - x^2 } - \frac{{x^2 }}
    {{\sqrt {1 - x^2 } }} = 2\sqrt {1 - x^2 } - \frac{1}
    {{\sqrt {1 - x^2 } }}.$

    Make-up: $\displaystyle \sqrt {1 - x^2 } = \frac{1}
    {2}\left[ {u'(x) + \frac{1}
    {{\sqrt {1 - x^2 } }}} \right].$

    Integrate $\displaystyle \int {\sqrt {1 - x^2 } \,dx} = \frac{1}
    {2}\left( {x\sqrt {1 - x^2 } + \arcsin x} \right) + k.$


    Consider the function $\displaystyle \varphi (x) = x\sqrt {x^2 - 1} \implies \varphi '(x) = \sqrt {x^2 - 1} + \frac{{x^2 }}
    {{\sqrt {x^2 - 1} }} = 2\sqrt {x^2 - 1} + \frac{1}
    {{\sqrt {x^2 - 1} }}.$

    Make-up & integrate

    $\displaystyle \int {\sqrt {x^2 - 1} \,dx} = \frac{1}
    {2}\left( {x\sqrt {x^2 - 1} - \int {\frac{1}
    {{\sqrt {x^2 - 1} }}\,dx} } \right)+k_1.$

    For the remaining integral, substitute $\displaystyle u=x+\sqrt{x^2-1},$

    $\displaystyle \int {\frac{1}
    {{\sqrt {x^2 - 1} }}\,dx} = \int {\frac{1}
    {u}\,du} = \ln \left| {x + \sqrt {x^2 - 1} } \right| + k_2.$

    And we happily get $\displaystyle \int {\sqrt {x^2 - 1} \,dx} = \frac{1}
    {2}\left( {x\sqrt {x^2 - 1} - \ln \left| {x + \sqrt {x^2 - 1} } \right|} \right) + k.$
    I understood everything except this step here:
    $\displaystyle \sqrt {1 - x^2 } - \frac{{x^2 }}
    {{\sqrt {1 - x^2 } }} = 2\sqrt {1 - x^2 } - \frac{1}
    {{\sqrt {1 - x^2 } }}.$

    Can you please explain it.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post
    I understood everything except this step here:
    $\displaystyle \sqrt {1 - x^2 } - \frac{{x^2 }}
    {{\sqrt {1 - x^2 } }} = 2\sqrt {1 - x^2 } - \frac{1}
    {{\sqrt {1 - x^2 } }}.$

    Can you please explain it.
    as in explain why it is true?

    $\displaystyle \sqrt{1 - x^2} - \frac {x^2}{\sqrt {1 - x^2}} = \sqrt{1 - x^2} - \frac {x^2 + 1 - 1}{\sqrt {1 - x^2}}$

    $\displaystyle = \sqrt{1 - x^2} + \frac {1 - x^2}{\sqrt {1 - x^2}} - \frac 1{\sqrt {1 - x^2}}$

    $\displaystyle = \sqrt{1 - x^2} + \sqrt{1 - x^2} - \frac 1{\sqrt {1 - x^2}}$

    $\displaystyle = 2 \sqrt{1 - x^2} - \frac 1{\sqrt {1 - x^2}}$
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  9. #9
    Super Member angel.white's Avatar
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    I see now, that's a smart way of doing business.
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