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Math Help - Integral help

  1. #1
    Newbie
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    Integral help

    Second question

    I am supposed to integrate this question by using trig substitution

    Integrate

    dx/(x^2 * (x^2 + 1) ^1/2)

    So I made my triangle, (x^2 + 1) is the hypotenuse, x 1 is adjacent to theta, and x is oposite of theta. Theta = to one angle

    So

    Tan (theta) = x/1 so
    x= tan(theta)
    dx = sec^2 theta
    sec (theta) = (x^2 + 1)^1/2

    So for my substituted integral I get

    (sec(theta))^2/[(tan(theta))^2(sec(theta))]

    So I get the one sec(theta) to cancel and I am left with

    (sec(theta))/[((tan(theta))^2]

    i can't seem to integrate it from there. I tried different trig identies, but nothing is working out.

    Thanks
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Why not using the facts \sec \theta =\frac{1}{\cos \theta },\,\tan \theta =\frac{\sin \theta }{\cos \theta } ?
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  3. #3
    Super Member

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    Hello, shoff14!

    Integrate using trig substitution: . \int \frac{dx}{x^2 \sqrt{x^2 + 1}}

    So: . x\:=\:\tan\theta \quad\Rightarrow\quad dx \:=\:\ sec^2\!\theta\,d\theta \quad\Rightarrow\quad \sqrt{x^2+1} \:=\:\\sec\theta

    Substitute: . \int\frac{\sec^2\!\theta\,d\theta}{\tan^2\!\theta\  sec\theta} \;=\;\int\frac{\sec\theta}{\tan^2\!\theta}\,d\thet  a

    i can't seem to integrate it from there.
    Change to sines and cosines . . .

    . . \frac{\sec\theta}{\tan^2\theta} \;=\;\frac{\dfrac{1}{\cos\theta}}{\dfrac{\sin^2\th  eta}{\cos^2\theta}} \;=\;\frac{\cos\theta}{\sin^2\!\theta} \;=\;\frac{1}{\sin\theta}\cdot\frac{\cos\theta}{\s  in\theta} \;=\;\csc\theta\cot\theta


    Now integrate: . \int\csc\theta\cot\theta\,d\theta . . . then back-substitute.

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  4. #4
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    thanks!
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