Inverse Trigonometric Integrals

The algebra sets the pattern, so read the whole integrand before reaching for a substitution. With only a constant multiple of $du$ on top, and with $a > 0$, a square root over $a^2 - u^2$ points to arcsine while a plain sum $a^2 + u^2$ in a denominator points to arctangent. A denominator of $u\sqrt{u^2 - a^2}$ gives arcsecant, but that form sits outside what AB and BC test. Each pattern also has a negated cofunction version, since $-\arccos(u/a) + C$ differentiates to the same integrand, which is why answer keys sometimes show arccosine.

The numerator decides as much as the denominator does. A factor of $u$ on top changes the answer entirely: $\int \frac{x\,dx}{x^2+1} = \frac{1}{2}\ln(x^2+1) + C$ by substitution rather than an arctangent, and $\int \frac{x\,dx}{\sqrt{1-x^2}} = -\sqrt{1-x^2} + C$ rather than an arcsine, so a numerator such as $x+1$ has to be split into its two pieces first. Completing the square reaches these patterns only when the leftover constant has the right sign, with $a$ taken as the positive square root: $x^2 + 6x + 13$ becomes $(x+3)^2 + 4$, an arctangent with $u = x+3$ and $a = 2$, while $x^2 - 6x + 5$ becomes $(x-3)^2 - 4$, which is $u^2 - a^2$ and calls for partial fractions instead.