1. ## Integrating correct substitution,,

Is this the right answer? I don't think it is somehow!

$\displaystyle \int e^x\sqrt{1-e^2x}dx$ $\displaystyle e^x=sinu$
$\displaystyle =\int e^x\sqrt{cos^2u}du$ $\displaystyle e^2x=sin^2u$
$\displaystyle =\int sinucosudx$ $\displaystyle x=lnsinu$
$\displaystyle dx=\frac{cosu}{sinu}du$
$\displaystyle =\int cos^2udu$
$\displaystyle =o.5\int cos2u+1du=$$\displaystyle \frac{sin2u}{4}+\frac{u}{2}+C$$\displaystyle =\frac{sin(2sin^-1(e^x))}{4}+\frac{sin^-1(e^x)}{2}+C$

2. Originally Posted by i_zz_y_ill
Is this the right answer? I don't think it is somehow!

$\displaystyle \int e^x\sqrt{1-e^2x}dx$
I used $\displaystyle e^x=sinu$
=\int e^x\sqrt{cos^2u}
=\int sinu cosudx
=\int cos^2udu
=o.5\int cos2u+1du
=\frac{sin2u}{4}+\frac{u}{2}+C
=\frac{sin(2sin^-1(e^x))}{4}+\frac{sin^-1(e^x)}{2}+C[/tex]
Hello,

I didn't read 'til the end because there are problems with the latex ><

the substitution is okay.

$\displaystyle e^x=\sin(u) \implies e^x dx=\cos(u) du \implies dx=\frac{\cos(u)}{e^x} ~ dx$
so the e^x will simplifiy with the one that is already in the integral.

PS : for the latex code, leave a space after \int, otherwise, they'll understand \intsinu as a single command, which doesn't exist

3. I did along the lines f what you said,my original post now has my working,,i think its wrong though!!! and can i not just substitute u^2=1-e^2x thats where im a bit confused i get a very simple answer when i do compared to my other method?

4. It's still very confusing

$\displaystyle e^x=\sin(u)$

Then $\displaystyle \dots=\int \cos(u) \sin(u) ~ du$

now, just substitute $\displaystyle y=\sin(u)$

and can i not just substitute u^2=1-e^2x
Yes you can

5. hmm i never got $\displaystyle \int sinucosudu$ i got $\displaystyle \int sinucosudx$ I think if you look at my first post i have the answer ,, i used $\displaystyle e^x=sinu$ as the substitution except the answer looks very messy''.

6. $\displaystyle \int{e^x}\sqrt{1-e^{2x}}dx\overbrace{\mapsto}^{u=e^x}\int\sqrt{1-u^2}du\overbrace{\mapsto}^{u=\sin(z)}\int\cos^2(z) dz$

7. oh right thanks,,,so that gives me
$\displaystyle =\frac{sin(2sin^-1(e^x))}{4}+\frac{sin^-1(e^x)}{2}+C$ which i cant simplify?

8. Originally Posted by i_zz_y_ill
oh right thanks,,,so that gives me
$\displaystyle =\frac{sin(2sin^-1(e^x))}{4}+\frac{sin^-1(e^x)}{2}+C$ which i cant simplify?
$\displaystyle \sin\left(2\arcsin\left(e^x\right)\right)=2\sin\le ft(\arcsin\left(e^x\right)\right)\cos\left(\arcsin \left(e^x\right)\right)=2e^x\sqrt{1-e^{2x}}$

9. I see where you get $\displaystyle 2sinarcsin(e^x)cos(arcsin(e^x)$ but I cannot see how you got to the next line,,,,and does the term $\displaystyle \frac{arcsin(e^x)}{2}$ also simplify?? thanks I really appreciate this

10. Originally Posted by oxrigby
I see where you get $\displaystyle 2sinarcsin(e^x)cos(arcsin(e^x)$ but I cannot see how you got to the next line,,,,and does the term $\displaystyle \frac{arcsin(e^x)}{2}$ also simplify?? thanks I really appreciate this
$\displaystyle \sin\left(\arcsin(f(x))\right)=f(x)$

$\displaystyle \cos\left(\arcsin(f(x))\right)=\sqrt{1-f(x)^2}$

And no, not to anything you would want.

11. how is $\displaystyle cos(arcsin(e^x)=\sqrt{1-exp(2x)}$?i cant see this