Integration

What you'll learn

How integration reverses differentiation, including why you need a constant of integration.
How to integrate powers, exponentials, logs and trigonometric functions.
How substitution, integration by parts and partial fractions help with harder integrals.
How integration is used for areas, parametric curves, differential equations and estimates.

1. The big idea: integration reverses differentiation

Integration is the opposite process to differentiation. If differentiating tells you the gradient function, integrating helps you recover the original function.

Integration reverses differentiation but gives a family of curves that differ by a constant.

Definition

Indefinite integral

An indefinite integral such as $dx\int f(x)\,dx$ means “find all functions whose derivative is $f (x)$ ”. The symbol $d x$ tells you the variable, and $+ c$ is the constant of integration because constants disappear when differentiated.

Key Idea

Reverse the differentiation rules

For $\neq -1$ , $dx=xn+1n+1+c\int x^n\,dx=\frac{x^{n+1}}{n+1}+c$ .
The special case is $dx=ln⁡∣x∣+c\int \frac{1}{x}\,dx=\ln|x|+c$ .
If the expression is a linear bracket like $a x + b$ , divide by the derivative of the bracket, so divide by $a$ .

Example

Basic algebraic integrals

Find
$dxI_1=\int \left(\sqrt{5x-3}+\frac{2}{(4x+1)^3}\right)\,dx$
Rewrite using powers, then apply the reverse chain rule:
$dx=215(5x−3)3/2−14(4x+1)−2+c\begin{aligned} I_1 &=\int \left((5x-3)^{1/2}+2(4x+1)^{-3}\right)\,dx \\ &=\frac{2}{15}(5x-3)^{3/2}-\frac{1}{4}(4x+1)^{-2}+c \end{aligned}$
For a definite log-style integral, find
$dxI_2=\int_0^1 \frac{6}{3x+2}\,dx$
The derivative of $3 x + 2$ is 3, so the numerator is twice that derivative:
$I2=[2ln⁡(3x+2)]01=2ln⁡5−2ln⁡2=2ln⁡(52)I_2=\left[2\ln(3x+2)\right]_0^1=2\ln5-2\ln2=2\ln\left(\frac{5}{2}\right)$

Common Mistake

Forgetting the hidden derivative

If you integrate $5x-3)^{1/2}$ as $23(5x−3)3/2\frac{2}{3}(5x-3)^{3/2}$ , differentiating gives five times too much. Always divide by the derivative of the bracket.

2. Trigonometric integrals and identities

For A-Level calculus, trigonometric angles are in radians unless told otherwise. You need to know the derivative facts so that you can reverse them confidently.

Useful identities include:

The double-angle identities show how squared trigonometric graphs can be rewritten before integration.

tan⁡2θ=sec⁡2θ−1sin⁡2x=1−cos⁡2x2cos⁡2x=1+cos⁡2x2\begin{aligned} \tan^2\theta &= \sec^2\theta-1 \\ \sin^2x &= \frac{1-\cos2x}{2} \\ \cos^2x &= \frac{1+\cos2x}{2} \end{aligned}

Tip

Differentiate to check

For trig integrals, a quick derivative check is very powerful. For example, differentiating $tan⁡(4x)\tan(4x)$ gives $4\sec^2(4x)$ , so integrating $sec^2(4x)$ must introduce a factor of $14\frac{1}{4}$ .

Example

Trig functions and a squared integral

Find
$dxI=\int\left(3\sec^2(2x)-4\operatorname{cosec}(5x)\cot(5x)\right)\,dx$
Use $ddxtan⁡(2x)=2sec⁡2(2x)\frac{d}{dx}\tan(2x)=2\sec^2(2x)$ and $ddxcosec⁡(5x)=−5cosec⁡(5x)cot⁡(5x)\frac{d}{dx}\operatorname{cosec}(5x)=-5\operatorname{cosec}(5x)\cot(5x)$ :
$I=32tan⁡(2x)+45cosec⁡(5x)+cI=\frac{3}{2}\tan(2x)+\frac{4}{5}\operatorname{cosec}(5x)+c$
Now find
$dxJ=\int_0^{\frac{\pi}{2}}\cos^2x\,dx$
Replace $cos^2x$ using the double-angle identity, then integrate:
$dx=[x2+sin⁡2x4]0π2=π4\begin{aligned} J&=\int_0^{\frac{\pi}{2}}\frac{1+\cos2x}{2}\,dx \\ &=\left[\frac{x}{2}+\frac{\sin2x}{4}\right]_0^{\frac{\pi}{2}} \\ &=\frac{\pi}{4} \end{aligned}$

