Reverse Differentiation Extended

Integrating $e^{ax+b}$, $\dfrac{1}{ax+b}$, $\sin(ax+b)$ and $\cos(ax+b)$ โ€” the reverse chain rule
9709 Paper 3 ยท ยง3.5 Integration ยท Week 6 Lesson 1
๐Ÿ’ก Lead In: The Rocket's Distance

The Scenario

A small rocket is launched vertically. Its upward velocity (in metres per second) at time $t$ seconds is $v(t) = e^{2t}$. The velocity grows exponentially โ€” the rocket accelerates rapidly.

To find how far the rocket has travelled, we must integrate the velocity: $\text{distance} = \displaystyle\int v(t)\, \text{d}t = \int e^{2t}\, \text{d}t$.

We already know that $\int e^x\, \text{d}x = e^x + C$. But our integral has $e^{2t}$, not $e^t$. The basic formula does not directly apply. A student guesses: โ€œJust write $e^{2t} + C$.โ€ But differentiating $e^{2t}$ gives $2e^{2t}$ (chain rule!), not $e^{2t}$ โ€” so the guess is twice too big.

The Question

We know $\dfrac{\text{d}}{\text{d}x}\!\left(e^{2x}\right) = 2e^{2x}$ (chain rule multiplies by 2). So to reverse this and find $\int e^{2x}\, \text{d}x$, we must divide by 2: $\int e^{2x}\, \text{d}x = \dfrac{1}{2}e^{2x} + C$.

What is the general rule for $\int e^{ax+b}\, \text{d}x$? And does the same โ€œdivide by $a$โ€ idea work for $\int \dfrac{1}{ax+b}\, \text{d}x$, $\int \sin(ax+b)\, \text{d}x$, and $\int \cos(ax+b)\, \text{d}x$?

๐Ÿ’ก The Bridge (click to reveal)

The chain rule says $\dfrac{\text{d}}{\text{d}x}\!\left[f(ax+b)\right] = a\,f'(ax+b)$. The factor of $a$ appears because the inner function $ax+b$ has derivative $a$.

To reverse this, we divide by $a$:

$$\int f(ax+b)\, \text{d}x = \frac{1}{a}\,F(ax+b) + C$$

where $F$ is the antiderivative of $f$. This single principle gives us all four formulas:

$$\int e^{ax+b}\, \text{d}x = \frac{1}{a}\,e^{ax+b} + C, \quad \int \frac{1}{ax+b}\, \text{d}x = \frac{1}{a}\ln|ax+b| + C$$

$$\int \sin(ax+b)\, \text{d}x = -\frac{1}{a}\cos(ax+b) + C, \quad \int \cos(ax+b)\, \text{d}x = \frac{1}{a}\sin(ax+b) + C$$

For the rocket: $\int e^{2t}\, \text{d}t = \frac{1}{2}e^{2t} + C$. Today we learn this โ€œreverse chain ruleโ€ in full.

๐Ÿ“– Key Concepts: The Reverse Chain Rule

Last week we learned the chain rule for differentiation: $\dfrac{\text{d}}{\text{d}x}\!\left[f(ax+b)\right] = a\,f'(ax+b)$. Today we reverse it. Since differentiation multiplies by $a$, integration must divide by $a$:

Master formula (reverse chain rule): $$\int f(ax+b)\, \text{d}x = \frac{1}{a}\,F(ax+b) + C \qquad (a \neq 0)$$

where $F$ is the antiderivative of $f$ (i.e. $F' = f$).

Key relation 1 โ€” $\int e^{ax+b}\, \text{d}x$:

Since $\dfrac{\text{d}}{\text{d}x}\!\left(e^{ax+b}\right) = a\,e^{ax+b}$, reversing gives:

$$\boxed{\;\int e^{ax+b}\, \text{d}x = \frac{1}{a}\,e^{ax+b} + C\;}\qquad(a \neq 0)$$
Key relation 2 โ€” $\int \dfrac{1}{ax+b}\, \text{d}x$:

Since $\dfrac{\text{d}}{\text{d}x}\!\left(\ln|ax+b|\right) = \dfrac{a}{ax+b}$, reversing gives:

$$\boxed{\;\int \frac{1}{ax+b}\, \text{d}x = \frac{1}{a}\ln|ax+b| + C\;}\qquad(a \neq 0)$$

The absolute value is needed because $\ln$ is only defined for positive arguments, but $ax+b$ can be negative.

