Anti-derivatives involving inverse trigonometric functions
Master anti-derivatives involving inverse trig functions for Year 12 Specialist Mathematics in Queensland (QCAA). These are the standard integrals that reverse the derivatives of arcsine, arccosine and arctangent, turning a surd or a sum-of-squares denominator into an inverse-trig function.
You will learn to recognise each standard form, find the value of \(a\), complete the square when the denominator is a quadratic, and evaluate definite integrals as exact multiples of pi — core integration techniques that underpin areas and volumes later in Unit 4.
Theory
Anti-derivatives involving inverse trig functions reverse the derivatives of arcsine, arccosine and arctangent in Year 12 Specialist Mathematics (QCAA, Queensland). The two standard forms are \(\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}}=\arcsin\dfrac{x}{a}+c\) and \(\displaystyle\int\dfrac{dx}{a^2+x^2}=\dfrac{1}{a}\arctan\dfrac{x}{a}+c\). This page shows how to recognise the form, find \(a\), and evaluate the definite integrals as exact multiples of \(\pi\).
These integrals are the reverse of the inverse-trig derivatives. Because \(\dfrac{d}{dx}\!\left[\arcsin\dfrac{x}{a}\right]=\dfrac{1}{\sqrt{a^2-x^2}}\), integrating that expression brings back \(\arcsin\dfrac{x}{a}\). Every result can be checked by differentiating it back to the integrand.
There are three standard forms. A surd denominator \(\sqrt{a^2-x^2}\) gives an arcsine (or, with a minus sign, an arccosine); a sum of squares denominator \(a^2+x^2\) gives an arctangent. The constant that plays the role of \(a^2\) is read straight off the integrand, so \(a=\sqrt{\text{that constant}}\).
The arctangent form carries an extra factor \(\dfrac{1}{a}\) out the front, while the arcsine form does not. When the \(x^2\) term has a coefficient, factor that constant out of the surd or denominator first; when the denominator is a full quadratic, complete the square to reach \((x-h)^2+a^2\).
A definite integral is evaluated by substituting the limits into the antiderivative. Because the inverse-trig functions return angles, these definite integrals come out as exact multiples of \(\pi\), such as \(\dfrac{\pi}{6}\), \(\dfrac{\pi}{4}\) or \(\dfrac{\pi}{8}\).
The two standard forms, with \(a>0\). A surd denominator gives an arcsine:
A sum-of-squares denominator gives an arctangent, with an extra factor \(\dfrac{1}{a}\):
For a quadratic denominator, complete the square first, then match the standard form:
Integrating to an inverse-trig form
- Identify the form: a surd \(\sqrt{a^2-x^2}\) in the denominator points to \(\arcsin\) (or \(\arccos\) if there is a leading minus); a sum of squares \(a^2+x^2\) points to \(\arctan\).
- Find \(a\): write the constant term as \(a^2\), so \(a=\sqrt{\text{constant}}\). If \(x^2\) has a coefficient, factor it out of the surd or denominator first; if the denominator is a full quadratic, complete the square to \((x-h)^2+a^2\).
- Write the antiderivative: \(\arcsin\dfrac{x}{a}+c\) for the surd, or \(\dfrac{1}{a}\arctan\dfrac{x}{a}+c\) for the sum of squares — remembering the \(\dfrac{1}{a}\) on the arctangent only.
- Finish: for a definite integral, substitute the upper and lower limits and simplify to an exact multiple of \(\pi\); otherwise leave \(+c\). Differentiate back to check.
Match the surd to \(\sqrt{a^2-x^2}\) and read off \(a\):
| \(25-x^2\) | \(=\) | \(5^2-x^2\) |
| \(a\) | \(=\) | \(5\) |
Apply the arcsine standard form (no front factor):
| \(\int\dfrac{1}{\sqrt{25-x^2}}\,dx\) | \(=\) | \(\arcsin\dfrac{x}{5}+c\) |
Check by differentiating back:
| \(\dfrac{d}{dx}\!\left[\arcsin\dfrac{x}{5}\right]\) | \(=\) | \(\dfrac{1}{\sqrt{25-x^2}}\ \checkmark\) |
\(\displaystyle\int\dfrac{1}{\sqrt{25-x^2}}\,dx=\arcsin\dfrac{x}{5}+c\).
