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.

NESA 2024 HSC-style practice: Evaluate _0^π/2 (1)/(sinθ + 1) \, dθ.

Use the substitution t = tan(θ)/(2), so sinθ = (2t)/(1 + t^2) and dθ = (2)/(1 + t^2)\,dt. When θ = 0, t = 0; when θ = (π)/(2), t = tan(π)/(4) = 1. Denominator: sinθ + 1 = (2t)/(1 + t^2) + 1 = (2t + 1 + t^2)/(1 + t^2) = ((t + 1)^2)/(1 + t^2). So the integrand times dθ becomes (1 + t^2)/((t + 1)^2) · (2)/(1 + t^2)\,dt = (2)/((t + 1)^2)\,dt. Therefore the integral = _0^1 2(t + 1)^-2\,dt = [(-2)/(t + 1)]_0^1 = -1 - (-2) = 1. Mark notes: 1 mark for the t = tan(θ)/(2) substitution and new limits, 1 mark for simplifying to (2)/((t + 1)^2), 1 mark for evaluating the definite integral to 1.

NESA 2022 HSC-style practice: Using the substitution t = tan(x)/(2), find (1)/(1 + cos x - sin x) \, dx.

With t = tan(x)/(2): cos x = (1 - t^2)/(1 + t^2), sin x = (2t)/(1 + t^2), and dx = (2)/(1 + t^2)\,dt. Rewrite the denominator over the common denominator 1 + t^2: 1 + cos x - sin x = ((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 int (2/(1 + t^2))/(2(1 - t)/(1 + t^2))\,dt = int (1)/(1 - t)\,dt = -ln|1 - t| + C. Substituting back t = tan(x)/(2): int = -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.

§-Syllabus dot point

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 with a reference triangle, 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 4.812 min answerUpdated 2026-06-20

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
Splitting a numerator to expose the derivative of the root
The t = tan(x/2) substitution
How exam questions ask about these substitutions

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. Three related techniques are examined: trigonometric substitutions that clear a square root of the form $\sqrt{a^2 - x^2}$ (and its relatives $\sqrt{a^2 + x^2}$ and $\sqrt{x^2 - a^2}$ ); completing the square to reduce a quadratic under a root or in a denominator to a standard inverse-trigonometric or logarithmic form; and the rationalising substitution $t = \tan\tfrac{x}{2}$ , which turns a rational function of $\sin x$ and $\cos x$ into an algebraic 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.

The reason this works is the Pythagorean identity: $a^2 - x^2 = a^2(1 - \sin^2\theta) = a^2\cos^2\theta$ , so $\sqrt{a^2 - x^2} = a\cos\theta$ (taking $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ so that cosine is non-negative and the square root stays positive). The substitution is engineered so that the awkward square root collapses to a single trig factor. After integrating in $\theta$ , convert back to $x$ using either the inverse relation $\theta = \arcsin\tfrac{x}{a}$ or, more reliably for mixed expressions, a reference right-triangle.

The reference right-triangle for converting back

The cleanest way to turn a $\theta$ -answer back into $x$ is to draw the substitution as a right-triangle. For $x = a\sin\theta$ , sine is opposite over hypotenuse, so put the opposite side equal to $x$ and the hypotenuse equal to $a$ . Pythagoras then gives the adjacent side as $\sqrt{a^2 - x^2}$ , and every trig function of $\theta$ can be read straight off the triangle.

Reading off the triangle, $\sin\theta = \dfrac{x}{a}$ , $\cos\theta = \dfrac{\sqrt{a^2 - x^2}}{a}$ , and $\tan\theta = \dfrac{x}{\sqrt{a^2 - x^2}}$ . So after integrating in $\theta$ you can rewrite any leftover $\cos\theta$ , $\sin 2\theta = 2\sin\theta\cos\theta$ , and similar terms purely in $x$ without wrestling with nested inverse functions. For the companion form $x = a\tan\theta$ you draw the same kind of triangle with opposite $x$ , adjacent $a$ , and hypotenuse $\sqrt{a^2 + x^2}$ .

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 (or under the root) is not already in this shape, complete the square first to expose it. For instance $x^2 + 4x + 13 = (x + 2)^2 + 9 = (x + 2)^2 + 3^2$ , 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 substitution $u = x + 2$ makes the standard form explicit, but with practice you can read it straight off the completed square. The same idea handles a quadratic under a root, turning it into the $\arcsin$ form, as in the 2023 HSC question above where $5 - 4x - x^2 = 9 - (x + 2)^2$ .

