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 notesdo 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 confidencethat 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.
Mark lesson as complete
Tick when you've finished the practice and review.