The final lesson of Further Integration. You will now stitch together every technique in MEX-C1, $u$-substitution, trig substitution, integration by parts, the IBP reduction loop, partial fractions, completing the square, and the $t = \tan(x/2)$ Weierstrass, into a single decision tree, then learn how to deploy them under HSC conditions: state the technique, show the substitution explicitly, change limits, and leave the answer in exact form.
+5 XP warm-up For each integrand, name the technique AND write the first line of working (no need to finish).
For each integrand, name the technique AND write the first line of working (no need to finish). (a) $\int \dfrac{x}{\sqrt{x^2 + 1}}\,dx$ (b) $\int x \sin x\,dx$ (c) $\int \dfrac{1}{(x-1)(x+2)}\,dx$ (d) $\int \dfrac{1}{\sqrt{9 - x^2}}\,dx$.
+5 XP to read The two habits that separate full marks from partial credit: classify the integrand before any algebra (state...
The two habits that separate full marks from partial credit: classify the integrand before any algebra (state the technique by name in the first line), and show the substitution explicitly ("Let $u = \ldots$; then $du = \ldots$"). Markers cannot give technique marks for working they cannot reconstruct.
The look-classify-state-execute decision tree: (1) look at the integrand; (2) classify into one of six standard forms; (3) state the technique by name; (4) execute. Most HSC questions reward steps 1–3 with method marks before any computation is checked.
Forms: rational · radical · log/exp · trig powers · trig irrationals · inverse
Every Extension 2 integrand falls into a small number of recognisable forms.
Every Extension 2 integrand falls into a small number of recognisable forms. Reading the integrand against the tree below picks the technique in seconds.
Worked through the hook: $\int \dfrac{1}{x^2 \sqrt{x^2 + 4}}\,dx$ has a $\sqrt{a^2 + x^2}$ radical $\Rightarrow$ trig sub $x = 2\tan\theta$. Opening line: "Let $x = 2\tan\theta$; then $dx = 2\sec^2\theta\,d\theta$ and $\sqrt{x^2 + 4} = 2\sec\theta$." That single line earns method marks; the algebra that follows is reward.
Decision tree: inner function $\to$ $u$-sub; radical $\to$ trig sub; product $\to$ IBP; rational $\to$ partial fractions; quadratic $\to$ complete square; trig rational $\to$ $t = \tan(x/2)$ · Standard radical pairs: $\sqrt{a^2 - x^2} \to a\sin\theta$; $\sqrt{a^2 + x^2} \to a\tan\theta$; $\sqrt{x^2 - a^2} \to a\sec\theta$
Pause, copy the complete Module 15 decision tree (six cases) and the three trig-sub radical pairs into your book.
Quick check: Which is the best opening technique for $\displaystyle\int \dfrac{x^2}{\sqrt{4 - x^2}}\,dx$?
We just saw the Module 15 decision tree: inner function → -sub; radical → trig sub; product → IBP; rational → partial...
We just saw the Module 15 decision tree: inner function → $u$-sub; radical → trig sub; product → IBP; rational → partial fractions; quadratic → complete square; rational trig → $t = \tan(x/2)$. That raises a question: what does a full-mark HSC response look like from the first line? This card answers it → state form, name substitution, show $u$ and $du$ (or $f, dg, df, g$) as a separate line, change limits if definite, give exact answer.
HSC markers reward a specific opening structure. Every integration question gets these four lines before any heavy computation:
Why this works. Markers can reconstruct your reasoning even if a later line has an arithmetic slip. Method marks are awarded for lines 1–3 independently of line 4. Silent algebra forfeits those marks.
Four-line opening: identify form $\to$ state substitution $\to$ transform $\to$ execute · State technique by name (IBP, $u$-sub, trig sub, partial fractions) · Show $u, du$ (or $f, dg, df, g$) as a separate line · For definite integrals: change limits in the transform line (or back-substitute at the end) · Answer in exact form, $\pi$, $\ln$, $e$, $\arctan$, surds, never decimals unless asked
Pause, copy the four-line HSC opening (identify form, state substitution, transform, execute), the requirement to show $u, du$ on a separate line, and the exact-form instruction ($\pi$, $\ln$, $e$, $\arctan$, surds) into your book.
