Evaluate :$\int_{-1}^{1} 2\sqrt{1-x^2} dx $ 
Evaluate: $$\int_{-1}^{1} 2\sqrt{1-x^2} dx $$

The answer is $\pi$
My attempt
$x = \sin(u), dx = \cos(u)du$
$$\int_{-1}^{1} 2 \sqrt{1-\sin^2(u)}\cos(u)du = \int_{-1}^{1} 2 \cos^2(u)du =\int_{-1}^{1} \frac{1}{2}(1+\cos(2u))du = \bigg(\frac{u}{2} + \frac{1}{2}\sin(2u) \bigg)\Bigg|_{-1}^{1}$$
confused how to proceed ?

Geometrically , the unit circle can be represented as $$x^2+y^2=1$$
so $$y=\pm \sqrt{1-x^2}$$ and your case $y=+ \sqrt{1-x^2}$

So $\int_{-1}^1\sqrt{1-x^2} dx $  is the area of a (upper )semi circle, which is $\frac{\pi}{2}$. So $$2 \int_{-1}^1 \sqrt{1-x^2}dx =2 \frac{\pi}{2}=\pi$$
