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Thread: integration problem no.3

  1. #1
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    integration problem no.3


    i stop here because the substitution method doesn't work
    Last edited by afeasfaerw23231233; Oct 26th 2007 at 11:34 AM.
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  2. #2
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    Hello, afeasfaerw23231233!

    With a quadratic form: . $\displaystyle x^2+a^2,\;\;x^2-a^2,\;\;a^2-x^2$
    . . a Trigonometric Substitution is recommended.


    Find $\displaystyle \int\frac{\sqrt{1-x^2}}{x^4}\,dx$

    Let $\displaystyle x \:=\:\sin\theta\quad\Rightarrow\quad dx\:=\:\cos\theta\,d\theta$
    . . and: .$\displaystyle \sqrt{1-x^2} \:=\:\sqrt{1-\sin^2\!\theta} \:=\:\sqrt{\cos^2\!\theta} \:=\:\cos\theta$

    Substitute: .$\displaystyle \int\frac{\cos\theta}{\sin^4\!\theta}(\cos\theta\, d\theta) \;=\;\int\frac{\cos^2\!\theta}{\sin^2\!\theta}\cdo t\frac{1}{\sin^2\!\theta}\,d\theta \;=\;\int\cot^2\!\theta\csc^2\!\theta\,d\theta$

    Let $\displaystyle u\,=\,\cot\theta\quad\Rightarrow\quad du \,=\,-\csc^2\!\theta\,d\theta$

    Substitute: .$\displaystyle -\int u^2\,du \;=\;-\frac{1}{3}u^3 + C$

    Back-substitute: .$\displaystyle -\frac{1}{3}\cot^3\!\theta + C$ .[1]


    Back-substitute again. .We have: .$\displaystyle \sin\theta \:= \:\frac{x}{1} \:= \:\frac{opp}{hyp}$

    $\displaystyle \theta$ is in a right triangle with: $\displaystyle opp = x,\;hyp = 1$
    Using Pythagorus, we have: .$\displaystyle adj = \sqrt{1-x^2}
    $
    . . Hence: .$\displaystyle \cot\theta \:=\:\frac{adj}{opp} \:=\:\frac{\sqrt{1-x^2}}{x} $

    Substitute into [1]: .$\displaystyle -\frac{1}{3}\left(\frac{\sqrt{1-x^2}}{x}\right)^3 + C \;=\;\boxed{-\frac{(1-x^2)^{\frac{3}{2}}}{3x^3}}$

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  3. #3
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    $\displaystyle \int {\frac{{\sqrt {1 - x^2 } }}
    {{x^4 }}\,dx}$

    Define $\displaystyle x=\frac1u,$ the integral becomes to

    $\displaystyle - \int {u\sqrt {u^2 - 1} \,du} = - \frac{{\left( {u^2 - 1} \right)^{3/2} }}
    {3} + k$

    Back substitute and we're done.
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