SOLUTION 1: $ \ \ $ To integrate $ \displaystyle{ \int {\sqrt{1-x^2} } \ dx } $ use the trig substitution
$$ x = \sin \theta $$
so that
$$ dx = \cos \theta \ d\theta $$
Substitute into the original problem, replacing all forms of $ x $, getting
$$ \displaystyle{ \int { \sqrt{1-x^2} } \ dx } = \displaystyle{
\int { \sqrt{1-\sin^2 \theta} } \ \cos \theta \ d\theta } $$
$$ = \displaystyle{ \int { \sqrt{\cos^2 \theta} \ \cos \theta \ } \ d\theta } $$
$$ = \displaystyle{ \int { \cos \theta \ \cos \theta \ } \ d\theta } $$
$$ = \displaystyle{ \int { \cos^2 \theta \ } \ du } $$
(Recall that $ \cos 2 \theta = 2 \cos^2 \theta -1 $ so
that $ \cos^2 \theta = \displaystyle \frac{1}{2}(1+ \cos 2 \theta ) $.)
$$ = \displaystyle{ \int { \frac{1}{2}(1+ \cos 2 \theta ) \ } \ du } $$
$$ = \displaystyle \frac{1}{2} \displaystyle{ \int { (1+ \cos 2 \theta ) \ } \ du } $$
$$ = \displaystyle \frac{1}{2} (\theta + \frac{1}{2} \sin 2 \theta ) + C $$
(Recall that $ \sin 2 \theta = 2 \sin \theta \cos \theta $.)
$$ = \displaystyle \frac{1}{2} (\theta + \frac{1}{2} 2 \sin \theta \cos \theta ) + C $$
$$ = \displaystyle \frac{1}{2} (\theta + \sin \theta \cos \theta ) + C $$
$\Bigg($ We need to write our final answer in terms of $x$.
Since
$$ \sin \theta = x = \displaystyle{ x \over 1 } = \displaystyle{ opposite \over hypotenuse } $$
it follows that $ \theta = \arcsin x $, and from the Pythagorean Theorem that
$$ \displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow $$
$$ (adjacent)^2 + (x)^2 = (1)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{1-x^2} \ \ \longrightarrow $$
$$ \cos \theta = \displaystyle{ adjacent \over hypotenuse }= \displaystyle{ \sqrt{1-x^2} \over 1 } = \sqrt{1-x^2} \ \Bigg) $$
$$ = \displaystyle \frac{1}{2} \Big(\arcsin x + x \sqrt{1-x^2} \Big) \ + C $$
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