Parametric equations and area I am not sure how to work this one out.
I am suppose to find the area of this parametric equation.
$$y = b\sin\theta, x = a\cos\theta$$
$$0 \leq 0 \leq 2\pi$$
I set up the equation in the memorized formula.
$$\int_0^{2\pi} \sqrt{1 + \left(\frac{b\cos\theta}{-a\sin\theta}\right)^2}d\theta$$
$$\int_0^{2\pi} \sqrt{1 + \frac{b^2\cos^2\theta}{a^2\sin^2\theta}}d\theta$$
$$\int_0^{2\pi} \sqrt{1 + \frac{b^2}{a^2}\cdot \csc^2\theta}d\theta$$
From here I am at a loss of what to do, I tried some trig subsitution but I do not think that will work and it only seems to complicate the problem.

Note that $\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=\cos^2\theta + \sin^2\theta=1.$ We then recognize that this describes an ellipse, which has area $\pi a b$.
EDIT: How to find the area of the ellipse.
There are many many proofs of this, but the easiest one you might find in a single-variable calculus course is as follows.
We will find the area of the ellipse in the first quadrant and quadruple it (we assume $x,y\geq0$ for what follows). Solve for $y$ in terms of $x$.
$$
1 = \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2
$$
$$
\iff  \left(\frac{y}{b}\right)^2= 1-\left(\frac{x}{a}\right)^2
$$
$$
\iff \frac{y}{b} = \sqrt{1-\left(\frac{x}{a}\right)^2}
$$
$$
y=b\sqrt{1-\left(\frac{x}{a}\right)^2}.
$$
Then we have
$$
\frac{A}{4} = \int_0^ab\sqrt{1-\left(\frac{x}{a}\right)^2}dx.
$$
Substitute $u = \frac{x}{a}$ so $dx=a\ du$ and the limits go from $u=0$ to $u=1$,
$$
\frac{A}{4} = ab\int_0^1\sqrt{1-u^2}du.
$$
At this point you could interpret the integral on the right as the the area of a quarter of a circle of radius 1, $\pi/4$. If not, we continue by making the trigonometric substitution $u = \sin t$, so $\sqrt{1-u^2}=\cos t$ and $du=\cos t\ dt$ where $t$ goes from $0$ to $\pi/2$,
$$
\frac{A}{4} = ab\int_0^{\pi/2} \cos^2t dt.
$$
Now we use the formula $\cos^2t = \frac{1}{2}(1+\cos(2t))$,
$$
\frac{A}{4} = \frac{ab}{2}\int_0^{\pi/2}(1+\cos(2t))dt
$$
$$
=\frac{ab}{2}\left(t+\frac{1}{2}\sin(2t) \right)_0^{\pi/2}
$$
$$
=\frac{ab}{2}\left(\left(\frac{\pi}{2}+0\right) - \left(0 + 0\right) \right)
$$
$$
\frac{\pi a b}{4}.
$$
Rearranging then gives us $A= \pi a b$.
