Integrating eˣ, Trig and (ax+b)ⁿ

Integrate exponentials, sine, cosine, sec², (ax+b)ⁿ and 1/(ax+b) by reversing the chain rule: divide by the number in front of x, then add + c.

In the exam for IGCSE Additional Maths 0606, Edexcel Further Pure Maths 4PM1 · 21 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Rewind the chain rule

You can now integrate powers of $x$ . But exam papers also ask you to integrate functions such as $e^{3 x}$ or $sin 2 x$ . One small idea handles all of them.

When you differentiate a function of $3 x$ , the chain rule multiplies by 3. So when you rewind it, you must divide by 3.

Example

$\frac{d}{d x} (e^{3 x}) = 3 e^{3 x}$ . The 3 appears because of the chain rule. So to rewind $e^{3 x}$ , you take $\frac{1}{3}$ of $e^{3 x}$ .

Formula

\int f (a x + b) d x = \frac{1}{a} F (a x + b) + c

$a$ — the number in front of $x$ inside the bracket. You divide by it.
$b$ — the number added inside. It just stays where it is.
$F$ — what the plain function integrates to. For example, $\int cos x d x = sin x$ , so when f is cos, F is sin.

Worked example

1
Check that $\int e^{3 x} d x = \frac{1}{3} e^{3 x} + c$ .
2
$\frac{d}{d x} (\frac{1}{3} e^{3 x}) = \frac{1}{3} \times 3 e^{3 x}$ — Differentiate. The chain rule brings out a 3.
3
$= e^{3 x}$ — The third and the 3 cancel. We are back where we started, so the answer is right.

Rule

This only works when the inside is linear: a number times $x$ , plus a number. For anything else, this rule does not apply.

Rule

Learn every result in this lesson. 0606 and 4037 give no calculus formulas. 4PM1 lists $sin a x$ , $cos a x$ and $e^{a x}$ as integrals to learn, and uses only the form $a x$ , with no $+ b$ .

Key idea

Differentiating multiplies by $a$ . Integrating divides by $a$ .

Exponentials

$e^{x}$ is special: its gradient is itself. So $e^{x}$ integrates to $e^{x}$ . With $a x + b$ in the power, divide by $a$ .

Formula

\int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + c

$a$ — the number in front of $x$ in the power. Divide by it.
$b$ — the number added in the power. It stays.

Worked example

1
Find $\int (4 e^{2 x} - e^{- x}) d x$ .
2
$4 e^{2 x} \to \frac{4}{2} e^{2 x} = 2 e^{2 x}$ — Here $a = 2$ . Divide the 4 by 2. The power does not change.
3
$- e^{- x} \to \frac{- 1}{- 1} e^{- x} = e^{- x}$ — Here $a = - 1$ . Dividing by −1 changes the sign.
4
$\int (4 e^{2 x} - e^{- x}) d x = 2 e^{2 x} + e^{- x} + c$ — Put the terms together and add + c.

Worked example

1
Find $\int \frac{e ^{2 x} + 3}{e ^{x}} d x$ .
2
$\frac{e ^{2 x}}{e ^{x}} + \frac{3}{e ^{x}} = e^{x} + 3 e^{- x}$ — Split the fraction. Dividing powers of e means subtracting: 2x − x = x. And 1 over $e^{x}$ is $e^{- x}$ .
3
$e^{x} + 3 e^{- x} \to e^{x} - 3 e^{- x}$ — Integrate each term. For $3 e^{- x}$ , divide by −1.
4
$\int \frac{e ^{2 x} + 3}{e ^{x}} d x = e^{x} - 3 e^{- x} + c$ — Add + c.

Common mistake

Changing the power. $\int e^{2 x} d x$ is $\frac{1}{2} e^{2 x} + c$ . The power stays $2 x$ . Do not use the power rule on $e$ .

Common mistake

Multiplying by $a$ instead of dividing. $4 e^{2 x}$ is the derivative of $2 e^{2 x}$ , not the integral.

Key idea

The letter e for this number comes from Leonhard Euler, a Swiss mathematician of the 1700s. It is the only base whose curve has a gradient equal to its own height.

Key idea

e stays e. Keep the power, divide by the number in front of $x$ .

Sine and cosine

Recall: $sin x$ differentiates to $cos x$ , and $cos x$ differentiates to $- sin x$ . Rewind these, and divide by $a$ .

Formula

\int cos (a x + b) d x = \frac{1}{a} sin (a x + b) + c

$a$ — the number in front of $x$ . Divide by it.

Formula

\int sin (a x + b) d x = - \frac{1}{a} cos (a x + b) + c

$-$ — the minus sign. $cos$ differentiates to $- sin$ , so to get back to $+ sin$ you need $- cos$ .

Memory trick

Differentiating goes round: sin → cos → −sin → −cos → sin. Integrating goes round the other way.