3. Integration by substitution

Definition

Substitution

A substitution changes the variable in an integral, usually from $x$ to $u$ , so that a complicated expression becomes simpler. It is the reverse of the chain rule.

For definite integrals, you can either change back to $x$ before substituting the limits, or change the limits into $u$ -values straight away. Do not mix the two methods.

Changing variables in a definite integral also changes the limits.

Example

Changing limits with $u = 1 + cos x$

Find
$dxI=\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\cos x}\,dx$
Let $u=1+cos⁡xu=1+\cos x$ , so $dudx=−sin⁡x\frac{du}{dx}=-\sin x$ and therefore $dx=−du\sin x\,dx=-du$ .
Change the limits: when $x = 0$ , $u = 2$ ; when $x=π2x=\frac{\pi}{2}$ , $u = 1$ .
Substitute carefully, including the minus sign:
$du=[ln⁡u]12=ln⁡2\begin{aligned} I&=\int_2^1 -\frac{1}{u}\,du \\ &=\int_1^2 \frac{1}{u}\,du \\ &=\left[\ln u\right]_1^2=\ln2 \end{aligned}$

Common Mistake

Mixing variables

If you change the limits into $u$ -values, finish the integral in $u$ . If you want to use the original $x$ -limits, change your answer back to $x$ first.

4. Integration by parts

Integration by parts is the product rule in reverse. Use it when your integrand is a product, especially involving $x$ , $ln⁡x\ln x$ , exponentials or trig functions.

Integration by parts comes from reversing the product rule and splitting a product into chosen parts.

dx\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx

Example

A product like $x sin x$

Let
$dxI=\int x\sin x\,dx$
Choose $u = x$ and $dvdx=sin⁡x\frac{dv}{dx}=\sin x$ , so $dudx=1\frac{du}{dx}=1$ and $v=−cos⁡xv=-\cos x$ .
Substitute into the formula:
$dx=−xcos⁡x+sin⁡x+c\begin{aligned} I&=x(-\cos x)-\int(-\cos x)(1)\,dx \\ &=-x\cos x+\int\cos x\,dx \\ &=-x\cos x+\sin x+c \end{aligned}$
A quick check is that differentiating $−xcos⁡x+sin⁡x-x\cos x+\sin x$ gives $xsin⁡xx\sin x$ .

5. Algebraic integration and partial fractions

Definition

Partial fractions

Partial fractions rewrite one complicated rational expression as a sum of simpler fractions. They are useful because terms like $1x+a\frac{1}{x+a}$ integrate to logarithms.

If the numerator has degree greater than or equal to the denominator, use algebraic division first. For repeated factors, include every power of the factor.

Example

Partial fractions before integrating

Decompose
$3x(x+1)(x−2)\frac{3x}{(x+1)(x-2)}$
Write
$3x(x+1)(x−2)=Ax+1+Bx−2\frac{3x}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}$
Multiply through by $(x + 1) (x - 2)$ and compare coefficients:
$3 x = A (x - 2) + B (x + 1)$
Match coefficients of $x$ and constants:
$B=2A+B=3,\qquad -2A+B=0 \Rightarrow A=1,\ B=2$
Integrate the simpler form:
$dx=ln⁡∣x+1∣+2ln⁡∣x−2∣+c\int\frac{3x}{(x+1)(x-2)}\,dx=\ln|x+1|+2\ln|x-2|+c$

Common Mistake

Vertical asymptotes

For indefinite integrals, use modulus signs in $ln⁡∣x−a∣\ln|x-a|$ . For definite integrals, check the interval does not cross a point where the denominator is zero.

