Year 12 Methods — Integration by Recognition (11H)

Today's lesson

So far, every antiderivative you've found has had a simple template — power, exponential, sine, cosine, $\tfrac{1}{x}$. But what about integrals like $\displaystyle\int x\cos(3x)\,dx$ that don't sit on any standard list? Today's trick: differentiate a related function and read off the integral.

Learning intentions

Recognise that integration is the reverse of differentiation: if $\dfrac{d}{dx}\bigl[F(x)\bigr] = g(x)$, then $\displaystyle\int g(x)\,dx = F(x) + c$
Differentiate a given function, then rearrange to find a related integral
Handle integrals that need a constant multiplier tidy-up (e.g. $\tfrac{1}{3}$ out the front)
Use this method on product-rule recognitions to integrate things like $x\cos(3x)$ and $xe^x$

Part 1 — The recognition idea

RECOGNITION RULE If $\dfrac{d}{dx}\bigl[F(x)\bigr] = g(x)$, then $$\int g(x)\,dx = F(x) + c.$$ In a "hence find" question, the examiner has done the differentiation work for you — your job is to rearrange.

💡 How to spot it. If the question gives you a function and says "Find $f'(x)$. Hence find $\int \ldots\,dx$", your $f'(x)$ from part (a) is going to look almost exactly like the integrand in part (b). Match them up, then fix any constants.

📺 Walkthrough: the recognition pattern — from $f(x)=\log_e(x^3+2)$ to $\int \tfrac{x^2}{x^3+2}\,dx$. Differentiate, then fix the constant.

Part 2 — Worked Example 1 (log-derivative pattern)

EXAMPLE 1

Let $f(x) = \log_e(x^3 + 2)$. (a) Find $f'(x)$. (b) Hence find $\displaystyle\int \dfrac{x^2}{x^3+2}\,dx$.

(a) Chain rule: $\dfrac{d}{dx}\log_e(u) = \dfrac{u'}{u}$ with $u = x^3+2$, $u' = 3x^2$.

$\therefore f'(x) = \dfrac{3x^2}{x^3+2}$

(b) Reverse part (a): $\displaystyle\int \dfrac{3x^2}{x^3+2}\,dx = \log_e(x^3+2) + c$.

But we want $\dfrac{x^2}{x^3+2}$ — that's $\tfrac{1}{3}$ of the integrand we just integrated. So pull the $\tfrac{1}{3}$ out:

$\displaystyle\int \dfrac{x^2}{x^3+2}\,dx = \dfrac{1}{3}\log_e(x^3+2) + c$

⚠️ Watch the constant. Almost every recognition question has a missing scalar — usually $\tfrac{1}{k}$ where $k$ comes from a chain-rule "inside derivative". Always sanity-check by differentiating your answer.

SIMILAR EXAMPLE 1

Let $f(x) = \log_e(3x^2 + 7)$. (a) Find $f'(x)$. (b) Hence evaluate $\displaystyle\int_0^2 \dfrac{x}{3x^2+7}\,dx$.

(a) $f'(x) = \dfrac{6x}{3x^2+7}$.

(b) Reverse: $\displaystyle\int \dfrac{6x}{3x^2+7}\,dx = \log_e(3x^2+7) + c$.

We want $\dfrac{x}{3x^2+7}$ — that's $\tfrac{1}{6}$ of the integrand above.

$\displaystyle\int_0^2 \dfrac{x}{3x^2+7}\,dx = \tfrac{1}{6}\bigl[\log_e(3x^2+7)\bigr]_0^2 = \tfrac{1}{6}\bigl(\log_e 19 - \log_e 7\bigr)$

$= \dfrac{1}{6}\log_e\!\left(\dfrac{19}{7}\right)$

Part 3 — Recognising the chain rule (worked walkthrough)

📺 Walkthrough: Example 1 line-by-line — differentiating $\log_e(x^3+2)$ and using the result to integrate $\dfrac{x^2}{x^3+2}$.

Part 4 — Worked Example 2 (product-rule pattern)

EXAMPLE 2

Let $f(x) = x\sin(3x)$. (a) Find $f'(x)$. (b) Hence find $\displaystyle\int_0^{\pi/6} x\cos(3x)\,dx$.

(a) Product rule: $f'(x) = \sin(3x) + x\cdot 3\cos(3x) = \sin(3x) + 3x\cos(3x)$.

(b) Reverse: $\displaystyle\int \bigl[\sin(3x) + 3x\cos(3x)\bigr]\,dx = x\sin(3x) + c$.

Split off the easy bit: $\displaystyle\int \sin(3x)\,dx = -\tfrac{1}{3}\cos(3x) + c$.

