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Homework Help: Integrals involving trig substitution

  1. Jul 18, 2010 #1
    1. The problem statement, all variables and given/known data

    I don't know how to solve these. How do you evaluate the integral of [tex]\frac{3dx}{\sqrt{3+X^2}}[/tex]? I know you have to set x=atan[tex]\theta[/tex].

    our a is [tex]\sqrt{3}[/tex] so x =[tex]\sqrt{3}[/tex]tan[tex]\theta[/tex]. That means

    I also made a right triangle using the information that tan[tex]\theta[/tex]=X/sqrt(3)

    So the two legs of the triangle are X and sqrt(3) and the hypotenuse is sqrt{X^2+3

    I try to solve using substitution but I can never get the right answer.

    The answer is supposed to be 3*ln(X + sqrt(3+X^2) + C

    I can't seem to get tags to work properly so just bear with the poor formatting.

    I have a test on this stuff tomorrow and this is a review problem. Sadly, my professor did not give us solutions; just answers.
  2. jcsd
  3. Jul 18, 2010 #2
    Yep, so we have:
    \int \frac{dx}{ \sqrt{x^2+3} }
    = \int \frac{\sqrt{3}\sec^2 \theta}{\sqrt{3(\tan^2\theta + 1)}} d\theta
    = \int \frac{\sqrt{3}\sec^2\theta}{\sqrt{3\sec^2\theta}} d\theta
    = \int \sec\theta d\theta
    = \int \sec\theta\,\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}\, d\theta

    EDIT: Idk if all that LaTeX is showing up correctly. If not, just click on it for the full code.
    EDIT#2: Thanks vela for fixing it. stupid brackets haha...
    Can you figure out the rest?
    Do a substitution on the denominator.

    The answer is indeed
    [tex] 3 \ ln|x + \sqrt{3+x^2}| \ + \ C [/tex].
    Last edited: Jul 19, 2010
  4. Jul 18, 2010 #3


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    Fixed your LaTeX. You were missing a few }'s.

    [tex]\int \frac{dx}{ \sqrt{x^2+3} }
    = \int \frac{\sqrt{3}\sec^2 \theta}{\sqrt{3(\tan^2\theta + 1)}} d\theta
    = \int \frac{\sqrt{3}\sec^2\theta}{\sqrt{3\sec^2\theta}} d\theta
    = \int \sec\theta d\theta
    = \int \sec\theta\,\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}\, d\theta[/tex]
  5. Jul 18, 2010 #4


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    There is another way:
    \int\frac{d\theta}{\cos\theta}=\int\frac{\cos\theta}{\cos^{2}\theta}d\theta =\int\frac{\cos\theta}{1-\sin^{2}\theta}d\theta
    Split up using partial fractions and you have a nice simple integral to do.
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