1. integration

I was doing this equation $\int e^x\aqrt{1-e^2x}dx$ and I used the substitution $e^x=sinu$ which gave me

$\frac{sin(arcsin(e^x))(cos(arcsin(e^x))}{2}+\frac{ arcsin(e^x)}{2}+C$
which simplifies to $\frac{e^x\sqrt{1-exp(2x)}}{2}+\frac{arcsin(e^x)}{2}+C$ but i cant see how the first term of this was obtained from the first line here. If some one could help youd be od great help!!!

2. Originally Posted by i_zz_y_ill
I was doing this equation $\int e^x\aqrt{1-e^2x}dx$ and I used the substitution $e^x=sinu$ which gave me

$\frac{sin(arcsin(e^x))(cos(arcsin(e^x))}{2}+\frac{ arcsin(e^x)}{2}+C$
which apparantly simplifies to $\frac{e^x\sqrt{1-exp(2x)}}{2}+\frac{arcsin(e^x)}{2}+C$ but i cant see how in my previous thread how this was done If some one could help youd be od great help!!!
$\int\sqrt{1-u^2}du$

If we let $u=\sin(z)$

Then we get

$\int\cos^2(z)dz$

$=\frac{1}{2}\int\bigg[1+\cos(2z)\bigg]dz$

$=\frac{1}{2}\bigg[z+\frac{1}{2}\sin(2z)\bigg]+C$(1)

So now seeing that

$\sin(2z)=2\sin(z)\cos(z)=2\sin(z)\sqrt{1-\sin^2(z)}$

We can rewrite (1) as

$\frac{1}{2}\bigg[z+\sin(z)\sqrt{1-\sin^2(z)}\bigg]+C$

Now backsubbing $\arcsin(u)=z$

Gives us

$\frac{1}{2}\bigg[\arcsin(u)+\sin\left(\arcsin(u)\right)\sqrt{1-\sin^2\left(\arcsin(u)\right)}\bigg]+C$

Now you really should know that

$\sin\left(\arcsin(z)\right)=z$

So using that we may further rewrite (1) as

$\frac{1}{2}\bigg[\arcsin(u)+u\sqrt{1-u^2}\bigg]+C$(2)

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

Now by the sub $m=e^x$

We obtain

$\int\sqrt{1-m^2}dm$

So now using (2) we see that this is equivalent to

$\frac{1}{2}\bigg[\arcsin(m)+m\sqrt{1-m^2}\bigg]+C$

Now backsubbing finally gives

$\boxed{\int{e^x\sqrt{1-e^{2x}}}dx=\frac{1}{2}\bigg[\arcsin\left(e^x\right)+e^x\sqrt{1-e^{2x}}\bigg]+C}$

That is about as drawn out and forward as I can be. If you still dont get it, I am not sure I can help