Worked example

1
Find $\int (6 cos 3 x - 4 sin 2 x) d x$ .
2
$6 cos 3 x \to \frac{6}{3} sin 3 x = 2 sin 3 x$ — Cosine integrates to sine. Divide by 3.
3
$- 4 sin 2 x$ $\to - 4 \times (- \frac{1}{2} cos 2 x) = 2 cos 2 x$ — Sine integrates to minus cosine, divided by 2. Two minus signs make a plus.
4
$\int (6 cos 3 x - 4 sin 2 x) d x$ $= 2 sin 3 x + 2 cos 2 x + c$ — Put together and add + c.

Worked example

1
A curve has $\frac{d y}{d x} = 6 sin 3 x$ and passes through $(\frac{π}{3}, 1)$ . Find the equation of the curve.
2
$y = - 2 cos 3 x + c$ — Sine gives minus cosine. 6 divided by 3 is 2.
3
$1 = - 2 cos π + c$ — Put in the point: $3 \times \frac{π}{3} = π$ .
4
$1 = - 2 (- 1) + c, c = - 1$ — $cos π = - 1$ , an exact value. No calculator needed.
5
$y = - 2 cos 3 x - 1$ — The full equation of the curve.

Common mistake

Getting the sign wrong. Check by differentiating: $- cos x$ gives $+ sin x$ .

Common mistake

Using degrees. In calculus, angles are always in radians. On the non-calculator paper, use exact values such as $cos π = - 1$ .

Key idea

Cosine to sine: no sign change. Sine to cosine: put a minus in front.

(ax + b) to a power

Rule

This step is for 0606 and 4037. In 4PM1, expand a bracket before you integrate.

You could expand $(2 x + 3)^{4}$ , but that is slow, and for $4 x + 1$ it is impossible. Instead, treat the bracket like $x$ and use the power rule. Then divide by $a$ .

Formula

\int (a x + b)^{n} d x = \frac{( a x + b ) ^{n + 1}}{a ( n + 1 )} + c

$n$ — the power. Any rational number except −1.
$a$ — the number in front of $x$ inside the bracket.

Worked example

1
Find $\int (2 x + 3)^{4} d x$ .
2
$(2 x + 3)^{5}$ — Add 1 to the power. Keep the bracket exactly as it is.
3
$\frac{( 2 x + 3 ) ^{5}}{5 \times 2}$ — Divide by the new power, 5, and by $a = 2$ .
4
$= \frac{( 2 x + 3 ) ^{5}}{10} + c$ — Tidy and add + c.

Worked example

1
Find $\int \frac{6}{( 3 x - 1 ) ^{3}} d x$ .
2
$6 (3 x - 1)^{- 3}$ — Bring the bracket up: the power becomes −3.
3
$\frac{6 ( 3 x - 1 ) ^{- 2}}{3 \times ( - 2 )}$ — Power −3 + 1 = −2. Divide by −2 and by $a = 3$ .
4
$= - (3 x - 1)^{- 2} + c = - \frac{1}{( 3 x - 1 ) ^{2}} + c$ — 6 ÷ (−6) = −1.

Worked example

1
A curve has $\frac{d y}{d x} = 4 x + 1$ and passes through $(2, 10)$ . Find the equation of the curve.
2
$y = \int (4 x + 1)^{\frac{1}{2}} d x$ — Write the root as a power.
3
$y = \frac{( 4 x + 1 ) ^{\frac{3}{2}}}{4 \times \frac{3}{2}} + c = \frac{( 4 x + 1 ) ^{\frac{3}{2}}}{6} + c$ — Power one half + 1 = 3/2. Divide by 3/2 and by 4.
4
$10 = \frac{9 ^{\frac{3}{2}}}{6} + c = \frac{27}{6} + c$ — Put in the point. $9^{\frac{3}{2}} = (9)^{3} = 27$ .
5
$c = 10 - \frac{9}{2} = \frac{11}{2}$ — Solve for c.
6
$y = \frac{( 4 x + 1 ) ^{\frac{3}{2}}}{6} + \frac{11}{2}$ — The full equation.

Mark schemes give a method mark for $k (a x + b)^{n + 1}$ with any number $k$ , then an accuracy mark for the right $k$ . So always write the bracket with its new power.

Common mistake

Forgetting to divide by $a$ . Differentiate your answer: if you do not get back exactly what you started with, a factor is missing.

Common mistake

Using this on a bracket that is not linear. $\int (x^{2} + 1)^{3} d x$ is NOT $\frac{( x ^{2} + 1 ) ^{4}}{4}$ divided by anything. Expand it first.

Key idea

Treat the bracket like $x$ : up one, divide by the new power, then divide by $a$ .

1 over (ax + b) gives ln

Rule

This step is for 0606 and 4037. 4PM1 does not ask you to integrate $\frac{1}{x}$ .

The power rule fails for the power −1, because you would divide by 0. You met the fix already: $\int \frac{1}{x} d x = ln x + c$ . The same idea works with a bracket, and you divide by $a$ .

