Calculus Exam Techniques
You know every integration technique — but exam marks slip away through small errors: a forgotten $+C$, a wrong sign from $\cos^{-1}$, or limits that weren't changed during substitution. This lesson makes those failure modes unforgettable, gives you a systematic checking strategy, and shows you how to verify any integral on the spot by differentiating.
A student claims $\displaystyle\int \dfrac{1}{\sqrt{1-x^2}}\,dx = \cos^{-1}x + C$. Before consulting any notes — do you think this is right or wrong, and how would you check it without a formula sheet?
Every integration exam question rewards two habits: spot the technique quickly (substitution, trig identity, inverse trig, or algebraic manipulation), then verify by differentiating your answer. These two moves together prevent nearly all mark-loss errors.
The differentiation check is the gold standard: differentiate your answer and confirm you recover the original integrand. It costs 30 seconds and catches sign errors, missing constants, and wrong techniques.
If $\displaystyle\int f(x)\,dx = F(x) + C$, then $F'(x) = f(x)$.
Key facts
- $\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}\,dx = \sin^{-1}\!\dfrac{x}{a}+C$ (not $\cos^{-1}$)
- Indefinite integrals always need $+C$; definite integrals evaluate at limits
- Substitution requires limits to be changed for definite integrals
Concepts
- Why differentiation is the fastest way to check an integration answer
- How sign errors arise in inverse trig derivatives
- Why failing to change limits in substitution gives a wrong numerical answer
Skills
- Verify any indefinite integral by differentiating the answer
- Correctly change limits when using substitution on definite integrals
- Identify and correct the three most common exam errors in integration
The most powerful exam tool is also the simplest: differentiate your answer. If $\displaystyle\int f(x)\,dx = F(x)+C$, then $F'(x)$ must equal $f(x)$. This takes 30 seconds and catches the majority of errors.
Answering the hook question: The student wrote $\displaystyle\int \dfrac{1}{\sqrt{1-x^2}}\,dx = \cos^{-1}x + C$.
Check: $\dfrac{d}{dx}(\cos^{-1}x) = \dfrac{-1}{\sqrt{1-x^2}} \neq \dfrac{1}{\sqrt{1-x^2}}$.
The sign is wrong. The correct answer is $\sin^{-1}x + C$, since $\dfrac{d}{dx}(\sin^{-1}x) = \dfrac{1}{\sqrt{1-x^2}}$.
The most powerful exam tool is also the simplest: differentiate your answer . If $\displaystyle\int f(x)\,dx = F(x)+C$, then $F'(x)$ must equal $f(x)$. This takes 30 seconds and catches the majority of errors.
Pause — copy the differentiation check procedure: compute $F'(x)$ after integrating and confirm $F'(x)=f(x)$; list three error types this catches into your book.
Quick check: A student writes $\displaystyle\int \dfrac{1}{\sqrt{1-x^2}}\,dx = -\cos^{-1}x + C$. Is this answer correct?
We just saw the differentiation check: if $\int f(x)\,dx=F(x)+C$, verify by computing $F'(x)$ and confirming it equals $f(x)$ — catching sign errors, missing chain-rule factors, and constant errors. That raises a question: for definite integrals using substitution, there are two approaches: back-substitute and use original limits, or convert limits to $u$-values — which is more reliable and why? This card answers it → converting limits avoids back-substitution entirely; use $u(a)$ and $u(b)$ as the new limits.
When applying substitution to a definite integral, you must convert the limits from $x$-values to $u$-values. Forgetting to do this — and then back-substituting at the end — is a common error that sometimes gives the right answer accidentally, but often does not.
Correct method: For $\displaystyle\int_a^b f(g(x))\,g'(x)\,dx$ with $u = g(x)$:
- New lower limit: $u = g(a)$
- New upper limit: $u = g(b)$
- Integrate entirely in terms of $u$, then evaluate directly — no back-substitution needed.
Example: Evaluate $\displaystyle\int_0^1 2x(x^2+1)^3\,dx$.
Let $u = x^2+1$, so $du = 2x\,dx$. When $x=0$, $u=1$. When $x=1$, $u=2$.
$= \displaystyle\int_1^2 u^3\,du = \left[\dfrac{u^4}{4}\right]_1^2 = \dfrac{16}{4} - \dfrac{1}{4} = \dfrac{15}{4}$.
When applying substitution to a definite integral , you must convert the limits from $x$-values to $u$-values. Forgetting to do this — and then back-substituting at the end — is a common error that sometimes gives the right answer accidentally,...
Pause — copy the limit-conversion rule for substitution: if $u=g(x)$ and $x\in[a,b]$, the new limits are $u\in[g(a),g(b)]$; include a worked example into your book.
Did you get this? True or false: when evaluating $\displaystyle\int_0^2 2x(x^2+3)^4\,dx$ with $u = x^2+3$, the new limits are $u=3$ and $u=7$.
Worked examples · 3 in a row, reveal as you go
A student claims $\displaystyle\int \dfrac{x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2} + C$. Verify this by differentiating.
Evaluate $\displaystyle\int_0^{\pi/2} \sin x \cos x\,dx$ using the substitution $u = \sin x$.