Key relation 3 โ€” $\int \sin(ax+b)\, \text{d}x$:

Since $\dfrac{\text{d}}{\text{d}x}\!\left(-\cos(ax+b)\right) = a\sin(ax+b)$, reversing gives:

$$\boxed{\;\int \sin(ax+b)\, \text{d}x = -\frac{1}{a}\cos(ax+b) + C\;}\qquad(a \neq 0)$$
Key relation 4 โ€” $\int \cos(ax+b)\, \text{d}x$:

Since $\dfrac{\text{d}}{\text{d}x}\!\left(\sin(ax+b)\right) = a\cos(ax+b)$, reversing gives:

$$\boxed{\;\int \cos(ax+b)\, \text{d}x = \frac{1}{a}\sin(ax+b) + C\;}\qquad(a \neq 0)$$

All four formulas follow the same pattern: integrate as usual, then divide by $a$. The sign on the trig formulas comes from the standard antiderivatives ($\int \sin x\, \text{d}x = -\cos x + C$, $\int \cos x\, \text{d}x = \sin x + C$).

Group Activity 1: Match the Antiderivative โ€” Integral to Result

In pairs, link every integral to its antiderivative. Click an integral card, then click the result you believe it pairs with โ€” a numbered line joins them. Match all six pairs, then press Check to score. Press Reset to shuffle and play again. Nobody clicks until both partners agree: first name the inner linear part $ax+b$ and its coefficient $a$, then say whether the factor $\dfrac{1}{a}$ is positive or carries a minus sign.

Linked: 0 / 6 Score: โ€“

Strategy tip: Every result is the standard antiderivative with one extra factor $\dfrac{1}{a}$, where $a$ is the coefficient of the inner linear part $ax+b$. Watch the signs: $\int\sin$ keeps the minus of $-\cos$, while $\int e^{-4x}$ picks up a minus from $\dfrac{1}{-4}$.

Individual Activity 2: Reverse-Chain Slip Hunt โ€” Find It and Fix It

Two worked integrals, each with exactly one wrong line. Work on your own: read the whole solution, predict what the faulty line should be, then click the line you believe is wrong and choose the corrected version from the options. One wrong line per round โ€” no guessing.

Round: 1 / 2 Score: 0 / 2

Group ยท Guided Worked Example 1: Integrating $\dfrac{1}{ax+b}$ Exercise 5A Q1

Find the following indefinite integrals.

(i) $\displaystyle\int \frac{3}{x} \, \text{d}x$ โ€ƒ (ii) $\displaystyle\int \frac{1}{4x} \, \text{d}x$ โ€ƒ (iii) $\displaystyle\int \frac{1}{x-5} \, \text{d}x$ โ€ƒ (iv) $\displaystyle\int \frac{1}{2x-9} \, \text{d}x$

๐Ÿ‘ Reveal Solution
Part (i): $\displaystyle\int \frac{3}{x}\, \text{d}x$

Factor out the 3: $\displaystyle\int \frac{3}{x}\, \text{d}x = 3\int \frac{1}{x}\, \text{d}x = 3\ln|x| + C$.

Here $a = 1$, $b = 0$, so $\frac{1}{a} = 1$. The 3 is a constant multiplier.

$\boxed{3\ln|x| + C}$

Part (ii): $\displaystyle\int \frac{1}{4x}\, \text{d}x$

Here $a = 4$, $b = 0$. Apply the formula:

$\displaystyle\int \frac{1}{4x}\, \text{d}x = \frac{1}{4}\ln|4x| + C$

$\boxed{\frac{1}{4}\ln|4x| + C}$

Part (iii): $\displaystyle\int \frac{1}{x-5}\, \text{d}x$

Here $a = 1$, $b = -5$. Apply the formula:

$\displaystyle\int \frac{1}{x-5}\, \text{d}x = \frac{1}{1}\ln|x-5| + C = \ln|x-5| + C$

$\boxed{\ln|x-5| + C}$

Part (iv): $\displaystyle\int \frac{1}{2x-9}\, \text{d}x$

Here $a = 2$, $b = -9$. Apply the formula:

$\displaystyle\int \frac{1}{2x-9}\, \text{d}x = \frac{1}{2}\ln|2x-9| + C$

$\boxed{\frac{1}{2}\ln|2x-9| + C}$

Individual ยท Guided Worked Example 2: Integrating $e^{ax+b}$ and simplifying Exercise 5A Q2

Find the following indefinite integrals.

