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Choose the trigonometric substitution $x=\sin u$ with $-\pi /2\le u\le \pi /2$. The cosine is nonnegative there, so $\sqrt{1-x^2}=\cos u$ and $dx=\cos udu$. The double-angle formula then givesSince $u=\arcsin x$ and $\sin 2u=2x\sqrt{1-x^2}$,The formula is an antiderivative on $(-1,1)$ and has continuous endpoint values at $\pm 1$.

A trigonometric substitution is integration by substitution using a trigonometric function to simplify an integrand, especially a square root. For example, $x=a\sin u$, with $a>0$ and $-\pi /2\le u\le \pi /2$, changes $\sqrt{a^2-x^2}$ to $a\cos u$ and $dx$ to $a\cos udu$. The interval restriction fixes the sign of the square root. Similarly $x=a\tan u$ simplifies $\sqrt{a^2+x^2}$. The transformed integral must include the differential and, for a definite integral, the new limits.