6. Areas, including parametric curves

A definite integral gives signed area. If the curve is above the $x$ -axis, the area is positive. If finding the area between two curves, integrate “upper curve minus lower curve”.

Definite integrals count area above the x-axis as positive and area below it as negative.

Definition

Parametric equations

Parametric equations describe $x$ and $y$ using a third variable, often $t$ . For area under a parametric curve, use $dtA=\int y\,dx=\int y(t)\frac{dx}{dt}\,dt$ .

For a parametric curve, the area element is the height y(t) times the horizontal change dx.

Example

Area under a parametric curve

A curve is given by $x=2sin⁡tx=2\sin t$ , $y=3sin⁡2ty=3\sin2t$ , where $0≤t≤π20\leq t\leq \frac{\pi}{2}$ . Since $x$ increases from 0 to 2, use the area formula directly.
Differentiate $x$ with respect to $t$ :
$dxdt=2cos⁡t\frac{dx}{dt}=2\cos t$
Substitute into $dtA=\int y\frac{dx}{dt}\,dt$ :
$dt\begin{aligned} A&=\int_0^{\frac{\pi}{2}}3\sin2t(2\cos t)\,dt \\ &=\int_0^{\frac{\pi}{2}}12\sin t\cos^2t\,dt \end{aligned}$
Use $u=cos⁡tu=\cos t$ , so $dtdu=-\sin t\,dt$ , with limits $u = 1$ to $u = 0$ :
$du=4A=\int_1^0 -12u^2\,du=\int_0^1 12u^2\,du=4$

Common Mistake

Negative parametric area

If $dxdt\frac{dx}{dt}$ is negative over the interval, $dt\int y\frac{dx}{dt}\,dt$ may give a negative signed area. Reverse the limits or take the positive area when the question asks for area.

7. Differential equations

Definition

Differential equation

A differential equation contains a derivative, such as $dydx\frac{dy}{dx}$ . It is separable if you can rearrange it so all the $y$ terms are with $d y$ and all the $x$ terms are with $d x$ .

When a rate is directly proportional to the amount present, you often get an exponential model such as $P=Ae^{kt}$ .

Example

Solving a separable differential equation

Solve $dydx=(y+2)2\frac{dy}{dx}=(y+2)^2$ , given that $y = - 1$ when $x = 0$ .
Separate the variables:
$dy=dx\frac{1}{(y+2)^2}\,dy=dx$
Integrate both sides:
$−1y+2=x+c-\frac{1}{y+2}=x+c$
Use the initial condition:
$\Rightarrow c=-1$
Rearrange into the form $y = f (x)$ :
$−1y+2=x−1⇒1y+2=1−x⇒y=11−x−2-\frac{1}{y+2}=x-1 \Rightarrow \frac{1}{y+2}=1-x \Rightarrow y=\frac{1}{1-x}-2$

8. The trapezium rule

The trapezium rule estimates an area by splitting it into trapezia. The vertical values used are called ordinates.

The trapezium rule replaces strips under a curve with trapezia whose parallel sides are the ordinates.

For width $h$ and ordinates $y0,y1,…,yny_0,y_1,\dots,y_n$ ,

dx≈h2(y0+yn+2(y1+y2+⋯+yn−1))\int_a^b y\,dx\approx \frac{h}{2}\left(y_0+y_n+2(y_1+y_2+\cdots+y_{n-1})\right)

Example

Using five ordinates

Estimate $dx\int_0^2 e^{x/2}\,dx$ using ordinates at $x = 0, 0.5, 1, 1.5, 2$ . The corresponding values are 1.000, 1.284, 1.649, 2.117 and 2.718.
There are four strips, so the strip width is 0.5.
Substitute into the trapezium rule:
$Estimate=0.52(1.000+2.718+2(1.284+1.649+2.117))=3.4545\begin{aligned} \text{Estimate} &=\frac{0.5}{2}\left(1.000+2.718+2(1.284+1.649+2.117)\right) \\ &=3.4545 \end{aligned}$
To three significant figures, the estimate is 3.45.

