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 a constant $a \neq 1$, complete the square when the denominator is a full quadratic, and use a linear substitution when the inside is $bx + c$.

This dot point is really a pattern-recognition test. The two target integrals are short to quote; the marks are for seeing that a given integrand is one of them in disguise. A square root and a minus sign point one way; a sum of squares with no root points the other. The deeper skill, and the one that separates a $3$-mark answer from a $1$-mark one, is reshaping a quadratic like $x^2 - 6x + 25$ into a completed square so the standard form appears. That reshaping is the same "shift the graph to the origin" move you use everywhere in the course, seen here through the integral.

The answer

The two standard antiderivatives

The $\arccos$ form, $\int \frac{-1}{\sqrt{1 - x^2}} \, dx = \arccos x + C$, exists but is rarely used: it only differs from the $\arcsin$ result by a sign, so the convention is to keep $\arcsin$ and carry any minus sign outside the integral.

The geometry is worth holding onto. The $\arctan$ integrand $\frac{1}{1 + x^2}$ is a bell-shaped curve, and the definite integral is the area under it. From $0$ to $1$ that area is exactly $\arctan 1 = \frac{\pi}{4}$, which is the value the $2020$ HSC asked for.

Generalising to a constant $a$

The general forms come straight from a substitution. For the $\arcsin$ form, substitute $u = \frac{x}{a}$ so $du = \frac{1}{a}\,dx$:

For the $\arctan$ form, substitute $x = a u$ (so $dx = a\,du$) and factor $a^2$ out of the denominator; one factor of $a$ cancels and one survives as the $\frac{1}{a}$:

That single surviving $\frac{1}{a}$ is the most-dropped piece in the topic, so it is worth seeing exactly where it comes from rather than memorising it as a rule.

Recognising the patterns

Three quick checks sort almost every integrand:

• Square root of (constant minus $x^2$) in the denominator, numerator a constant: an $\arcsin$. Example $\frac{1}{\sqrt{9 - x^2}}$.
• Constant plus $x^2$ in the denominator (no root), numerator a constant: an $\arctan$. Example $\frac{1}{9 + x^2}$.
• Numerator is the derivative of the denominator: this is a logarithm, $\ln|\text{denominator}|$, not an inverse trig. Example $\frac{2x}{x^2 + 9}$ integrates to $\ln(x^2 + 9)$, not an $\arctan$.

That last check is the crucial fork: $\frac{1}{x^2 + 9}$ is an $\arctan$, but $\frac{2x}{x^2 + 9}$ (numerator $= \frac{d}{dx}(x^2 + 9)$) is a log. Always glance at the numerator before committing.

Completing the square, stage by stage

When the denominator is a full quadratic such as $x^2 - 6x + 25$, it is not in the form $a^2 + x^2$ because of the linear $-6x$ term. Completing the square removes the linear term and exposes the standard form. The four panels track the integrand $\frac{1}{x^2 - 6x + 25}$ through the method.

Stage 1, spot that the integrand is off-centre. Graphing $y = \frac{1}{x^2 - 6x + 25}$ shows a bell whose peak is at $x = 3$, not at the $y$-axis. A bell centred away from the axis is the visual signal that the denominator hides a completed square $(x - h)^2 + k$ with $h \neq 0$; here $h = 3$.

Stage 2, complete the square to read off $a$. Take half the coefficient of $x$, square it, and rewrite: $x^2 - 6x + 25 = (x - 3)^2 - 9 + 25 = (x - 3)^2 + 16$. Now the denominator is in the form $(x - 3)^2 + a^2$ with $a^2 = 16$, so $a = 4$. The completed square also confirms the peak position $x = 3$ from the diagram.

Stage 3, substitute $u = x - 3$ to centre the bell. Let $u = x - 3$, so $du = dx$. This shifts the graph three units left, moving the peak onto the axis, and turns the integral into $\int \frac{1}{u^2 + 16}\,du$, which is the pure $\arctan$ form with $a = 4$. The substitution is the algebraic version of sliding the curve to the origin.

Stage 4, integrate and undo the substitution. Apply the $\arctan$ form with $a = 4$: $\int \frac{1}{u^2 + 16}\,du = \frac{1}{4}\arctan\frac{u}{4} + C$. The antiderivative is a scaled $\arctan$ that flattens towards horizontal asymptotes. Finally replace $u = x - 3$ to return to the original variable:

The $\arcsin$ version of the method is identical except the completed square sits under a square root, and a leading minus sign on $x^2$ must be factored out first. For $\sqrt{4 + 2x - x^2}$, write $4 + 2x - x^2 = -(x^2 - 2x) + 4 = -((x - 1)^2 - 1) + 4 = 5 - (x - 1)^2$, so the integrand becomes $\frac{1}{\sqrt{5 - (x - 1)^2}}$; substitute $u = x - 1$ and apply the $\arcsin$ form with $a = \sqrt{5}$.

Linear substitution shortcut

When the inside is linear, $bx + c$, you can quote the adjusted forms directly (the $\frac{1}{b}$ comes from $du = b\,dx$):

How exam questions ask about inverse-trig integrals

The wording flags how much reshaping is needed before the standard form appears:

• "Evaluate $\int \frac{1}{\sqrt{1 - x^2}}\,dx$" or "$\int \frac{1}{1 + x^2}\,dx$" (constants only). A direct quote of the standard antiderivative; $1$ to $2$ marks. Watch the $\frac{1}{a}$ when $a \neq 1$.
• "Evaluate $\int_0^{1/2} \ldots$ giving an exact answer." A definite integral: find the antiderivative, substitute the bounds, and keep exact values such as $\frac{\pi}{6}$ and $\frac{\pi}{4}$.
• "Find $\int \frac{1}{x^2 + bx + c}\,dx$" with a linear term. Complete the square first, then substitute and apply the $\arctan$ form; usually $3$ marks, one of them for the completed square.
• "Find $\int \frac{1}{\sqrt{\ldots}}\,dx$" with a quadratic under the root. Complete the square (factoring out any $-x^2$) to reach $\sqrt{a^2 - (x - h)^2}$, then the $\arcsin$ form.
• "Find $\int \frac{px + q}{x^2 + bx + c}\,dx$" (linear numerator). Split it: write the numerator as a multiple of the denominator's derivative plus a constant. The first piece is a log, the second an $\arctan$ after completing the square.
• A definite integral with limits outside $[-a, a]$ for an $\arcsin$ form. A trap or a prompt to notice the integrand is undefined there; check the domain before integrating.