Did you get this? True or false: in an HSC integration question, method marks are awarded for stating the technique and writing the substitution explicitly, even if the final arithmetic is wrong.
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle\int \frac{1}{x^2 \sqrt{x^2 + 4}}\,dx$ using a trigonometric substitution.
Evaluate $\displaystyle\int \frac{1}{(x - 1)(x^2 + 1)}\,dx$ in exact form.
Let $I_n = \displaystyle\int_0^{\pi/2} \sin^n x\,dx$ for $n \geq 0$. Show that $I_n = \dfrac{n-1}{n} I_{n-2}$, then evaluate $I_5$.
Fill the gap: For $\int \dfrac{dx}{\sqrt{a^2 - x^2}}$ use the trig substitution $x = $ ; for $\int \dfrac{dx}{a^2 + x^2}$ use $x = $ .
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for $\int \dfrac{dx}{\sqrt{x^2 - 9}}$, the correct trig substitution is $x = 3\sec\theta$.
Activities · practice with the ideas
Activities · practice with the ideas
For each integrand, write ONLY the opening line (technique + substitution). Do not solve. (a) $\int x^2 \cos x\,dx$ (b) $\int \dfrac{dx}{x^2 + 6x + 13}$ (c) $\int \dfrac{dx}{\sqrt{16 - x^2}}$.
Evaluate $\displaystyle\int_0^{1} \dfrac{1}{(x+1)(x+2)}\,dx$ in exact form.
Use IBP to evaluate $\displaystyle\int_0^{\pi/2} x \cos x\,dx$ in exact form.
Evaluate $\displaystyle\int_0^{2} \dfrac{1}{\sqrt{4 - x^2}}\,dx$ using $x = 2\sin\theta$.
Use the reduction $I_n = \tfrac{n-1}{n}I_{n-2}$ with $I_0 = \pi/2$ to evaluate $\displaystyle\int_0^{\pi/2}\sin^6 x\,dx$.
Odd one out: Three of these integrals call for trig substitution as the first move. Which one does NOT?
Earlier you drafted the opening line for $\int \dfrac{1}{x^2\sqrt{x^2+4}}\,dx$, naming the technique, choosing the substitution, and predicting the transformed integral.
The correct opening, "Let $x = 2\tan\theta$; then $dx = 2\sec^2\theta\,d\theta$ and $\sqrt{x^2 + 4} = 2\sec\theta$", is a full method mark, before any further algebra. The lesson of Module 15 is that integration is recognition: the integrand tells you the technique, and the marks live in stating it cleanly. Carry the decision tree and the four-line opening into every HSC question.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Evaluate $\displaystyle\int \dfrac{x + 3}{x^2 + 4x + 5}\,dx$. State the technique explicitly. (3 marks)
Q2. Evaluate $\displaystyle\int_0^{1} x \arctan x\,dx$ in exact form using integration by parts. (4 marks)
Q3. Let $I_n = \displaystyle\int_0^{1} x^n e^{-x}\,dx$ for $n \geq 0$. Show that $I_n = -e^{-1} + nI_{n-1}$, then evaluate $I_3$ in exact form. (4 marks)
Activity answers:
1. (a) IBP, $f = x^2$, $dg = \cos x\,dx \Rightarrow df = 2x\,dx$, $g = \sin x$. (b) Complete the square: $x^2 + 6x + 13 = (x+3)^2 + 4$; sub $u = x + 3$, giving $\int du/(u^2 + 4)$. (c) Trig sub $x = 4\sin\theta$, $dx = 4\cos\theta\,d\theta$.
