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 confidencethat 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.