What is sign of the square root?

sqrt(a^2cos^2θ) = a|cosθ|. Choose the range of θ so cosine is non-negative, otherwise a sign error creeps in.

NSWMaths Extension 2Syllabus dot point

How do trigonometric and rationalising substitutions transform integrals containing square roots or awkward rational functions into standard forms?

Evaluate integrals using trigonometric substitution and the t = tan(x/2) substitution, including completing the square to reach standard inverse-trigonometric and logarithmic forms

A focused answer to the HSC Maths Extension 2 dot point on substitution integration. Trigonometric substitutions for square roots, completing the square to standard forms, and the t = tan(x/2) substitution for rational trigonometric integrals, with verified worked examples.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Trigonometric substitution for square roots
Completing the square to reach standard forms
The t = tan(x/2) substitution

What this dot point is asking

NESA wants you to evaluate integrals that are not immediately standard by choosing a substitution that simplifies the integrand. Trigonometric substitutions clear square roots of the form $\sqrt{a^2 - x^2}$ and similar; completing the square reduces quadratics under roots or in denominators to standard inverse-trigonometric or logarithmic forms; and the $t = \tan\tfrac{x}{2}$ substitution converts a rational function of $\sin x$ and $\cos x$ into a rational function of $t$ .

Trigonometric substitution for square roots

An integrand containing $\sqrt{a^2 - x^2}$ becomes a clean trigonometric expression under the substitution

x = a\sin\theta, \qquad dx = a\cos\theta\,d\theta.

Then $a^2 - x^2 = a^2(1 - \sin^2\theta) = a^2\cos^2\theta$ , so $\sqrt{a^2 - x^2} = a\cos\theta$ (taking $\theta$ in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ so cosine is non-negative). The square root vanishes, and the integral becomes a trigonometric one. After integrating, convert back to $x$ using a right-triangle or the inverse relation $\theta = \arcsin\tfrac{x}{a}$ .

The related forms $\sqrt{a^2 + x^2}$ and $\sqrt{x^2 - a^2}$ are handled in the same spirit, but the syllabus emphasises the $\sqrt{a^2 - x^2}$ case with $x = a\sin\theta$ .

Completing the square to reach standard forms

Two standard integrals from the reference sheet are

\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan\frac{x}{a} + C, \qquad \int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C.

When the quadratic in the denominator is not in this shape, complete the square first. For instance $x^2 + 4x + 13 = (x + 2)^2 + 9$ , so

\int \frac{dx}{x^2 + 4x + 13} = \int \frac{dx}{(x+2)^2 + 3^2} = \frac{1}{3}\arctan\frac{x + 2}{3} + C.

The same technique with a substitution $u = x + 2$ makes the standard form explicit.

The t = tan(x/2) substitution

For an integral that is a rational function of $\sin x$ and $\cos x$ , the Weierstrass substitution $t = \tan\tfrac{x}{2}$ converts everything to a rational function of $t$ . The identities are

\sin x = \frac{2t}{1 + t^2}, \qquad \cos x = \frac{1 - t^2}{1 + t^2}, \qquad dx = \frac{2}{1 + t^2}\,dt.

After substituting, the integrand becomes algebraic and is typically finished with partial fractions or a standard form. Convert back using $t = \tan\tfrac{x}{2}$ .

Worked example

Evaluate $\displaystyle\int_0^{1} \sqrt{4 - x^2}\,dx$

Use $x = 2\sin\theta$ , so $dx = 2\cos\theta\,d\theta$ and $\sqrt{4 - x^2} = 2\cos\theta$ .

Change limits: when $x = 0$ , $\sin\theta = 0$ so $\theta = 0$ ; when $x = 1$ , $\sin\theta = \tfrac{1}{2}$ so $\theta = \tfrac{\pi}{6}$ .

\int_0^{1}\sqrt{4 - x^2}\,dx = \int_0^{\pi/6} (2\cos\theta)(2\cos\theta)\,d\theta = \int_0^{\pi/6} 4\cos^2\theta\,d\theta.

Use $\cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta)$ :

= \int_0^{\pi/6} 2(1 + \cos 2\theta)\,d\theta = \Big[\,2\theta + \sin 2\theta\,\Big]_0^{\pi/6}.

Evaluate: at $\theta = \tfrac{\pi}{6}$ , $2\theta = \tfrac{\pi}{3}$ and $\sin\tfrac{\pi}{3} = \tfrac{\sqrt{3}}{2}$ ; at $\theta = 0$ both terms are $0$ . So the integral equals

\frac{\pi}{3} + \frac{\sqrt{3}}{2}.