Splitting a numerator to expose the derivative of the root

A frequent exam pattern is $\displaystyle\int \frac{\text{linear}}{\sqrt{\text{quadratic}}}\,dx$ . The trick is to split the linear numerator into a piece proportional to the derivative of the quadratic plus a constant. The derivative piece integrates directly to a square root (since $\tfrac{d}{dx}\sqrt{q} = \tfrac{q'}{2\sqrt{q}}$ ), and the constant piece is finished by completing the square to an $\arcsin$ . This is exactly the structure of the 2023 HSC question, and recognising it saves you from attempting a full trig substitution where a clever split is faster.

The t = tan(x/2) substitution

For an integral that is a rational function of $\sin x$ and $\cos x$ , the rationalising substitution $t = \tan\tfrac{x}{2}$ (the Weierstrass, or t-formula, substitution) converts everything to a rational function of $t$ , which can then be finished with partial fractions or a standard form. 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. Convert back at the end using $t = \tan\tfrac{x}{2}$ . For a definite integral, change the limits to $t$ -values instead (for example $x = \tfrac{\pi}{2}$ gives $t = \tan\tfrac{\pi}{4} = 1$ ), which avoids back-substitution entirely.

How exam questions ask about these substitutions

"Evaluate $\int \sqrt{a^2 - x^2}\,dx$ " or an area under a semicircle. Use $x = a\sin\theta$ , reduce to $\int \cos^2\theta\,d\theta$ , and apply the double-angle identity.
"Find $\int \dfrac{dx}{\sqrt{a^2 - x^2}}$ / $\int \dfrac{dx}{a^2 + x^2}$ " off the reference sheet. Quote the standard $\arcsin$ or $\arctan$ form directly.
A quadratic under a root or in a denominator that is not standard. Complete the square first, then read off the $\arcsin$ or $\arctan$ .
"Find $\int \dfrac{\text{linear}}{\sqrt{\text{quadratic}}}\,dx$ ." Split the numerator into (derivative of the quadratic) plus a constant; one part gives a square root, the other an $\arcsin$ .
"Using the substitution $t = \tan\tfrac{x}{2}$ , find ..." (a rational function of $\sin x$ , $\cos x$ ). Apply the three t-identities, simplify to an algebraic integral, then finish and back-substitute.
"Hence find the exact value ..." (definite). Change limits to the new variable, integrate, evaluate; "exact" means surds, $\pi$ , $\ln$ stay, no decimals.

Exam technique

State the substitution and its differential on the first line ("let $x = a\sin\theta$ , so $dx = a\cos\theta\,d\theta$ "), and immediately show the square root collapsing ( $\sqrt{a^2 - x^2} = a\cos\theta$ ); that is usually the first mark. For a definite integral either change the limits to the new variable and integrate straight through, or keep the original limits and back-substitute, but never mix the two. When you must return a trig-substitution answer to $x$ , draw the reference triangle and read off each ratio rather than juggling inverse functions. When a question names the substitution, using it is compulsory, a correct answer by another route can be marked down. Keep the modulus inside any $\ln$ , choose $\theta$ so the square root is non-negative, and keep $+C$ on indefinite answers.

Evaluate

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

(trig substitution)

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

The square root $\sqrt{4 - x^2}$ has the form $\sqrt{a^2 - x^2}$ with $a = 2$ , which signals the substitution $x = 2\sin\theta$ . The Pythagorean identity $1 - \sin^2\theta = \cos^2\theta$ then eliminates the root entirely.

Apply the trigonometric substitution: Set $x = 2\sin\theta$ , so $dx = 2\cos\theta\,d\theta$ . The square root simplifies because $4 - x^2 = 4(1 - \sin^2\theta) = 4\cos^2\theta$ , giving $\sqrt{4 - x^2} = 2\cos\theta$ (taking $\theta \in [0, \tfrac{\pi}{2}]$ so cosine is non-negative).

Convert the limits of integration: Change the limits to match the new variable. When $x = 0$ : $\sin\theta = 0$ , so $\theta = 0$ . When $x = 1$ : $\sin\theta = \tfrac{1}{2}$ , so $\theta = \tfrac{\pi}{6}$ . Now substitute everything:

\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 the double-angle identity to integrate: The power $\cos^2\theta$ cannot be integrated directly, but the identity $\cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta)$ converts it to a standard form:

\int_0^{\pi/6} 4\cos^2\theta\,d\theta = \int_0^{\pi/6} 2(1 + \cos 2\theta)\,d\theta = \Big[\,2\theta + \sin 2\theta\,\Big]_0^{\pi/6}.

