Module 8 Synthesis & Exam Technique
You've covered substitution, trig identities, inverse trig integrals, completing the square, and harder techniques across 19 lessons. This final lesson builds the decision tree that connects them all: given any integral, how do you choose the right method — fast, under exam pressure? That strategic fluency is what Module 8 is really testing.
Three integrals — three different techniques. Without working them out fully, record your first instinct for which method applies to each: (a) $\displaystyle\int \dfrac{3x^2}{x^3+1}\,dx$, (b) $\displaystyle\int \cos^2\!x\,dx$, (c) $\displaystyle\int \dfrac{1}{\sqrt{9-x^2}}\,dx$.
Every Module 8 integration question can be classified in under 10 seconds using this four-question scan. Run through it during reading time and the technique is chosen before you write a word.
- Is the numerator the derivative of the denominator (or nearly so)? → Use substitution.
- Is there $\sin^2$ or $\cos^2$? → Apply the double-angle identity first.
- Does it match $\dfrac{1}{\sqrt{a^2-x^2}}$ or $\dfrac{a}{a^2+x^2}$? → Inverse trig formula.
- Is the denominator a quadratic that doesn't factor nicely? → Complete the square, then inverse trig.
Key facts
- The complete set of Module 8 standard integral forms and when each applies
- The four-question decision scan for technique selection
- How completing the square converts a quadratic integrand to an inverse trig form
Concepts
- Why mixed integrals require a hierarchy of technique checks rather than a single rule
- How each technique in Module 8 generalises from a core idea (chain rule, trig identity, or standard form)
- Why memorising the $+C$ and $\frac{1}{a}$ factors is non-negotiable in HSC marking
Skills
- Classify any Module 8 integral and select the correct technique in under 15 seconds
- Solve multi-step integration problems combining techniques across the module
- Write complete, correctly formatted solutions that earn full HSC marks
Here are the three hook integrals classified by the decision scan:
- (a) $\displaystyle\int \dfrac{3x^2}{x^3+1}\,dx$ — numerator $3x^2$ is the derivative of $x^3+1$. Technique: substitution with $u=x^3+1$. Answer: $\ln|x^3+1|+C$.
- (b) $\displaystyle\int \cos^2\!x\,dx$ — even power of cosine. Technique: double-angle identity $\cos^2\!x=\tfrac{1}{2}(1+\cos 2x)$. Answer: $\tfrac{x}{2}+\tfrac{\sin 2x}{4}+C$.
- (c) $\displaystyle\int \dfrac{1}{\sqrt{9-x^2}}\,dx$ — matches $\dfrac{1}{\sqrt{a^2-x^2}}$ with $a=3$. Technique: inverse trig. Answer: $\sin^{-1}\!\dfrac{x}{3}+C$.
Decision tree: (1) look for $f'(x)g(f(x))$ → substitution; (2) look for $a^2\pm x^2$ form → inverse-trig standard; (3) otherwise complete the square.
Pause — copy the decision tree as a three-branch flowchart: substitution structure check → radical form check → complete the square into your book.
Quick check: Which technique should be used first for $\displaystyle\int \dfrac{1}{x^2+6x+13}\,dx$?
We just saw the decision scan: check for substitution structure $f'(x)g(f(x))$; if not, check for $\sqrt{a^2-x^2}$ ($\to\sin^{-1}$) or $a^2+x^2$ ($\to\tan^{-1}$); if not, complete the square. That raises a question: when the denominator is a non-standard quadratic and completing the square still doesn't give $u^2+a^2$ immediately, what is the one remaining algebraic manoeuvre? This card answers it → factor out the leading coefficient of $x^2$ before completing the square.
When the denominator is a quadratic that does not immediately match $a^2+x^2$, complete the square to reduce it to that form, then substitute to apply the $\tan^{-1}$ formula.
Example: $\displaystyle\int \dfrac{1}{x^2+6x+13}\,dx$.
Complete the square: $x^2+6x+13 = (x+3)^2+4$.
Let $u = x+3$, $du = dx$:
The same approach works for $\displaystyle\int \dfrac{1}{\sqrt{a^2-(x+p)^2}}\,dx$ — complete the square, shift by $p$, and apply the $\sin^{-1}$ formula.
When the denominator is a quadratic that does not immediately match $a^2+x^2$, complete the square to reduce it to that form, then substitute to apply the $\tan^{-1}$ formula.
Pause — copy the general completing-the-square routine for integration: factor out leading coefficient, complete the square, substitute $u=x+p$, apply $\tan^{-1}$ or $\sin^{-1}$ formula into your book.
Did you get this? True or false: $x^2+4x+9$ completes to $(x+2)^2+5$.
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle\int_0^1 \dfrac{x}{x^2+4}\,dx$.
Find $\displaystyle\int \dfrac{1}{x^2+4x+8}\,dx$.
