Integrate functions whose antiderivative involves $\arcsin$, $\arccos$ or $\arctan$

What this dot point is asking

NESA wants you to recognise integrands of the form $\frac{1}{\sqrt{a^2 - x^2}}$, $\frac{1}{a^2 + x^2}$ and their variants, and write the antiderivative in terms of $\arcsin$ or $\arctan$. You should be able to handle constants, completing the square, and linear substitutions.

The answer

Standard antiderivatives

(Usually we use $\arcsin$ rather than $\arccos$ because the negative sign is awkward; both work.)

Generalising to a constant IMATH_13

Derivation: in the first, substitute $u = \frac{x}{a}$. In the second, substitute $x = a u$ and note $dx = a \, du$, then factor.

Recognising the patterns

The structure $\sqrt{\text{constant} - x^2}$ in the denominator with nothing else fancy in the numerator points to $\arcsin$.

The structure $\text{constant} + x^2$ in the denominator, again with no other troublesome factor, points to $\arctan$.

If the numerator is a multiple of the derivative of the denominator (for $\arctan$) or the derivative of the inside (for $\arcsin$), the integral is straightforward.

Completing the square

If the denominator is $\sqrt{4 + 2 x - x^2}$ or $x^2 + 4 x + 13$, complete the square first to fit a standard form.

$4 + 2 x - x^2 = -(x^2 - 2 x) + 4 = -(x - 1)^2 + 5$, so the integrand becomes $\frac{1}{\sqrt{5 - (x - 1)^2}}$. Substitute $u = x - 1$ and apply the $\arcsin$ pattern.

$x^2 + 4 x + 13 = (x + 2)^2 + 9$. Substitute $u = x + 2$ to get $\frac{1}{u^2 + 9}$, then apply the $\arctan$ pattern.

Linear substitution shortcut

$\int \frac{1}{\sqrt{a^2 - (b x + c)^2}} \, dx = \frac{1}{b} \arcsin \frac{b x + c}{a} + C$.

$\int \frac{1}{a^2 + (b x + c)^2} \, dx = \frac{1}{a b} \arctan \frac{b x + c}{a} + C$.