2. Partial fractions: $\dfrac{1}{(x+1)(x+2)} = \dfrac{1}{x+1} - \dfrac{1}{x+2}$. $\int_0^1\left[\dfrac{1}{x+1} - \dfrac{1}{x+2}\right]dx = [\ln(x+1) - \ln(x+2)]_0^1 = (\ln 2 - \ln 3) - (0 - \ln 2) = 2\ln 2 - \ln 3 = \ln(4/3)$.
3. $f = x$, $dg = \cos x\,dx \Rightarrow df = dx$, $g = \sin x$. $\int_0^{\pi/2} x\cos x\,dx = [x\sin x]_0^{\pi/2} - \int_0^{\pi/2}\sin x\,dx = \tfrac{\pi}{2} - 1$.
4. $x = 2\sin\theta$, $dx = 2\cos\theta\,d\theta$, $\sqrt{4 - x^2} = 2\cos\theta$. Limits $0 \to \pi/2$. $\int_0^{\pi/2}(2\cos\theta)/(2\cos\theta)\,d\theta = \int_0^{\pi/2} d\theta = \pi/2$. (Or recognise as $[\arcsin(x/2)]_0^2 = \pi/2$.)
5. $I_6 = \tfrac{5}{6}I_4 = \tfrac{5}{6}\cdot\tfrac{3}{4}I_2 = \tfrac{5}{6}\cdot\tfrac{3}{4}\cdot\tfrac{1}{2}I_0 = \tfrac{5}{6}\cdot\tfrac{3}{4}\cdot\tfrac{1}{2}\cdot\tfrac{\pi}{2} = \dfrac{15\pi}{96} = \dfrac{5\pi}{32}$.
Q1 (3 marks): Split numerator: $x + 3 = \tfrac{1}{2}(2x + 4) + 1$ [1]. So $\int \dfrac{x+3}{x^2+4x+5}\,dx = \tfrac{1}{2}\int \dfrac{2x+4}{x^2+4x+5}\,dx + \int \dfrac{dx}{(x+2)^2 + 1} = \tfrac{1}{2}\ln(x^2 + 4x + 5) + \arctan(x + 2) + C$ [2, one for $\ln$ piece, one for $\arctan$ piece].
Q2 (4 marks): IBP with $f = \arctan x$, $dg = x\,dx \Rightarrow df = dx/(1+x^2)$, $g = x^2/2$ [1]. $\int_0^1 x\arctan x\,dx = \left[\tfrac{x^2}{2}\arctan x\right]_0^1 - \tfrac{1}{2}\int_0^1 \dfrac{x^2}{1+x^2}\,dx = \tfrac{\pi}{8} - \tfrac{1}{2}\int_0^1\!\left[1 - \dfrac{1}{1+x^2}\right]dx$ [2]. $= \tfrac{\pi}{8} - \tfrac{1}{2}[x - \arctan x]_0^1 = \tfrac{\pi}{8} - \tfrac{1}{2}\!\left(1 - \tfrac{\pi}{4}\right) = \tfrac{\pi}{4} - \tfrac{1}{2}$ [1].
Q3 (4 marks): IBP with $f = x^n$, $dg = e^{-x}\,dx \Rightarrow df = nx^{n-1}\,dx$, $g = -e^{-x}$. $I_n = [-x^n e^{-x}]_0^1 + n\int_0^1 x^{n-1}e^{-x}\,dx = -e^{-1} + n I_{n-1}$ [1]. $I_0 = \int_0^1 e^{-x}\,dx = 1 - e^{-1}$ [1]. $I_1 = -e^{-1} + 1 \cdot (1 - e^{-1}) = 1 - 2e^{-1}$; $I_2 = -e^{-1} + 2(1 - 2e^{-1}) = 2 - 5e^{-1}$ [1]; $I_3 = -e^{-1} + 3(2 - 5e^{-1}) = 6 - 16e^{-1} = 6 - \tfrac{16}{e}$ [1].
Sit the module quiz, then close the lesson off.
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