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 $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(adjacent)^2 + (x)^2 = (1)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{1-x^2} \ \ \longrightarrow$$ $$\cos \theta = \displaystyle{ adjacent \over hypotenuse }= \displaystyle{ \sqrt{1-x^2} \over 1 } = \sqrt{1-x^2} \ \Bigg)$$ $$= \displaystyle \frac{1}{2} \Big(\arcsin x + x \sqrt{1-x^2} \Big) \ + C$$