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$