Integrating correct substitution,,

**i_zz_y_ill** · November 5th 2008, 11:11 AM

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

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

**Moo** · November 5th 2008, 11:15 AM

Originally Posted by i_zz_y_ill

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

$\int e^x\sqrt{1-e^2x}dx$
I used $e^x=sinu$
=\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.

$e^x=\sin(u) \implies e^x dx=\cos(u) du \implies dx=\frac{\cos(u)}{e^x} ~ dx$
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

**i_zz_y_ill** · November 5th 2008, 11:27 AM

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?

**Moo** · November 5th 2008, 11:42 AM

It's still very confusing

$e^x=\sin(u)$

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

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

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

Yes you can

**i_zz_y_ill** · November 5th 2008, 12:24 PM

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

**Mathstud28** · November 5th 2008, 12:47 PM

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

**i_zz_y_ill** · November 5th 2008, 12:53 PM

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

**Mathstud28** · November 5th 2008, 12:59 PM

Originally Posted by i_zz_y_ill

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

$\sin\left(2\arcsin\left(e^x\right)\right)=2\sin\le ft(\arcsin\left(e^x\right)\right)\cos\left(\arcsin \left(e^x\right)\right)=2e^x\sqrt{1-e^{2x}}$

**oxrigby** · November 5th 2008, 01:09 PM

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

**Mathstud28** · November 5th 2008, 01:13 PM

Originally Posted by oxrigby

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