Antidifferentiation to inverse-circular forms
Master antidifferentiation to inverse-circular forms in Year 12 VCE Specialist Mathematics. 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-circular function. It sits in the Calculus area of study of the VCE Mathematics Study Design (VCAA), within the Integral calculus topic of Unit 4.
You will learn to recognise each standard form \(\int\dfrac{dx}{\sqrt{a^2-x^2}}\) and \(\int\dfrac{dx}{a^2+x^2}\), 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. 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 \(\dfrac{\pi}{4}\).