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Evaluate $\int \sqrt{1-x^2} d x$.
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Let $x=\sin u$ so $d x=\cos u d u$. Then
$\int \sqrt{1-x^2} d x=\int \sqrt{1-\sin ^2 u} \cos u d u=\int \sqrt{\cos ^2 u} \cos u d u .$
We would like to replace $\sqrt{\cos ^2 u}$ by $\cos u$, but this is valid only if $\cos u$ is positive, since $\sqrt{\cos ^2 u}$ is positive. Consider again the substitution $x=\sin u$. We could just as well think of this as $u=\arcsin x$. If we do, then by the definition of the arcsine, $-\pi / 2 \leq u \leq \pi / 2$, so $\cos u \geq 0$. Then we continue:
\begin{aligned} \int \sqrt{\cos ^2 u} \cos u d u &=\int \cos ^2 u d u=\int \frac{1+\cos 2 u}{2} d u=\frac{u}{2}+\frac{\sin 2 u}{4}+C \\ &=\frac{\arcsin x}{2}+\frac{\sin (2 \arcsin x)}{4}+C . \end{aligned}
This is a perfectly good answer, though the term $\sin (2 \arcsin x)$ is a bit unpleasant. It is possible to simplify this. Using the identity $\sin 2 x=2 \sin x \cos x$, we can write $\sin 2 u=$ $2 \sin u \cos u=2 \sin (\arcsin x) \sqrt{1-\sin ^2 u}=2 x \sqrt{1-\sin ^2(\arcsin x)}=2 x \sqrt{1-x^2}$. Then the full antiderivative is
$\frac{\arcsin x}{2}+\frac{2 x \sqrt{1-x^2}}{4}=\frac{\arcsin x}{2}+\frac{x \sqrt{1-x^2}}{2}+C .$
by Platinum (141,884 points)

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