Did I Substitute Correctly in My Integral Calculation?

[alba_ei]

Homework Statement

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

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 I've 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?

 

[ChaoticLlama]

Wayy to complicated

\[<br /> \begin{array}{l}<br /> \int {\frac{{e^x }}{{\sqrt {4 - e^{2x} } }}} dx \\ <br /> {\rm{let }}u = e^x \\ <br /> {\rm{ }}du = e^x dx \\ <br /> \int {\frac{{du}}{{\sqrt {4 - u^2 } }}} \\ <br /> {\rm{let }}u = 2\sin \theta \\ <br /> {\rm{ }}du = 2\cos \theta d\theta \\ <br /> \end{array}<br /> \]<br />

What would you do from here?

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

 

[alba_ei]

Wayy to complicated

\[<br /> \begin{array}{l}<br /> \int {\frac{{e^x }}{{\sqrt {4 - e^{2x} } }}} dx \\ <br /> {\rm{let }}u = e^x \\ <br /> {\rm{ }}du = e^x dx \\ <br /> \int {\frac{{du}}{{\sqrt {4 - u^2 } }}} \\ <br /> {\rm{let }}u = 2\sin \theta \\ <br /> {\rm{ }}du = 2\cos \theta d\theta \\ <br /> \end{array}<br /> \]<br />

What would you do from here?

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

 

[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.