(i) $\displaystyle\int e^{3x} \, \text{d}x$ โ€ƒ (ii) $\displaystyle\int e^{-4x} \, \text{d}x$ โ€ƒ (iii) $\displaystyle\int e^{-x/3} \, \text{d}x$ โ€ƒ (iv) $\displaystyle\int \frac{10}{e^{5x}} \, \text{d}x$ โ€ƒ (v) $\displaystyle\int \frac{e^{3x} + 4}{e^{2x}} \, \text{d}x$

๐Ÿ‘ Reveal Solution
Part (i): $\displaystyle\int e^{3x}\, \text{d}x$

$a = 3$, $b = 0$. $\frac{1}{a} = \frac{1}{3}$.

$\boxed{\frac{1}{3}e^{3x} + C}$

Part (ii): $\displaystyle\int e^{-4x}\, \text{d}x$

$a = -4$, $b = 0$. $\frac{1}{a} = \frac{1}{-4} = -\frac{1}{4}$.

$\boxed{-\frac{1}{4}e^{-4x} + C}$

Part (iii): $\displaystyle\int e^{-x/3}\, \text{d}x$

$a = -\frac{1}{3}$, $b = 0$. $\frac{1}{a} = \frac{1}{-1/3} = -3$.

$\boxed{-3\,e^{-x/3} + C}$

Part (iv): $\displaystyle\int \frac{10}{e^{5x}}\, \text{d}x$

Rewrite: $\frac{10}{e^{5x}} = 10e^{-5x}$. Now $a = -5$, so $\frac{1}{a} = -\frac{1}{5}$.

$\displaystyle\int 10e^{-5x}\, \text{d}x = 10 \cdot \left(-\frac{1}{5}\right)e^{-5x} + C = -2e^{-5x} + C$

$\boxed{-2e^{-5x} + C}$

Part (v): $\displaystyle\int \frac{e^{3x} + 4}{e^{2x}}\, \text{d}x$

Split the fraction: $\frac{e^{3x}+4}{e^{2x}} = \frac{e^{3x}}{e^{2x}} + \frac{4}{e^{2x}} = e^{x} + 4e^{-2x}$.

Now integrate each term:

$\displaystyle\int e^{x}\, \text{d}x = e^{x} + C_1$ โ€‚ (here $a = 1$, so $\frac{1}{a} = 1$)

$\displaystyle\int 4e^{-2x}\, \text{d}x = 4 \cdot \left(-\frac{1}{2}\right)e^{-2x} = -2e^{-2x}$