Formula

\int \frac{1}{a x + b} d x = \frac{1}{a} ln (a x + b) + c

$a$ — the number in front of $x$ on the bottom. Divide by it.
$a x + b$ — must be positive here, because ln only works on positive numbers.

Why? Differentiate $ln (a x + b)$ with the chain rule: you get $\frac{a}{a x + b}$ . So rewinding $\frac{1}{a x + b}$ needs $\frac{1}{a}$ in front.

Worked example

1
Find $\int \frac{6}{2 x + 5} d x$ .
2
$6 \times \frac{1}{2 x + 5}$ — The 6 waits in front.
3
$6 \times \frac{1}{2} ln (2 x + 5)$ — One over a linear bracket gives ln of the bracket. Divide by $a = 2$ .
4
$= 3 ln (2 x + 5) + c$ — Tidy and add + c.

Worked example

1
Find $\int \frac{4}{1 - 2 x} d x$ , for $x < \frac{1}{2}$ .
2
$a = - 2$ — The number in front of $x$ is −2. Watch the sign.
3
$4 \times \frac{1}{- 2} ln (1 - 2 x)$ — ln of the bracket, divided by −2.
4
$= - 2 ln (1 - 2 x) + c$ — Tidy and add + c. The condition $x < \frac{1}{2}$ keeps the bracket positive.

Common mistake

Using ln when the bracket is squared. $\int \frac{1}{( 2 x + 1 ) ^{2}} d x$ is a power −2, so use the power rule: $- \frac{1}{2 ( 2 x + 1 )} + c$ .

Common mistake

Losing the minus sign when $a$ is negative, as in $1 - 2 x$ .

Key idea

Only the power −1 gives ln. Leave ln answers exact on the non-calculator paper.

sec² and using identities

Rule

The sec² part is for 0606 and 4037 only. 4PM1 does not ask you to integrate $sec^{2}$ .

You know $\frac{d}{d x} (tan x) = sec^{2} x$ . Rewind it, and divide by $a$ .

Formula

\int sec^{2} (a x + b) d x = \frac{1}{a} tan (a x + b) + c

$a$ — the number in front of $x$ . Divide by it.

Worked example

1
Find $\int 4 sec^{2} (2 x + 1) d x$ .
2
$4 \times \frac{1}{2} tan (2 x + 1)$ — sec² rewinds to tan. Divide by $a = 2$ .
3
$= 2 tan (2 x + 1) + c$ — Tidy and add + c.

Some functions have no rule of their own, such as $tan^{2} x$ or $sin^{2} x$ . Use an identity to turn them into functions you can integrate.

Worked example

1
0606 and 4037: find $\int tan^{2} 3 x d x$ .
2
$tan^{2} 3 x = sec^{2} 3 x - 1$ — From $sec^{2} A = 1 + tan^{2} A$ , given in the List of formulas.
3
$\int (sec^{2} 3 x - 1) d x$ — Now both terms have a rule.
4
$= \frac{1}{3} tan 3 x - x + c$ — Divide by 3 for the sec² term. The −1 becomes −x.

Worked example

1
4PM1: find $\int 4 sin^{2} x d x$ .
2
$cos 2 x = 1 - 2 sin^{2} x$ — Put $A = B = x$ in $cos (A + B)$ from the formula sheet: $cos 2 x = cos^{2} x - sin^{2} x$ . Then use $cos^{2} x = 1 - sin^{2} x$ .
3
$4 sin^{2} x = 2 - 2 cos 2 x$ — Rearrange: $2 sin^{2} x = 1 - cos 2 x$ , then double it.
4
$\int (2 - 2 cos 2 x) d x = 2 x - sin 2 x + c$ — 2 becomes 2x. Cosine gives sine, divided by 2.

Common mistake

Writing $\int tan^{2} x d x = \frac{t a n ^{3} x}{3}$ . The power rule only works on powers of $x$ or of a linear bracket, not on powers of tan or sin.

Key idea

No rule for it? Change it with an identity first, then integrate.

Which method? Read the question

Look at the function. What you see tells you what to do.

You see	You do
$e^{a x + b}$	Keep the power, divide by $a$
$cos (a x + b)$	$\frac{1}{a} sin (a x + b)$
$sin (a x + b)$	$- \frac{1}{a} cos (a x + b)$
$(a x + b)^{n}$ (0606 / 4037)	Up one, divide by the new power and by $a$
$\frac{1}{a x + b}$ (0606 / 4037)	$\frac{1}{a} ln (a x + b)$
$sec^{2} (a x + b)$ (0606 / 4037)	$\frac{1}{a} tan (a x + b)$
$tan^{2} x$ (0606 / 4037), $sin^{2} x$ or $cos^{2} x$ (4PM1)	Use an identity first
A gradient and a point	Integrate with + c, then put in the point

Common mistake

Multiplying by $a$ out of differentiation habit. When you integrate, you divide.

Key idea

Every rule in this lesson is the chain rule run backwards. Differentiate your answer to check it.

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