Tough definite integration For a curve given by:
$x=e^{-t}\cos{2t}$, $y=\sin t$
R is the region bounding this curve, the x axis and the y axis (y-intercept is point a and x-intercept is point b). 


*

*Find the exact coordinates of a and b

*By considering $\int_0^ax~dy$, show that the exact area of region R is $\frac{3}{10}[\sqrt2e^{\frac{-3\pi}{4}}-e^{-\pi}]$
Alright, sadly I am extremely unfamiliar with parametric equations, so I don't even know how to do part 1. I assumed that for the y-intercept you just let $\sin t =0$ but that seems to give me the wrong answer.   I resorted to typing it into a graphical calculator, and I did get the answers a=0.707 and b=1. 0.707 appears to be $\frac{\sqrt2}{2}$. I only did this so that I could proceed with the question, this is obviously not a viable answer.
On to the next part. Firstly, I rewrote the equation in terms of x and y, using the fact that $\cos(2\arcsin y) = 1-2y^2$ 
$y=\arcsin t \implies x=e^{-\arcsin y}-2y^2e^{-\arcsin y}$
Since we need the area of region R, I then considered 
$\int_0^{\frac{\sqrt2}{2}}e^{-\arcsin y}-2y^2e^{-\arcsin y}~dy$
I then tried a u-substitution, where $u=\arcsin y$. Hence $y=\sin u$ and $dy=\cos u du$
The limits (which as I have mentioned earlier I am not sure about) then just become $\frac{\pi}{4}$ and 0. Hence we have:
$\int_0^{\frac{\pi}{4}}[e^{-u}-2e^{-u}\sin ^2u]\cos u ~du$. Which I have no idea how to integrate.
In conclusion: If indeed this is the right method (i doubt it), how do I proceed?
Otherwise, I am really hoping that there is a much simpler method along the way, perhaps avoiding the $\arcsin y$.
Thank you very much!
UPDATE I know understand how to find the x and y intercepts, which still leaves the integral. 
 A: The following figure shows a part of the curve in question:

As you can see there are many regions "bounded by the curve and the axes" (not all of them visible in the figure). Most likely the triangular region $R$ with corners $(0,0)$, $(1,0)$, and $(0,{1\over\sqrt{2}})$ is meant.
When a region is given by  parametric representations of its boundary arcs the simplest way to compute its area is to use one of Green's formulas
$${\rm area}(R)=\left\{\eqalign{&\int_{\partial R} x\>dy,\cr &-\int_{\partial R}y\>dx,\cr &{1\over2}\int_{\partial R}(x\>dy-y\>dx)\ .\cr}\right.$$
Note that we have to integrate  counterclockwise along the full boundary of $R$. Since $\partial R$ consists of two segments on the axes and the arc
$$\gamma:\quad t\mapsto\bigl(e^{-t}\cos(2t),\sin t\bigr)\qquad \biggl(0\leq t\leq{\pi\over4}\biggr)$$
of your curve we use the first of the above formulas, because along the axis segments we have either $x(t)\equiv0$ or $\dot y(t)\equiv0$, so that these segments give no contribution to the total boundary integral. It then follows that
$${\rm area}(R)=\int_\gamma x\>dy=\int_0^{\pi/4}x(t)\>\dot y(t)\ dt=\int_0^{\pi/4}e^{-t}\cos(2t)\> \cos t \ dt={1\over10}\bigl(3+e^{-\pi/4}\bigr)\ .$$
While we are at it we can compute the area of the little loop $L$ to the left of the $x$-axis. This loop is bounded by the single arc
$$\gamma':\quad t\mapsto\bigl(e^{-t}\cos(2t),\sin t\bigr)\qquad \biggl({\pi\over 4}\leq t\leq{3\pi\over4}\biggr)\ ,$$
but $\gamma'$ goes the wrong way around. Therefore we obtain
$${\rm area}(L)=\int_{\partial L}x\>dy=-\int_{\gamma'}x\>dy=-\int_{\pi/4}^{3\pi/4}e^{-t}\cos(2t)\> \cos t \ dt={e^{-3\pi/4}\over 5\sqrt{2}}\bigl(e^{\pi/2}-3\bigr)\ .$$