Combining: $\boxed{e^{x} - 2e^{-2x} + C}$

โœ๏ธ Concept Quiz
Score: 0 / 4
1. The integral $\displaystyle\int e^{ax+b}\, \text{d}x$ (with $a \neq 0$) equals:
$a \cdot e^{ax+b} + C$
$\dfrac{1}{a}e^{ax+b} + C$
$e^{ax+b} + C$
$\dfrac{1}{b}e^{ax+b} + C$
Since $\frac{\text{d}}{\text{d}x}\!\left(e^{ax+b}\right) = a\,e^{ax+b}$ (chain rule multiplies by $a$), reversing means dividing by $a$: $\int e^{ax+b}\, \text{d}x = \frac{1}{a}e^{ax+b} + C$. Verify: $\frac{\text{d}}{\text{d}x}\!\left(\frac{1}{a}e^{ax+b}\right) = \frac{1}{a} \cdot a \cdot e^{ax+b} = e^{ax+b}$. The trap $a \cdot e^{ax+b}$ multiplies instead of dividing.
2. The integral $\displaystyle\int \frac{1}{ax+b}\, \text{d}x$ (with $a \neq 0$) equals:
$\ln|ax+b| + C$
$\dfrac{1}{a}\ln|ax+b| + C$
$a\ln|ax+b| + C$
$\dfrac{1}{a}\ln(ax+b) + C$
Since $\frac{\text{d}}{\text{d}x}\!\left(\ln|ax+b|\right) = \frac{a}{ax+b}$, reversing gives $\int \frac{1}{ax+b}\, \text{d}x = \frac{1}{a}\ln|ax+b| + C$. The absolute value is essential because $ax+b$ can be negative. The trap $\ln(ax+b)$ (without absolute value) is only valid when $ax+b > 0$. The trap $a\ln|ax+b|$ multiplies by $a$ instead of dividing.
3. The integral $\displaystyle\int \sin(ax+b)\, \text{d}x$ (with $a \neq 0$) equals:
$\dfrac{1}{a}\cos(ax+b) + C$
$-\dfrac{1}{a}\cos(ax+b) + C$
$-\dfrac{1}{a}\sin(ax+b) + C$
$-\cos(ax+b) + C$
The basic integral is $\int \sin x\, \text{d}x = -\cos x + C$ (note the minus sign!). Applying the reverse chain rule (divide by $a$): $\int \sin(ax+b)\, \text{d}x = -\frac{1}{a}\cos(ax+b) + C$. Verify: $\frac{\text{d}}{\text{d}x}\!\left(-\frac{1}{a}\cos(ax+b)\right) = -\frac{1}{a}\cdot(-a)\sin(ax+b) = \sin(ax+b)$. The trap $\frac{1}{a}\cos(ax+b)$ (without the minus) forgets that the antiderivative of $\sin$ is $-\cos$, not $+\cos$.
4. To find $\displaystyle\int \cos(3x-\pi)\, \text{d}x$, you should:
$3\sin(3x-\pi) + C$
$\dfrac{1}{3}\sin(3x-\pi) + C$
$-\dfrac{1}{3}\sin(3x-\pi) + C$
$\sin(3x-\pi) + C$
Here $a = 3$, $b = -\pi$. The basic integral is $\int \cos x\, \text{d}x = \sin x + C$. Applying the reverse chain rule: $\int \cos(3x-\pi)\, \text{d}x = \frac{1}{3}\sin(3x-\pi) + C$. Verify: $\frac{\text{d}}{\text{d}x}\!\left(\frac{1}{3}\sin(3x-\pi)\right) = \frac{1}{3}\cdot 3\cos(3x-\pi) = \cos(3x-\pi)$. The trap $3\sin(3x-\pi)$ multiplies by $a$; the trap $-\frac{1}{3}\sin(3x-\pi)$ applies the sine rule's minus sign to cosine (cosine's antiderivative has NO minus sign).
๐ŸŽ‰
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๐Ÿ“ More Practice

Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.

Exercise 5A Q3(i)(ii) Standard

Find the following definite integrals. Where appropriate give your answers to 3 significant figures.

(i) $\displaystyle\int_{0}^{4} 4e^{2x} \, \text{d}x$ โ€ƒ (ii) $\displaystyle\int_{1}^{3} \frac{4}{2x+1} \, \text{d}x$

๐Ÿ’ก Hint

For (i): $\int 4e^{2x}\, \text{d}x = 4 \cdot \frac{1}{2}e^{2x} = 2e^{2x}$. Evaluate at 0 and 4.

For (ii): $\int \frac{4}{2x+1}\, \text{d}x = 4 \cdot \frac{1}{2}\ln|2x+1| = 2\ln|2x+1|$. Evaluate at 1 and 3.

๐Ÿ‘ Reveal Solution
Part (i): $\displaystyle\int_{0}^{4} 4e^{2x}\, \text{d}x$

Find the antiderivative: $\int 4e^{2x}\, \text{d}x = 4 \cdot \frac{1}{2}e^{2x} = 2e^{2x}$.

Evaluate: $\left[2e^{2x}\right]_{0}^{4} = 2e^{8} - 2e^{0} = 2e^{8} - 2$.

Numerically: $2(2980.958) - 2 = 5959.916 \approx 5960$ (3 s.f.).