Find the error in this working and correct it: $\displaystyle\int_1^2 \dfrac{1}{x^2+4}\,dx = \tan^{-1}(x^2+4)\Big|_1^2 = \cdots$
Fill the gap: $\displaystyle\int \dfrac{1}{9+x^2}\,dx = \dfrac{1}{3}\tan^{-1}\!\dfrac{x}{$$} + C$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\displaystyle\int \dfrac{1}{4+x^2}\,dx = \dfrac{1}{2}\tan^{-1}\!\dfrac{x}{2}+C$.
Activities · practice with the ideas
Verify by differentiation that $\displaystyle\int x\sqrt{x^2+1}\,dx = \dfrac{1}{3}(x^2+1)^{3/2}+C$.
Evaluate $\displaystyle\int_0^1 \dfrac{2x}{x^2+4}\,dx$ using the substitution $u = x^2+4$. Show the limit change clearly.
Identify the error and find the correct value: a student writes $\displaystyle\int_0^1 \dfrac{1}{\sqrt{4-x^2}}\,dx = \sin^{-1}(x) \Big|_0^1 = \dfrac{\pi}{2}$.
Evaluate $\displaystyle\int_0^1 \dfrac{3}{9+x^2}\,dx$. Leave your answer in exact form.
A student evaluates $\displaystyle\int_1^3 2x(x^2-1)^5\,dx$ and forgets to change the limits. What limits do they incorrectly use in $u$, and what are the correct $u$-limits? ($u = x^2-1$)
Odd one out: Three of these are correct indefinite integrals. Which one is NOT?
Earlier you considered whether $\displaystyle\int \dfrac{1}{\sqrt{1-x^2}}\,dx = \cos^{-1}x + C$ was correct.
The differentiation check reveals the answer: $\dfrac{d}{dx}(\cos^{-1}x) = \dfrac{-1}{\sqrt{1-x^2}}$, which carries the wrong sign. The correct antiderivative is $\sin^{-1}x + C$. Note that $-\cos^{-1}x + C$ is also valid (and equivalent, since $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$), but $\cos^{-1}x + C$ alone is wrong. The differentiation check is the one technique that catches this immediately.
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. Verify by differentiation that $\displaystyle\int \dfrac{2x}{\sqrt{x^2+5}}\,dx = 2\sqrt{x^2+5}+C$. (2 marks)
Q2. Evaluate $\displaystyle\int_0^2 \dfrac{x}{\sqrt{9-x^2}}\,dx$ using the substitution $u = 9-x^2$. Show the limit change clearly. (3 marks)
Q3. Evaluate $\displaystyle\int_0^2 \dfrac{1}{x^2+4}\,dx$. Leave your answer in exact form. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $\dfrac{d}{dx}(x^2+1)^{3/2}\!\cdot\!\tfrac{1}{3} = \tfrac{1}{3}\!\cdot\!\tfrac{3}{2}(x^2+1)^{1/2}\!\cdot\!2x = x\sqrt{x^2+1}$ ✓
2. $u=x^2+4$, $du=2x\,dx$; limits: $x=0\Rightarrow u=4$, $x=1\Rightarrow u=5$. $\displaystyle\int_4^5\dfrac{1}{u}\,du=\ln5-\ln4=\ln\!\tfrac{5}{4}$.
3. Error: should be $\sin^{-1}\!\dfrac{x}{2}$, not $\sin^{-1}x$. Correct: $\left[\sin^{-1}\!\dfrac{x}{2}\right]_0^1 = \sin^{-1}\!\tfrac{1}{2}-0=\dfrac{\pi}{6}$.
4. $\displaystyle\int_0^1\dfrac{3}{9+x^2}\,dx = \left[\tan^{-1}\!\dfrac{x}{3}\right]_0^1 = \tan^{-1}\!\tfrac{1}{3}$.
5. Incorrect limits: $u=1$ to $u=3$ (same as $x$-limits). Correct: $x=1\Rightarrow u=0$; $x=3\Rightarrow u=8$.
Q1 (2 marks): $\dfrac{d}{dx}(2\sqrt{x^2+5}) = 2\cdot\dfrac{1}{2}(x^2+5)^{-1/2}\cdot2x = \dfrac{2x}{\sqrt{x^2+5}}$ [1] $= $ integrand [1]. ✓
Q2 (3 marks): $u=9-x^2$, $du=-2x\,dx$ [1]; limits $x=0\Rightarrow u=9$, $x=2\Rightarrow u=5$ [1]; $\displaystyle\int_9^5\dfrac{-1}{2\sqrt{u}}\,du = \left[-\sqrt{u}\right]_9^5 = -\sqrt{5}+3 = 3-\sqrt{5}$ [1].
Q3 (3 marks): $a=2$; antiderivative is $\dfrac{1}{2}\tan^{-1}\!\dfrac{x}{2}$ [1]; $\left[\dfrac{1}{2}\tan^{-1}\!\dfrac{x}{2}\right]_0^2 = \dfrac{1}{2}\!\left(\tan^{-1}1 - 0\right)$ [1] $= \dfrac{1}{2}\cdot\dfrac{\pi}{4} = \dfrac{\pi}{8}$ [1].
Five timed questions on identifying and correcting integration errors. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering calculus technique questions. Lighter alternative to the boss.
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