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. 
 A: 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.
A: Hint: Try making the substitution $x=2\sin(u)$ and find the expressions for $dx$ and $u$. In general, in the integrals of the form $\displaystyle\int\sqrt{a^2-x^2}\,dx$ you can make the substitution $x=a\sin(u)$.
A: This integral requires a trigonometric substitution. Let:
$$x = 2\sin(\theta) \quad \mathtt{and} \quad dx = 2\cos(\theta) d\theta$$
Plugging this in gives:
$$\int_{-1}^1 \sqrt{4-x^2} dx = \int_{-\frac{\pi}{6}}^\frac{\pi}{6} \sqrt{4-4\sin^2(\theta)} 2\cos(\theta) d\theta = 4\int_{-\frac{\pi}{6}}^\frac{\pi}{6} \sqrt{1-\sin^2(\theta)}\cos(\theta) d\theta$$
$$= 4\int_{-\frac{\pi}{6}}^\frac{\pi}{6}  \cos^2(\theta) d\theta = 2\int_{-\frac{\pi}{6}}^\frac{\pi}{6} (\cos(2\theta) + 1) d\theta = 2\Big[ \frac{\sin(2\theta)}{2} + \theta \Big] \Big|_{-\frac{\pi}{6}}^\frac{\pi}{6} = \sqrt{3} + \frac{2\pi}{3}$$
Notice that I reevaluated the bounds according to the trigonometric substitution. Also to integrate $\cos^2(\theta)$ I used a double angle formula.
A: OK so you now have three clear explanations how to do this using trig substitution, but you might also find it interesting to consider the integral as the area of part of the inside of the circle $x^2+y^2=4$