$\boxed{2e^{8} - 2 \approx 5960\text{ (3 s.f.)}}$

Part (ii): $\displaystyle\int_{1}^{3} \frac{4}{2x+1}\, \text{d}x$

Find the antiderivative: $\int \frac{4}{2x+1}\, \text{d}x = 4 \cdot \frac{1}{2}\ln|2x+1| = 2\ln|2x+1|$.

Evaluate: $\left[2\ln|2x+1|\right]_{1}^{3} = 2\ln(7) - 2\ln(3) = 2\ln\!\left(\frac{7}{3}\right)$.

Numerically: $2\ln\!\left(\frac{7}{3}\right) = 2(0.8473) = 1.6946 \approx 1.69$ (3 s.f.).

$\boxed{2\ln\!\left(\frac{7}{3}\right) \approx 1.69\text{ (3 s.f.)}}$

Exercise 5A Q3(iii)(iv) Medium

Find the following definite integrals. Where appropriate give your answers to 3 significant figures.

(iii) $\displaystyle\int_{-1}^{1} \left(e^{x} + e^{-x}\right) \, \text{d}x$ โ€ƒ (iv) $\displaystyle\int_{-2}^{1} e^{3x-2} \, \text{d}x$

๐Ÿ’ก Hint

For (iii): integrate each term separately. $\int e^{x}\, \text{d}x = e^{x}$ and $\int e^{-x}\, \text{d}x = -e^{-x}$ (here $a = -1$).

For (iv): here $a = 3$, $b = -2$, so $\int e^{3x-2}\, \text{d}x = \frac{1}{3}e^{3x-2}$. Evaluate at $-2$ and $1$.

๐Ÿ‘ Reveal Solution
Part (iii): $\displaystyle\int_{-1}^{1} \left(e^{x} + e^{-x}\right)\, \text{d}x$

Integrate each term: $\int e^{x}\, \text{d}x = e^{x}$ (here $a=1$) and $\int e^{-x}\, \text{d}x = -e^{-x}$ (here $a=-1$, so $\frac{1}{a}=-1$).

So: $\displaystyle\int_{-1}^{1} \left(e^{x} + e^{-x}\right)\, \text{d}x = \left[e^{x} - e^{-x}\right]_{-1}^{1}$

$= \left(e^{1} - e^{-1}\right) - \left(e^{-1} - e^{1}\right) = e - \frac{1}{e} - \frac{1}{e} + e = 2e - \frac{2}{e}$

Numerically: $2(2.71828) - 2(0.36788) = 5.43656 - 0.73576 = 4.70080 \approx 4.70$ (3 s.f.).

$\boxed{2e - \frac{2}{e} \approx 4.70\text{ (3 s.f.)}}$

Part (iv): $\displaystyle\int_{-2}^{1} e^{3x-2}\, \text{d}x$

Here $a = 3$, $b = -2$. So $\int e^{3x-2}\, \text{d}x = \frac{1}{3}e^{3x-2}$.

Evaluate: $\left[\frac{1}{3}e^{3x-2}\right]_{-2}^{1} = \frac{1}{3}\left(e^{3(1)-2} - e^{3(-2)-2}\right) = \frac{1}{3}\left(e^{1} - e^{-8}\right)$.

Numerically: $\frac{1}{3}(2.71828 - 0.000335) = \frac{1}{3}(2.71794) = 0.90598 \approx 0.906$ (3 s.f.).

$\boxed{\frac{1}{3}\left(e - e^{-8}\right) \approx 0.906\text{ (3 s.f.)}}$

Exercise 5A Q7 Medium

A curve is such that $\dfrac{\text{d}y}{\text{d}x} = e^{2x} - 2e^{-x}$. The point $(0, 1)$ lies on the curve.

(i) Find the equation of the curve.

(ii) The curve has one stationary point. Find the $x$ coordinate of this point and determine whether it is a maximum or a minimum point.

(9709 Paper 2 Q6 November 2005)

๐Ÿ’ก Hint

For (i): integrate $\frac{\text{d}y}{\text{d}x}$ to get $y$. $\int e^{2x}\, \text{d}x = \frac{1}{2}e^{2x}$ and $\int 2e^{-x}\, \text{d}x = -2e^{-x}$. Use the point $(0, 1)$ to find $C$.

