How should I go about solving this definite integral? The integral is:
$$\int_{-1}^1\sqrt{4-x^2}dx$$
I'm having difficulty figuring out how to go about this. I attempted to use u-substitution, both by substituting $u$ for $\sqrt{4-x^2}$ entirely, and then just $x^2$, but I quickly realized that neither of those work. I've just been introduced to integrals, so I'm not the best at figuring out what to do with them, yet. 

Let $x = 2 \sin \theta$. Then $dx = 2\cos \theta \; d\theta$. When $x = -1, \theta = -\frac{\pi}{6}$ and when $x = 1, \theta = \frac{\pi}{6}$. Therefore
$$\int_{-1}^1 \sqrt{4-x^2} \; dx = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \sqrt{4 - 4 \sin^2 \theta} \; 2\cos \theta \; d\theta = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} 4\cos^2 \theta \; d\theta.$$
Now you can use, say the double angle formula,
$$2\cos^2\theta - 1 = \cos 2\theta$$
to obtain
$$\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} 4\cos^2 \theta \; d\theta = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} 2(1+\cos 2\theta) \; d\theta = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} (1+\cos \theta) \; d\theta.$$
I suppose the remaining steps are easy.

Added: When you see $a^2-x^2$ and $a^2+x^2$, you can always substitute $x = a\sin \theta$ and $x = \tan \theta$ respectively to see if it helps.
