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Is this integral correct?

  1. May 3, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \frac{e^{3x}}{\sqrt{4 - e^{2x}} \, dx [/tex]

    3. The attempt at a solution
    Let [tex] u = e^x [/tex]. Then [tex] du = e^x \, dx [/tex]. So [tex] u^2 = e^{2x}[/tex]. We have

    [tex]\int \frac{u^2}{\sqrt{4 - u^2)} \, du [/tex]
    We make the substitution [tex]u = 2 \sin\theta [/tex]. [tex]du = 2 \cos\theta \, d\theta [/tex]. Then

    [tex]\int \frac{4\sin^2\theta}{2\cos\theta)} 2\cos\theta \, d\theta [/tex]
    [tex]=4 \int \sin^2\theta \, d\theta [/tex]
    [tex]=2 \int (1 -\cos 2\theta) \, d\theta [/tex]
    [tex]=2 \theta -\sin 2\theta = 2 \theta -2\sin\theta \cos\theta + C [/tex]
    Changing variables

    [tex]=2 \arcsin \frac{e^x}{2} -\frac{e^x}{2}\sqrt{4 - e^{2x}} + C [/tex]

    But, my proffesor say that ive an error at the first step when i took u^2 as e^2x he said that i should take u as u^2= e^3x, the answer is right he said that the error was cover by another error and that is how i get the answer right. Is anything bad in the way i solved the integral or why my proffesor said that?
    Last edited: May 3, 2007
  2. jcsd
  3. May 3, 2007 #2
    Wayy to complicated

    \int {\frac{{e^x }}{{\sqrt {4 - e^{2x} } }}} dx \\
    {\rm{let }}u = e^x \\
    {\rm{ }}du = e^x dx \\
    \int {\frac{{du}}{{\sqrt {4 - u^2 } }}} \\
    {\rm{let }}u = 2\sin \theta \\
    {\rm{ }}du = 2\cos \theta d\theta \\

    What would you do from here?

    P.S. u^2= e^3x wouldn't help anything here
    Last edited: May 3, 2007
  4. May 3, 2007 #3
    Its [tex] \int \frac{{e^{3x} }{\sqrt {4 - e^{2x} } dx [/tex]
  5. May 4, 2007 #4


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    Homework Helper

    Nope, your answer is totally correct. And there's no error in your work. There are several ways to go about solving an integral, and yours is one of them. :)
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