For (ii): set $\frac{\text{d}y}{\text{d}x} = 0$. This gives $e^{2x} = 2e^{-x}$, so $e^{3x} = 2$. Use the second derivative test to classify.

๐Ÿ‘ Reveal Solution
Part (i): Find the equation of the curve

Integrate: $y = \int \left(e^{2x} - 2e^{-x}\right)\, \text{d}x = \frac{1}{2}e^{2x} + 2e^{-x} + C$.

(Note: $\int 2e^{-x}\, \text{d}x = 2 \cdot \frac{1}{-1}e^{-x} = -2e^{-x}$, so subtracting gives $+2e^{-x}$.)

Use $(0, 1)$: $1 = \frac{1}{2}e^{0} + 2e^{0} + C = \frac{1}{2} + 2 + C = \frac{5}{2} + C$.

So $C = 1 - \frac{5}{2} = -\frac{3}{2}$.

$\boxed{y = \frac{1}{2}e^{2x} + 2e^{-x} - \frac{3}{2}}$

Part (ii): Find the stationary point and classify it

Set $\frac{\text{d}y}{\text{d}x} = 0$: $e^{2x} - 2e^{-x} = 0$, so $e^{2x} = 2e^{-x}$.

Multiply both sides by $e^{x}$: $e^{3x} = 2$.

Take $\ln$: $3x = \ln 2$, so $x = \frac{1}{3}\ln 2 \approx 0.231$.

Second derivative: $\frac{\text{d}^{2}y}{\text{d}x^{2}} = 2e^{2x} + 2e^{-x}$.

Since $e^{2x} > 0$ and $e^{-x} > 0$ for all $x$, we have $\frac{\text{d}^{2}y}{\text{d}x^{2}} > 0$ always.

$\boxed{x = \frac{1}{3}\ln 2, \text{ and it is a minimum point.}}$

Exercise 5A Q9 Challenging

Find the exact value of the constant $k$ for which $\displaystyle\int_{1}^{k} \frac{1}{2x - 1} \, \text{d}x = 1$.

(9709 Paper 3 Q1 November 2007)

๐Ÿ’ก Hint 1

First find the antiderivative: $\int \frac{1}{2x-1}\, \text{d}x = \frac{1}{2}\ln|2x-1|$.

Then evaluate at the limits 1 and $k$ and set the result equal to 1.

๐Ÿ’ก Hint 2

You should get: $\frac{1}{2}\left(\ln|2k-1| - \ln|2(1)-1|\right) = 1$.

Simplify $\ln|2(1)-1| = \ln 1 = 0$.

So $\frac{1}{2}\ln(2k-1) = 1$ (assuming $k > \frac{1}{2}$ so $2k-1 > 0$).

Solve for $k$ by exponentiating both sides.

๐Ÿ‘ Reveal Solution
Step 1: Find the antiderivative

Here $a = 2$, $b = -1$. So $\int \frac{1}{2x-1}\, \text{d}x = \frac{1}{2}\ln|2x-1|$.

Step 2: Evaluate the definite integral

$\displaystyle\int_{1}^{k} \frac{1}{2x-1}\, \text{d}x = \left[\frac{1}{2}\ln|2x-1|\right]_{1}^{k} = \frac{1}{2}\left(\ln|2k-1| - \ln|2(1)-1|\right)$

$= \frac{1}{2}\left(\ln|2k-1| - \ln 1\right) = \frac{1}{2}\ln|2k-1|$

Step 3: Set equal to 1 and solve

$\frac{1}{2}\ln|2k-1| = 1$

$\ln|2k-1| = 2$

$|2k-1| = e^{2}$

Since the integral is defined on $[1, k]$ and $2x - 1 > 0$ at $x = 1$, we need $k > \frac{1}{2}$ (in fact $k \geq 1$ for the lower limit to be valid), so $2k - 1 > 0$.

$2k - 1 = e^{2}$

$k = \frac{e^{2} + 1}{2}$

Final Answer

$\boxed{k = \dfrac{e^{2} + 1}{2}}$

Verification: $\frac{1}{2}\ln\!\left(2 \cdot \frac{e^{2}+1}{2} - 1\right) = \frac{1}{2}\ln(e^{2}) = \frac{1}{2} \cdot 2 = 1$. โœ“

๐ŸŽซ

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