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Math Help - 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; October 26th 2007 at 11:34 AM.
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  2. #2
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    Hello, afeasfaerw23231233!

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


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

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

    Substitute: . \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  u\,=\,\cot\theta\quad\Rightarrow\quad du \,=\,-\csc^2\!\theta\,d\theta

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

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


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

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

    Substitute into [1]: . -\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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    \int {\frac{{\sqrt {1 - x^2 } }}<br />
{{x^4 }}\,dx}

    Define x=\frac1u, the integral becomes to

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

    Back substitute and we're done.
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