(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Homework Help: Is this integral correct?

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