Numerical check: $\tfrac{\pi}{3} \approx 1.047$ and $\tfrac{\sqrt{3}}{2} \approx 0.866$ , giving about $1.913$ , which is a reasonable area under the curve from $0$ to $1$ .

Common traps

Forgetting to change the limits: With a definite integral and a substitution, convert the limits to the new variable, or convert back to $x$ before substituting the original limits. Do not mix the two.
Dropping $dx$ in the substitution: Replacing $x = a\sin\theta$ also replaces $dx$ with $a\cos\theta\,d\theta$ . Omitting the derivative factor gives the wrong integral.
Sign of the square root: $\sqrt{a^2\cos^2\theta} = a|\cos\theta|$ . Choose the range of $\theta$ so cosine is non-negative, otherwise a sign error creeps in.
Mis-stating the t-substitution identities: It is $\cos x = \frac{1 - t^2}{1 + t^2}$ , not $\frac{1 + t^2}{1 - t^2}$ , and $dx = \frac{2}{1 + t^2}dt$ . A wrong identity derails the whole problem.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 HSC3 marksEvaluate integral from 0 to pi/2 of 1/(sin q + 1) dq.

Show worked answer →

Use the substitution t = tan(q/2), so sin q = 2t/(1 + t^2) and dq = 2 dt/(1 + t^2). When q = 0, t = 0; when q = pi/2, t = tan(pi/4) = 1.

Denominator: sin q + 1 = 2t/(1 + t^2) + 1 = (2t + 1 + t^2)/(1 + t^2) = (t + 1)^2/(1 + t^2).

So the integrand times dq becomes
[ (1 + t^2)/(t + 1)^2 ] . [ 2/(1 + t^2) ] dt = 2/(t + 1)^2 dt.

Therefore the integral = integral from 0 to 1 of 2 (t + 1)^(-2) dt = [ -2/(t + 1) ] from 0 to 1 = -2/2 - (-2/1) = -1 + 2 = 1.

Mark notes: 1 mark for the t = tan(q/2) substitution and new limits, 1 mark for simplifying to 2/(t + 1)^2, 1 mark for evaluating the definite integral to 1.

2023 HSC3 marksFind integral of (1 - x)/sqrt(5 - 4x - x^2) dx.

Show worked answer →

Split the numerator so part of it matches the derivative of the expression under the root. Note d/dx (5 - 4x - x^2) = -4 - 2x, and -4 - 2x = 2(-2 - x). Write

1 - x = (-2 - x) + 3.

So integral of (1 - x)/sqrt(5 - 4x - x^2) dx = integral of (-2 - x)/sqrt(5 - 4x - x^2) dx + integral of 3/sqrt(5 - 4x - x^2) dx.

First integral: since d/dx sqrt(5 - 4x - x^2) = (-2 - x)/sqrt(5 - 4x - x^2), this is exactly sqrt(5 - 4x - x^2).

Second integral: complete the square, 5 - 4x - x^2 = 9 - (x + 2)^2. Then integral of 3/sqrt(9 - (x + 2)^2) dx = 3 arcsin((x + 2)/3).

Combining: integral = sqrt(5 - 4x - x^2) + 3 arcsin((x + 2)/3) + C.

Mark notes: 1 mark for splitting 1 - x to expose the derivative -2 - x, 1 mark for the square-root term, 1 mark for completing the square and the 3 arcsin((x + 2)/3) term.

2022 HSC3 marksUsing the substitution t = tan(x/2), find integral of 1/(1 + cos x - sin x) dx.

Show worked answer →

With t = tan(x/2): cos x = (1 - t^2)/(1 + t^2), sin x = 2t/(1 + t^2), and dx = 2 dt/(1 + t^2).

Rewrite the denominator over the common denominator 1 + t^2:
1 + cos x - sin x = (1 + t^2)/(1 + t^2) + (1 - t^2)/(1 + t^2) - 2t/(1 + t^2)
= (1 + t^2 + 1 - t^2 - 2t)/(1 + t^2) = (2 - 2t)/(1 + t^2) = 2(1 - t)/(1 + t^2).

So the integral becomes
integral of [ 2/(1 + t^2) ] / [ 2(1 - t)/(1 + t^2) ] dt = integral of 1/(1 - t) dt = -ln|1 - t| + C.

Substituting back t = tan(x/2):

integral = -ln|1 - tan(x/2)| + C.

Mark notes: 1 mark for the standard t = tan(x/2) substitutions, 1 mark for simplifying the denominator to 2(1 - t)/(1 + t^2), 1 mark for integrating to -ln|1 - tan(x/2)| + C.