Match the denominator to \(a^2+x^2\) and read off \(a\):
| \(49+x^2\) | \(=\) | \(7^2+x^2\) |
| \(a\) | \(=\) | \(7\) |
Apply the arctangent standard form — remember the \(\dfrac{1}{a}\) front factor:
| \(\int\dfrac{1}{49+x^2}\,dx\) | \(=\) | \(\dfrac{1}{7}\arctan\dfrac{x}{7}+c\) |
\(\displaystyle\int\dfrac{1}{49+x^2}\,dx=\dfrac{1}{7}\arctan\dfrac{x}{7}+c\).
Integrate first (\(\arcsin\) form, \(a=3\)):
| \(\int_0^{3/2}\dfrac{1}{\sqrt{9-x^2}}\,dx\) | \(=\) | \(\left[\arcsin\dfrac{x}{3}\right]_0^{3/2}\) |
Substitute the upper and lower limits:
| \(=\) | \(\arcsin\dfrac{3/2}{3}-\arcsin 0\) | |
| \(=\) | \(\arcsin\dfrac{1}{2}-0\) | |
| \(=\) | \(\dfrac{\pi}{6}\) |
\(\displaystyle\int_0^{3/2}\dfrac{1}{\sqrt{9-x^2}}\,dx=\dfrac{\pi}{6}\).
Complete the square on the denominator:
| \(x^2+2x+10\) | \(=\) | \((x+1)^2+10-1\) |
| \(=\) | \((x+1)^2+9\) |
Now it is the arctangent form with \((x-h)^2+a^2\), so \(a=3\):
| \((x+1)^2+9\) | \(=\) | \((x+1)^2+3^2\) |
| \(a\) | \(=\) | \(3\) |
Apply the arctangent form (shift \(x\to x+1\), front factor \(\dfrac{1}{3}\)):
| \(\int\dfrac{1}{x^2+2x+10}\,dx\) | \(=\) | \(\dfrac{1}{3}\arctan\dfrac{x+1}{3}+c\) |
\(\displaystyle\int\dfrac{1}{x^2+2x+10}\,dx=\dfrac{1}{3}\arctan\dfrac{x+1}{3}+c\).
Common pitfalls
Frequently asked questions
What is the integral of 1 over the square root of a squared minus x squared?
It is \(\arcsin\dfrac{x}{a}+c\). A surd denominator \(\sqrt{a^2-x^2}\) integrates to an arcsine, with no factor out the front.
What is the integral of 1 over a squared plus x squared?
It is \(\dfrac{1}{a}\arctan\dfrac{x}{a}+c\). A sum-of-squares denominator integrates to an arctangent, and it carries an extra factor \(\dfrac{1}{a}\).
How do I find the value of \(a\)?
The constant in the integrand equals \(a^2\), so \(a=\sqrt{\text{that constant}}\). For \(\sqrt{9-x^2}\), \(a^2=9\) and \(a=3\); for \(25+x^2\), \(a=5\).
When do I get an arcsine and when an arctangent?
A square-root (surd) denominator gives an arcsine (or arccosine if there is a leading minus). A denominator that is a sum of squares, with no surd, gives an arctangent.
How do I integrate when the denominator is a quadratic like \(x^2+2x+10\)?
Complete the square to write it as \((x-h)^2+a^2\). Here \(x^2+2x+10=(x+1)^2+9\), so the integral is \(\dfrac{1}{3}\arctan\dfrac{x+1}{3}+c\).
Why do these definite integrals give answers with \(\pi\) in them?
The antiderivatives are inverse-trig functions, which return angles in radians. Evaluating them at the limits gives exact angles such as \(\dfrac{\pi}{6}\) or \(\dfrac{\pi}{4}\).