Evaluate at the limits: 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 definite integral is

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

As a numerical check: $\tfrac{\pi}{3} \approx 1.047$ and $\tfrac{\sqrt{3}}{2} \approx 0.866$ , giving about $1.913$ , a reasonable area under $\sqrt{4 - x^2}$ from $x = 0$ to $x = 1$ .

Final answer: $\displaystyle\int_0^{1} \sqrt{4 - x^2}\,dx = \dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2}$ .

Evaluate

\displaystyle\int \frac{dx}{(4 + x^2)^{3/2}}

using

x = 2\tan\theta

and the reference triangle

Evaluate $\displaystyle\int \frac{dx}{(4 + x^2)^{3/2}}$ using $x = 2\tan\theta$ and the reference triangle

Here the relevant form is $\sqrt{a^2 + x^2}$ with $a = 2$ , so the matching substitution is $x = a\tan\theta$ . This worked example also shows the triangle doing the back-substitution.

Apply the tangent substitution: Set $x = 2\tan\theta$ , so $dx = 2\sec^2\theta\,d\theta$ . Then $4 + x^2 = 4(1 + \tan^2\theta) = 4\sec^2\theta$ , so $(4 + x^2)^{3/2} = (4\sec^2\theta)^{3/2} = 8\sec^3\theta$ .

Simplify the integrand to a single trig function:

\int \frac{dx}{(4 + x^2)^{3/2}} = \int \frac{2\sec^2\theta}{8\sec^3\theta}\,d\theta = \frac{1}{4}\int \frac{1}{\sec\theta}\,d\theta = \frac{1}{4}\int \cos\theta\,d\theta.

Integrate in $\theta$ :

\frac{1}{4}\int \cos\theta\,d\theta = \frac{1}{4}\sin\theta + C.

Convert back to $x$ with the reference triangle: Since $x = 2\tan\theta$ , draw a right-triangle with opposite $x$ and adjacent $2$ ; the hypotenuse is $\sqrt{4 + x^2}$ . Reading off, $\sin\theta = \dfrac{x}{\sqrt{4 + x^2}}$ . Therefore

\int \frac{dx}{(4 + x^2)^{3/2}} = \frac{1}{4}\cdot\frac{x}{\sqrt{4 + x^2}} + C = \frac{x}{4\sqrt{4 + x^2}} + C.

Final answer: $\displaystyle\int \frac{dx}{(4 + x^2)^{3/2}} = \frac{x}{4\sqrt{4 + x^2}} + C$ .

Edge case: a rational trig integral with the

t = \tan\tfrac{x}{2}

substitution,

\displaystyle\int \frac{dx}{1 + \sin x}

Evaluate $\displaystyle\int \frac{dx}{1 + \sin x}$ with the $t = \tan\tfrac{x}{2}$ substitution

When the integrand is a rational function of $\sin x$ (or $\cos x$ ) with nothing to cancel, the $t$ -substitution rationalises it.

Apply the three t-identities: Let $t = \tan\tfrac{x}{2}$ , so $\sin x = \dfrac{2t}{1 + t^2}$ and $dx = \dfrac{2}{1 + t^2}\,dt$ . The denominator becomes

1 + \sin x = 1 + \frac{2t}{1 + t^2} = \frac{1 + t^2 + 2t}{1 + t^2} = \frac{(1 + t)^2}{1 + t^2}.

Simplify to an algebraic integral: Dividing $dx$ by the denominator, the $1 + t^2$ factors cancel:

\int \frac{dx}{1 + \sin x} = \int \frac{1 + t^2}{(1 + t)^2}\cdot\frac{2}{1 + t^2}\,dt = \int \frac{2}{(1 + t)^2}\,dt.

Integrate by the power rule and back-substitute:

\int 2(1 + t)^{-2}\,dt = \frac{-2}{1 + t} + C = \frac{-2}{1 + \tan\frac{x}{2}} + C.

Final answer: $\displaystyle\int \frac{dx}{1 + \sin x} = -\frac{2}{1 + \tan\frac{x}{2}} + C$ .

