calculus - Finding the area of the top half of a circle

Alright, I'm trying to calculate the area of the top half of a circle of radius $a$. Here's what I did so far:

$$\int_{-a}^a \sqrt{(a^2 - x^2) }dx$$

So I wrote $x$ as $a \cdot \sin \theta$:

$$\int_{-a}^a \sqrt{(a^2 - a^2\sin^2 \theta )}$$

$$\int_{-a}^a a \sqrt{( 1 - \sin \theta^2)}$$

$$\int_{-a}^a [a \cdot \cos \theta]$$

$$2 \sin(a) a$$

The problem is that my textbook states that the area is actually:

$$\frac{\pi a^2}{2}$$

I've done this calculation over and over and I'm sure there are no mistakes, so what is going on here?

Answer

$$x=a\sin t\implies dx=a\cos t\,dt$$

and from here

$$\int_{-a}^a\sqrt{a^2-x^2}\,dx=a\int_{-\frac\pi2}^\frac\pi2\sqrt{1-\sin^2 t}\,a\cos t\,dt=a^2\int_{-\frac\pi2}^\frac\pi2\cos^2t=$$

$$=\left.\frac{a^2}2(t+\cos t\sin t)\right|_{-\frac\pi2}^{\frac\pi2}=\frac{a^2\pi}2$$