$\therefore \displaystyle\int 3x\cos(3x)\,dx = x\sin(3x) - \!\left(-\tfrac{1}{3}\cos(3x)\right) + c = x\sin(3x) + \tfrac{1}{3}\cos(3x) + c$

Divide by $3$: $\displaystyle\int x\cos(3x)\,dx = \tfrac{1}{3}x\sin(3x) + \tfrac{1}{9}\cos(3x) + c$.

Evaluate on $[0, \tfrac{\pi}{6}]$: $\left[\tfrac{1}{3}x\sin(3x) + \tfrac{1}{9}\cos(3x)\right]_0^{\pi/6}$

$=\!\left(\tfrac{1}{3}\cdot\tfrac{\pi}{6}\cdot\sin\tfrac{\pi}{2} + \tfrac{1}{9}\cos\tfrac{\pi}{2}\right) - \!\left(0 + \tfrac{1}{9}\right)$

$= \tfrac{\pi}{18}\cdot 1 + 0 - \tfrac{1}{9}$

$=\dfrac{\pi}{18} - \dfrac{1}{9} = \dfrac{\pi - 2}{18}$

📺 Walkthrough: Example 2 in detail — product-rule recognition then evaluation on $[0, \tfrac{\pi}{6}]$ to land at $\tfrac{\pi-2}{18}$.

SIMILAR EXAMPLE 2

Let $f(x) = xe^x$. (a) Find $f'(x)$. (b) Hence find $\displaystyle\int xe^x\,dx$.

(a) Product rule: $f'(x) = e^x + xe^x$.

(b) Reverse: $\displaystyle\int (e^x + xe^x)\,dx = xe^x + c$.

$\displaystyle\int e^x\,dx = e^x + c$, so $\displaystyle\int xe^x\,dx = xe^x - e^x + c$

$= (x-1)e^x + c$

Part 5 — Setting-out template

RECIPE

Differentiate the function given in part (a). Show every chain/product step.
Reverse the result: $\int f'(x)\,dx = f(x) + c$.
Compare your reversed integrand to the target integrand in part (b). Look for a missing scalar.
If the target has extra easy pieces (e.g. a stand-alone $\sin(3x)$), integrate those separately and subtract.
Sanity check: differentiate your final answer and confirm you get back the target integrand.

Practice 5.1 — chain-rule recognition.

Let $f(x) = e^{x^2+1}$. Find $f'(x)$, hence find $\displaystyle\int xe^{x^2+1}\,dx$.
Let $f(x) = \sin(x^2)$. Find $f'(x)$, hence find $\displaystyle\int x\cos(x^2)\,dx$.
Let $f(x) = (x^2+1)^5$. Find $f'(x)$, hence find $\displaystyle\int x(x^2+1)^4\,dx$.

a) $f'(x) = 2x\,e^{x^2+1}$, so $\displaystyle\int 2x\,e^{x^2+1}\,dx = e^{x^2+1} + c$. Hence $\displaystyle\int xe^{x^2+1}\,dx = \dfrac{1}{2}e^{x^2+1} + c$.
b) $f'(x) = 2x\cos(x^2)$. Hence $\displaystyle\int x\cos(x^2)\,dx = \dfrac{1}{2}\sin(x^2) + c$.
c) $f'(x) = 10x(x^2+1)^4$. Hence $\displaystyle\int x(x^2+1)^4\,dx = \dfrac{1}{10}(x^2+1)^5 + c$.

Practice 5.2 — product-rule recognition.

Let $f(x) = x^2 e^x$. (a) Find $f'(x)$. (b) Hence find $\displaystyle\int (x^2 + 2x)e^x\,dx$.

(a) $f'(x) = 2xe^x + x^2 e^x = (x^2 + 2x)e^x$.
(b) Reverse directly: $\displaystyle\int (x^2 + 2x)e^x\,dx = x^2 e^x + c$.

Part 6 — Quick quiz (5 min)

Working program — Cambridge §11H

After the quiz, open Chapter 11 and work through these:

Exercise	Set work
11H — Integration by recognition	Q1, Q3, Q5, Q7, Q9
VCAA 2013 Exam 1	Hence-style integration question (see notes)

Tomorrow we look at 11J — Average value of a function. Bring your CAS.

Exit ticket — write in your book

Before you pack up, in your exercise book:

State the recognition rule in one sentence: "If the derivative of $F(x)$ is …, then the integral of … is …"
$f(x) = e^{3x}$. Find $f'(x)$. Hence find $\displaystyle\int e^{3x}\,dx$.
Why does almost every "hence find" recognition question end with a fractional constant out the front of the integral?