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.
Choosing the wrong trig substitution: Use $x = a\sin\theta$ for $\sqrt{a^2 - x^2}$ but $x = a\tan\theta$ for $\sqrt{a^2 + x^2}$ . Matching the wrong one leaves a square root that will not simplify.
Mis-stating the t-substitution identities: It is $\cos x = \dfrac{1 - t^2}{1 + t^2}$ , not $\dfrac{1 + t^2}{1 - t^2}$ , and $dx = \dfrac{2}{1 + t^2}\,dt$ . 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

\displaystyle\int_0^{\pi/2} \frac{1}{\sin\theta + 1} \, d\theta

Show worked answer →

Use the substitution $t = \tan\frac{\theta}{2}$ , so $\sin\theta = \frac{2t}{1 + t^2}$ and $d\theta = \frac{2}{1 + t^2}\,dt$ . When $\theta = 0$ , $t = 0$ ; when $\theta = \frac{\pi}{2}$ , $t = \tan\frac{\pi}{4} = 1$ .

Denominator: $\sin\theta + 1 = \frac{2t}{1 + t^2} + 1 = \frac{2t + 1 + t^2}{1 + t^2} = \frac{(t + 1)^2}{1 + t^2}$ .

So the integrand times $d\theta$ becomes $\frac{1 + t^2}{(t + 1)^2} \cdot \frac{2}{1 + t^2}\,dt = \frac{2}{(t + 1)^2}\,dt$ .

Therefore the integral $= \int_0^1 2(t + 1)^{-2}\,dt = \left[\frac{-2}{t + 1}\right]_0^1 = -1 - (-2) = 1$ .

Mark notes: 1 mark for the $t = \tan\frac{\theta}{2}$ substitution and new limits, 1 mark for simplifying to $\frac{2}{(t + 1)^2}$ , 1 mark for evaluating the definite integral to $1$ .

2023 HSC3 marksFind

\displaystyle\int \frac{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 $\frac{d}{dx}(5 - 4x - x^2) = -4 - 2x = 2(-2 - x)$ . Write

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

So $\int \frac{1 - x}{\sqrt{5 - 4x - x^2}}\,dx = \int \frac{-2 - x}{\sqrt{5 - 4x - x^2}}\,dx + \int \frac{3}{\sqrt{5 - 4x - x^2}}\,dx$ .

First integral: since $\frac{d}{dx}\sqrt{5 - 4x - x^2} = \frac{-2 - x}{\sqrt{5 - 4x - x^2}}$ , this is exactly $\sqrt{5 - 4x - x^2}$ .

Second integral: complete the square, $5 - 4x - x^2 = 9 - (x + 2)^2$ . Then $\int \frac{3}{\sqrt{9 - (x + 2)^2}}\,dx = 3\arcsin\frac{x + 2}{3}$ .

Combining: $\int = \sqrt{5 - 4x - x^2} + 3\arcsin\frac{x + 2}{3} + C$ .

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 $3\arcsin\frac{x + 2}{3}$ term.

2022 HSC3 marksUsing the substitution

t = \tan\frac{x}{2}

, find

\displaystyle\int \frac{1}{1 + \cos x - \sin x} \, dx

Show worked answer →

With $t = \tan\frac{x}{2}$ : $\cos x = \frac{1 - t^2}{1 + t^2}$ , $\sin x = \frac{2t}{1 + t^2}$ , and $dx = \frac{2}{1 + t^2}\,dt$ .

Rewrite the denominator over the common denominator $1 + t^2$ :
$1 + \cos x - \sin x = \frac{(1 + t^2) + (1 - t^2) - 2t}{1 + t^2} = \frac{2 - 2t}{1 + t^2} = \frac{2(1 - t)}{1 + t^2}$ .

So the integral becomes $\int \frac{2/(1 + t^2)}{2(1 - t)/(1 + t^2)}\,dt = \int \frac{1}{1 - t}\,dt = -\ln|1 - t| + C$ .

Substituting back $t = \tan\frac{x}{2}$ : $\int = -\ln\left|1 - \tan\frac{x}{2}\right| + C$ .

Mark notes: 1 mark for the standard $t = \tan\frac{x}{2}$ substitutions, 1 mark for simplifying the denominator to $\frac{2(1 - t)}{1 + t^2}$ , 1 mark for integrating to $-\ln\left|1 - \tan\frac{x}{2}\right| + C$ .