Tip

Overestimate or underestimate

Trapezia use straight chords. For a concave-up curve, the chords lie above the curve, so the estimate is an overestimate; for a concave-down curve, it is an underestimate.

Chord positions explain why the trapezium rule overestimates concave-up curves and underestimates concave-down curves.

Exam technique

In the exam

For every indefinite integral, include $+ c$ unless the question is definite.
Differentiate your answer quickly to check missing scale factors.
In substitution, change the limits or change back to the original variable, but do not mix methods.
For areas, decide whether you need signed area, positive area, or upper curve minus lower curve.
Keep exact answers exact: simplify logs using laws of logarithms only at the end.

Self review

Check yourself

Why does $dx\int(2x+1)^5\,dx$ need a factor involving the derivative of $2 x + 1$ ?
In the substitution $u=1+cos⁡xu=1+\cos x$ , what happens to the limits $x = 0$ and $x=π2x=\frac{\pi}{2}$ ?
Which technique would you try first for $dx\int x\ln x\,dx$ , and why?

Integration

What you'll learn

How integration reverses differentiation, including why you need a constant of integration.
How to integrate powers, exponentials, logs and trigonometric functions.
How substitution, integration by parts and partial fractions help with harder integrals.
How integration is used for areas, parametric curves, differential equations and estimates.

1. The big idea: integration reverses differentiation

Integration is the opposite process to differentiation. If differentiating tells you the gradient function, integrating helps you recover the original function.

Integration reverses differentiation but gives a family of curves that differ by a constant.

Definition

Indefinite integral

Key Idea

Reverse the differentiation rules

For $\neq -1$ , $dx=xn+1n+1+c\int x^n\,dx=\frac{x^{n+1}}{n+1}+c$ .
The special case is $dx=ln⁡∣x∣+c\int \frac{1}{x}\,dx=\ln|x|+c$ .
If the expression is a linear bracket like $a x + b$ , divide by the derivative of the bracket, so divide by $a$ .

Example

Basic algebraic integrals

Find
$dxI_1=\int \left(\sqrt{5x-3}+\frac{2}{(4x+1)^3}\right)\,dx$
Rewrite using powers, then apply the reverse chain rule:
$dx=215(5x−3)3/2−14(4x+1)−2+c\begin{aligned} I_1 &=\int \left((5x-3)^{1/2}+2(4x+1)^{-3}\right)\,dx \\ &=\frac{2}{15}(5x-3)^{3/2}-\frac{1}{4}(4x+1)^{-2}+c \end{aligned}$
For a definite log-style integral, find
$dxI_2=\int_0^1 \frac{6}{3x+2}\,dx$
The derivative of $3 x + 2$ is 3, so the numerator is twice that derivative:
$I2=[2ln⁡(3x+2)]01=2ln⁡5−2ln⁡2=2ln⁡(52)I_2=\left[2\ln(3x+2)\right]_0^1=2\ln5-2\ln2=2\ln\left(\frac{5}{2}\right)$

Common Mistake

Forgetting the hidden derivative

If you integrate $5x-3)^{1/2}$ as $23(5x−3)3/2\frac{2}{3}(5x-3)^{3/2}$ , differentiating gives five times too much. Always divide by the derivative of the bracket.

2. Trigonometric integrals and identities

For A-Level calculus, trigonometric angles are in radians unless told otherwise. You need to know the derivative facts so that you can reverse them confidently.

Useful identities include:

The double-angle identities show how squared trigonometric graphs can be rewritten before integration.

tan⁡2θ=sec⁡2θ−1sin⁡2x=1−cos⁡2x2cos⁡2x=1+cos⁡2x2\begin{aligned} \tan^2\theta &= \sec^2\theta-1 \\ \sin^2x &= \frac{1-\cos2x}{2} \\ \cos^2x &= \frac{1+\cos2x}{2} \end{aligned}

