Trigonometric and rationalising substitutions in HSC Mathematics Extension 2: substitutions for square-root integrands, completing the square, and 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. 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

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

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

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

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.