# Trigonometric Substitution

1. Oct 9, 2008

### protivakid

1. The problem statement, all variables and given/known data

$$\int$$ $$\frac{e^{3x}dx}{\sqrt{1-e^{2x}}}$$

2. Relevant equations

3. The attempt at a solution

Alright so I am able to do other similar problems fine, I think it is the "e" that is throwing me off as well as the fact that the "x" is in the exponent. I started the problem as so...

sin$$\vartheta$$=$$\frac{e^{\sqrt{x}}}{\sqrt{1}}$$

cos$$\vartheta$$d$$\vartheta$$=$$\frac{e^{\sqrt{x}}dx}{\sqrt{1}}$$

Am I off to the right start and if so can I have some helpful hints as what to do next? Thanks guys.

By the way, those symbols next to Sin & Cos are Theta, I am still learning how to use this forum, sorry.

2. Oct 9, 2008

### grief

I think you should have sin(theta) = e^x, cos(theta) =sqrt(1-e^2x). Remember that for any n, (e^nx) =(e^x)^n

3. Oct 10, 2008

### rocomath

Re-write it.

$$\int\frac{(e^x)^3dx}{\sqrt{1-(e^x)^2}}$$

$$e^x=\sin xdx$$

Take the natural log, then it's derivative.

4. Oct 10, 2008

### protivakid

Alright so taking your advice I set sin(theta) to e^x dx, and cos(theta) to sqrt(1-e^2x). That then gave me sin^3(theta)/cos((theta). I set u=sin(theta)d(theta) and du=cos(theta) which gave me u^3du^-1. That then became u^2/2 which is sin^2/2. My final answer U then got from that was e^2x/2 + c. Does that sound correct, if not please advise me and I am sorry for misunderstanding your help.

5. Oct 10, 2008

### rocomath

Looks wrong.

6. Oct 10, 2008

### Kreizhn

You cannot have a du in the denominator. Think about what this would means in terms of the Riemann sum and you'll see that it doesn't make any sense. Instead, consider an expansion of sin(x)³. That is

$$\frac{sin^3(x)}{cos(x)} = sin(x) \frac{1-cos^2(x)}{cos(x)} = tan(x) - sin(x)cos(x)$$ You can integrate this.

7. Oct 10, 2008

### protivakid

I am trying to set up a triangle for visual aid, is the following correct...

sin=e^x
cos=sqrt(1-e^2x)
tan= (e^x)/(sqrt(1-e^2x)

Thanks guys, greatly appreciated.

8. Oct 10, 2008

### Kreizhn

Yes, those are correct, though you're missing a bracket in tan

9. Oct 10, 2008

### protivakid

Thanks, i'll try to take it from here but I don't think there is too much left to do.

10. Oct 10, 2008

### Kreizhn

I didn't read all the way back,and that was my mistake, but did you remember to substitute for dx in the integral as well?

If $e^x = \sin\theta$ then $e^x dx = \cos\theta d\theta$ which implies that $dx = \mathrm{cot}\theta d\theta$. That should actually make the integral quite easy