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Math Help - Trig substitution

  1. #1
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    Trig substitution

    I've tried to substitute x in but seem to get the integration wrong.

    1. \int \frac{x^3}{\sqrt{x^2+9}} dx let x=3tan\theta

    2. \int \frac{1}{x^2\sqrt{25-x^2}}dx let x=5sin\theta
    Last edited by ThePerfectHacker; July 23rd 2006 at 02:43 PM.
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  2. #2
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    Your choice of substitution is good. Don't forget dx. That can be overlooked

    For instance, #2. x=5sin{\theta}\;\  dx=5cos{\theta}d{\theta}

    \int{(25sin^{2}{\theta})\sqrt{25(1-sin^{2}{\theta})}5cos{\theta}d{\theta}}

    625\int{sin^{2}{\theta}cos^{2}{\theta}d{\theta}
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  3. #3
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    Hello, c_323_h!

    Don't be discouraged . . . the first one takes forever
    . . and there are a dozen places to make errors (I've made them all).


    1)\;\;\int \frac{x^3}{\sqrt{x^2+9}}\,dx\qquad\text{Let }x = 3\tan\theta

    Let: x = 3\tan\theta\quad\Rightarrow\quad dx = 3\sec^2\!\theta\,d\theta
    . . and \sqrt{x^2 + 9}\:=\:\sqrt{9\tan^2\!\theta + 9} \:=\:\sqrt{9(\tan^2\!\theta + 1)} \:= \sqrt{9\sec^2\!\theta} \:= \:3\sec\theta

    Substitute: . \int\frac{(3\tan\theta)^3}{3\sec\theta}(3\sec^2\! \theta\,d\theta) \;= \;27\!\int\tan^3\!\theta\sec\theta\,d\theta

    We have: . 27\!\int\tan^2\!\theta(\sec\theta\tan\theta\,d \theta) \;= \;27\!\int(\sec\theta - 1)(\sec\theta\tan\theta\,d\theta)

    Let: u = \sec\theta\quad\Rightarrow\quad du = \sec\theta\tan\theta\,d\theta

    Substitute: . 27\!\int(u^2 - 1)\,du \;= \;27\left(\frac{u^3}{3} - u\right) + C \;= \;9u(u^2 - 3) + C

    Back-substitute
    . . Since u = \sec\theta, we have: . 9\sec\theta(\sec^2\!\theta - 3) + C

    Back-substitute: Since \tan\theta = \frac{x}{3}, then \sec\theta = \frac{\sqrt{x^2+9}}{3}

    We have: . 9\cdot\frac{\sqrt{x^2+9}}{3}\left[\left(\frac{\sqrt{x^2+9}}{3}\right)^2 - 3\right] + C \;= 3\sqrt{x^2 + 9}\left(\frac{x^2 + 9}{9} - 3\right) + C

    . . = \;3\sqrt{x^2+9}\left(\frac{x^2+9-27}{9}\right) + C \;= \;\frac{1}{3}\sqrt{x^2+9}\left(x^2 - 18\right) + C

    I need a nap . . .

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  4. #4
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    Quote Originally Posted by c_323_h
    <br />
\int \frac{1}{x^2\sqrt{25-x^2}}dx<br />
    I do these problems in a formal way.
    Ignore this, if you do not understand I do not want to confuse you.
    ------
    Let,
    u=\sin^{-1}(5x)
    Then,
    \frac{du}{dx}=\frac{1}{\sqrt{25-x^2}}--->Denominator.
    Express integrand as,
    \int \frac{1}{x^2}\cdot \frac{1}{\sqrt{25-x^2}}dx
    Also, we have,
    \sin u=5x
    Thus,
    \csc u=\frac{1}{5x}
    Thus,
    \csc^2 u=\frac{1}{25x^2}
    Thus,
    \frac{1}{x^2}=25\csc^2 u
    Thus, you finally have,
    \int 25\csc^2u \frac{du}{dx}dx=25\int \csc^2udu=-25\cot u+C
    Substitute back,
    -25\cot (\sin^{-1}(5x))+C
    Simplify the inverse trigonometric with trigonometric, to get,
    -25\cdot \frac{\sqrt{1-25x^2}}{5x}+C
    Thus,
    -\frac{5\sqrt{1-25x^2}}{x}+C
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