<s>[INST] 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 ?
 [/INST] Before evaluating the definite integral, you need to make the following back-substitution:
$$
u=\arcsin{x}
$$
And use this trigonometric identity ($-\frac{\pi}{2}\le u\le \frac{\pi}{2} \implies \cos{u}\ge0$):
$$
\sin{2u}=2\sin{u}\cos{u}=2\sin{(\arcsin{x})}\sqrt{1-\sin^2{(\arcsin{x})}}=2x\sqrt{1-x^2}
$$
You also forgot that you have a $2$ in front of your integral:
$$
2\left[\frac{\arcsin{x}}{2}+x\sqrt{1-x^2}\right]_{-1}^{1}=\\
2\left(\frac{\arcsin{(1)}}{2}+1\cdot\sqrt{1-1^2}-\frac{\arcsin{(-1)}}{2}-1\cdot\sqrt{1-(-1)^2}\right)=\\
2\left(\frac{\pi}{4}+0+\frac{\pi}{4}-0\right)=2\frac{2\pi}{4}=\frac{4\pi}{4}=\pi.
$$
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