Evaluate $\displaystyle\int_0^{\pi/4} \sin^2\!x\,dx$.
Fill the gap: Completing the square: $x^2+8x+25 = (x+4)^2 + $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\displaystyle\int \cos^2\!x\,dx = \dfrac{x}{2}+\dfrac{\sin 2x}{4}+C$.
Activities · practice with the ideas
Classify each integral by technique (do NOT evaluate): (a) $\displaystyle\int \dfrac{\cos x}{\sin^2\!x}\,dx$, (b) $\displaystyle\int \dfrac{1}{x^2-2x+5}\,dx$, (c) $\displaystyle\int \sin^2\!3x\,dx$.
Evaluate $\displaystyle\int_0^{\pi/2} \cos^2\!x\,dx$ using the double-angle identity.
Find $\displaystyle\int \dfrac{1}{x^2-2x+5}\,dx$. (Hint: complete the square first.)
Evaluate $\displaystyle\int_0^1 \dfrac{1}{\sqrt{4-x^2}}\,dx$. Leave your answer in exact form.
A student writes $\displaystyle\int \sin^2\!x\,dx = \dfrac{\sin^3\!x}{3}+C$. Identify the error and state the correct answer.
Odd one out: Three of these integrals should be solved using substitution. Which one requires a different technique?
At the start you classified three integrals by technique. The answers: (a) substitution with $u=x^3+1$ → $\ln|x^3+1|+C$; (b) double-angle identity → $\tfrac{x}{2}+\tfrac{\sin 2x}{4}+C$; (c) inverse trig formula → $\sin^{-1}\!\tfrac{x}{3}+C$.
Module 8 is ultimately about having a fast, reliable decision process. The techniques themselves are mechanical once you've identified which one applies. What changes between a Band 4 and Band 6 response is the speed and confidence with which the right technique is selected — and whether the answer is verified before moving on.
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 $\displaystyle\int_0^{\pi/2} \cos^2\!x\,dx$. (2 marks)
Q2. Find $\displaystyle\int \dfrac{1}{x^2-2x+5}\,dx$. (3 marks)
Q3. Evaluate $\displaystyle\int_0^1 \dfrac{1}{\sqrt{4-x^2}}\,dx$. Leave your answer in exact form. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1a. $\dfrac{\cos x}{\sin^2 x}$: substitution $u=\sin x$, $du=\cos x\,dx \Rightarrow \int u^{-2}\,du = -\dfrac{1}{\sin x}+C$. 1b. $\dfrac{1}{x^2-2x+5}$: complete the square → $\dfrac{1}{(x-1)^2+4}$, use $\tan^{-1}$ form. 1c. $\sin^2\!3x$: double-angle identity $\sin^2\!3x = \tfrac{1}{2}(1-\cos 6x)$.
2. $\displaystyle\int_0^{\pi/2}\!\tfrac{1+\cos 2x}{2}\,dx = \left[\tfrac{x}{2}+\tfrac{\sin 2x}{4}\right]_0^{\pi/2} = \dfrac{\pi}{4}$.
3. $x^2-2x+5=(x-1)^2+4$; let $u=x-1$: $\dfrac{1}{2}\tan^{-1}\!\dfrac{x-1}{2}+C$.
4. $\left[\sin^{-1}\!\dfrac{x}{2}\right]_0^1 = \sin^{-1}\!\tfrac{1}{2} = \dfrac{\pi}{6}$.
5. Error: $\dfrac{\sin^3\!x}{3}$ would require $\int\!\sin^2\!x\cos\!x\,dx$ (substitution). Correct: apply $\sin^2\!x=\tfrac{1}{2}(1-\cos 2x)$; answer $\tfrac{x}{2}-\tfrac{\sin 2x}{4}+C$.
Q1 (2 marks): $\cos^2\!x = \tfrac{1}{2}(1+\cos 2x)$ [1]; $\left[\tfrac{x}{2}+\tfrac{\sin 2x}{4}\right]_0^{\pi/2} = \dfrac{\pi}{4}$ [1].
Q2 (3 marks): Complete the square: $x^2-2x+5=(x-1)^2+4$ [1]; let $u=x-1$, integral $= \displaystyle\int\dfrac{du}{u^2+4}$ [1]; $= \dfrac{1}{2}\tan^{-1}\!\dfrac{x-1}{2}+C$ [1].
Q3 (3 marks): Standard form $\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\!\tfrac{x}{a}$ with $a=2$ [1]; antiderivative $\sin^{-1}\!\tfrac{x}{2}$ [1]; $\sin^{-1}\!\tfrac{1}{2}-\sin^{-1}0=\dfrac{\pi}{6}$ [1].
Five mixed integration questions spanning the full Module 8. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering Module 8 synthesis questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.