# Is this integral correct?

1. May 3, 2007

### alba_ei

1. The problem statement, all variables and given/known data
$$\int \frac{e^{3x}}{\sqrt{4 - e^{2x}} \, dx$$

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

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

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

$$=2 \arcsin \frac{e^x}{2} -\frac{e^x}{2}\sqrt{4 - e^{2x}} + C$$

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. May 3, 2007

### ChaoticLlama

Wayy to complicated

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

What would you do from here?

P.S. u^2= e^3x wouldn't help anything here

Last edited: May 3, 2007
3. May 3, 2007

### alba_ei

Its $$\int \frac{{e^{3x} }{\sqrt {4 - e^{2x} } dx$$

4. May 4, 2007

### VietDao29

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. :)
Congratulations.

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