QCE Specialist Mathematics - Unit 4 - Integration techniques
Inverse Trigonometric Integrals | QCE Specialist Mathematics
Learn QCE Specialist inverse trigonometric functions, their derivatives and integrals that produce arcsine and arctangent forms.
Updated 2026-05-18 - 4 min read
QCAA official coverage - Specialist Mathematics 2025 v1.4
Exact syllabus points covered
- Understand and use the inverse trigonometric functions: arcsine, arccosine and arctangent.
- Use the derivative of the inverse trigonometric functions arcsine, arccosine and arctangent: $\frac{d}{dx}\sin^{-1}\left(\frac{x}{a}\right)=\frac1{\sqrt{a^2-x^2}}$, $\frac{d}{dx}\cos^{-1}\left(\frac{x}{a}\right)=-\frac1{\sqrt{a^2-x^2}}$ and $\frac{d}{dx}\tan^{-1}\left(\frac{x}{a}\right)=\frac{a}{a^2+x^2}$.
- Integrate expressions of the form $\pm\frac1{\sqrt{a^2-x^2}}$ and $\frac{a}{a^2+x^2}$, including $\int \frac1{\sqrt{a^2-x^2}}\,dx=\sin^{-1}\left(\frac{x}{a}\right)+c$, $\int \frac{-1}{\sqrt{a^2-x^2}}\,dx=\cos^{-1}\left(\frac{x}{a}\right)+c$ and $\int \frac{a}{a^2+x^2}\,dx=\tan^{-1}\left(\frac{x}{a}\right)+c$.
Inverse trigonometric functions undo trigonometric functions, but only after the original function has been restricted to a sensible domain. This is why $\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$ each have specific ranges.
Inverse trig functions
The main inverse functions are:
$ \sin^{-1}x,\quad \cos^{-1}x,\quad \tan^{-1}x. $
Their standard ranges are:
$ \sin^{-1}x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right], $
$ \cos^{-1}x\in[0,\pi], $
$ \tan^{-1}x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right). $
This range restriction is what makes the inverse a function rather than many possible answers.
This is also why inverse notation is not the same as reciprocal notation. $\sin^{-1}x$ means arcsine, not $\frac{1}{\sin x}$. The reciprocal sine function is $\cosec x$.
Graphs, domains and ranges
The inverse trig graphs come from restricting the original trig graphs so that each output has only one input. For $\sin x$, the chosen branch is:
$ -\frac{\pi}{2}\le x\le \frac{\pi}{2}. $
Reflecting that branch in the line $y=x$ gives $y=\sin^{-1}x$. Its domain is $[-1,1]$ and its range is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.
For $y=\cos^{-1}x$, the original cosine branch is restricted to $0\le x\le \pi$. That gives domain $[-1,1]$ and range $[0,\pi]$. For $y=\tan^{-1}x$, the tangent branch is restricted to $-\frac{\pi}{2}<x<\frac{\pi}{2}$, so the inverse has domain all real numbers and horizontal asymptotes:
$ y=\frac{\pi}{2}\quad\text{and}\quad y=-\frac{\pi}{2}. $
Derivatives that create integral formulas
The key derivatives are:
$ \frac{d}{dx}\sin^{-1}x=\frac1{\sqrt{1-x^2}}, $
$ \frac{d}{dx}\cos^{-1}x=-\frac1{\sqrt{1-x^2}}, $
$ \frac{d}{dx}\tan^{-1}x=\frac1{1+x^2}. $
With scaling, the common integral forms are:
$ \int \frac1{\sqrt{a^2-x^2}}\,dx=\sin^{-1}\left(\frac{x}{a}\right)+c $
$ \int \frac{-1}{\sqrt{a^2-x^2}}\,dx=\cos^{-1}\left(\frac{x}{a}\right)+c $
and
$ \int \frac{a}{a^2+x^2}\,dx=\tan^{-1}\left(\frac{x}{a}\right)+c. $
These formulas come from differentiating the inverse trig functions with a scaled input:
$ \frac{d}{dx}\sin^{-1}\left(\frac{x}{a}\right)=\frac1{\sqrt{a^2-x^2}}. $
That derivation is useful because it helps you remember where the $a$ belongs.
Scaling traps
The parameter in an inverse trig derivative is easy to put in the wrong place. Work from the chain rule if you are unsure.
For example:
$ \frac{d}{dx}\sin^{-1}(2x) =\frac{2}{\sqrt{1-4x^2}} =\frac{1}{\sqrt{\frac14-x^2}}. $
Both forms are equivalent. The second form shows that this is the same as differentiating $\sin^{-1}\left(\frac{x}{1/2}\right)$.
For arctangent:
$ \frac{d}{dx}\tan^{-1}(3x) =\frac{3}{1+9x^2}. $
This can also be matched to:
$ \frac{d}{dx}\tan^{-1}\left(\frac{x}{a}\right)=\frac{a}{a^2+x^2} $
with $a=\frac13$.
Completing the square
Sometimes the denominator is not already in the right form. Completing the square can reveal it.
For example:
$ 6x-x^2=9-(x-3)^2. $
So
$ \int \frac1{\sqrt{6x-x^2}}\,dx = \int \frac1{\sqrt{9-(x-3)^2}}\,dx. $
Let $u=x-3$. Then:
$ \int \frac1{\sqrt{9-u^2}}\,du =\sin^{-1}\left(\frac{u}{3}\right)+c. $
Substitute back:
$ \sin^{-1}\left(\frac{x-3}{3}\right)+c. $
Completing the square also helps with arctangent forms. For example:
$ x^2-4x+13=(x-2)^2+9. $
So:
$ \int\frac{1}{x^2-4x+13}\,dx =\int\frac{1}{(x-2)^2+3^2}\,dx. $
Let $u=x-2$. Since the numerator is $1$ rather than $3$, factor the constant carefully:
$ \int\frac{1}{u^2+3^2}\,du =\frac13\tan^{-1}\left(\frac{u}{3}\right)+c. $
Thus:
$ \frac13\tan^{-1}\left(\frac{x-2}{3}\right)+c. $
For cosine inverse, remember that:
$ \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x. $
That means an integral that could be written using arcsine can sometimes be written using arccosine as well, with the sign and constant adjusted. For indefinite integrals, two expressions that differ only by a constant are the same family of antiderivatives.
Matching the form
The hardest part is often not the integration; it is reshaping the expression. A denominator like $x^2+6x+13$ is not immediately $a^2+x^2$, but completing the square gives:
$ x^2+6x+13=(x+3)^2+4. $
Then use $u=x+3$ and $a=2$. For a square-root form, completing the square usually aims for:
$ a^2-(x-h)^2. $
If the expression becomes $(x-h)^2-a^2$, it is not an arcsine pattern in this syllabus section.
For definite integrals, check the domain as well as the algebra. An arcsine result such as $\sin^{-1}\left(\frac{x}{a}\right)$ only makes sense for $-\!a\le x\le a$ in the real-valued setting. If the bounds fall outside the valid interval, revisit the original integrand before evaluating.
Why the range matters
Inverse trigonometric notation carries a built-in range. For example, $\sin^{-1}\left(\frac12\right)=\frac{\pi}{6}$, not every angle whose sine is $\frac12$. That is what makes inverse trig functions usable as functions in calculus.
In applied questions, the principal value may still need interpretation. If a physical angle must be obtuse, the calculator's inverse sine output might be a reference angle, and the geometry of the question decides whether a supplementary angle is needed.