Tip

Differentiate to check

Example

Trig functions and a squared integral

Find
$dxI=\int\left(3\sec^2(2x)-4\operatorname{cosec}(5x)\cot(5x)\right)\,dx$
Use $ddxtan⁡(2x)=2sec⁡2(2x)\frac{d}{dx}\tan(2x)=2\sec^2(2x)$ and $ddxcosec⁡(5x)=−5cosec⁡(5x)cot⁡(5x)\frac{d}{dx}\operatorname{cosec}(5x)=-5\operatorname{cosec}(5x)\cot(5x)$ :
$I=32tan⁡(2x)+45cosec⁡(5x)+cI=\frac{3}{2}\tan(2x)+\frac{4}{5}\operatorname{cosec}(5x)+c$
Now find
$dxJ=\int_0^{\frac{\pi}{2}}\cos^2x\,dx$
Replace $cos^2x$ using the double-angle identity, then integrate:
$dx=[x2+sin⁡2x4]0π2=π4\begin{aligned} J&=\int_0^{\frac{\pi}{2}}\frac{1+\cos2x}{2}\,dx \\ &=\left[\frac{x}{2}+\frac{\sin2x}{4}\right]_0^{\frac{\pi}{2}} \\ &=\frac{\pi}{4} \end{aligned}$

3. Integration by substitution

Definition

Substitution

A substitution changes the variable in an integral, usually from $x$ to $u$ , so that a complicated expression becomes simpler. It is the reverse of the chain rule.

For definite integrals, you can either change back to $x$ before substituting the limits, or change the limits into $u$ -values straight away. Do not mix the two methods.

Changing variables in a definite integral also changes the limits.

Example

Changing limits with $u = 1 + cos x$

Find
$dxI=\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\cos x}\,dx$
Let $u=1+cos⁡xu=1+\cos x$ , so $dudx=−sin⁡x\frac{du}{dx}=-\sin x$ and therefore $dx=−du\sin x\,dx=-du$ .
Change the limits: when $x = 0$ , $u = 2$ ; when $x=π2x=\frac{\pi}{2}$ , $u = 1$ .
Substitute carefully, including the minus sign:
$du=[ln⁡u]12=ln⁡2\begin{aligned} I&=\int_2^1 -\frac{1}{u}\,du \\ &=\int_1^2 \frac{1}{u}\,du \\ &=\left[\ln u\right]_1^2=\ln2 \end{aligned}$

Common Mistake

Mixing variables

If you change the limits into $u$ -values, finish the integral in $u$ . If you want to use the original $x$ -limits, change your answer back to $x$ first.

4. Integration by parts

Integration by parts is the product rule in reverse. Use it when your integrand is a product, especially involving $x$ , $ln⁡x\ln x$ , exponentials or trig functions.

Integration by parts comes from reversing the product rule and splitting a product into chosen parts.

dx\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx

Example

A product like $x sin x$

Let
$dxI=\int x\sin x\,dx$
Choose $u = x$ and $dvdx=sin⁡x\frac{dv}{dx}=\sin x$ , so $dudx=1\frac{du}{dx}=1$ and $v=−cos⁡xv=-\cos x$ .
Substitute into the formula:
$dx=−xcos⁡x+sin⁡x+c\begin{aligned} I&=x(-\cos x)-\int(-\cos x)(1)\,dx \\ &=-x\cos x+\int\cos x\,dx \\ &=-x\cos x+\sin x+c \end{aligned}$
A quick check is that differentiating $−xcos⁡x+sin⁡x-x\cos x+\sin x$ gives $xsin⁡xx\sin x$ .

5. Algebraic integration and partial fractions

Definition

Partial fractions

Partial fractions rewrite one complicated rational expression as a sum of simpler fractions. They are useful because terms like $1x+a\frac{1}{x+a}$ integrate to logarithms.

If the numerator has degree greater than or equal to the denominator, use algebraic division first. For repeated factors, include every power of the factor.

Example

Partial fractions before integrating

Decompose
$3x(x+1)(x−2)\frac{3x}{(x+1)(x-2)}$
Write
$3x(x+1)(x−2)=Ax+1+Bx−2\frac{3x}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}$
Multiply through by $(x + 1) (x - 2)$ and compare coefficients:
$3 x = A (x - 2) + B (x + 1)$
Match coefficients of $x$ and constants:
$B=2A+B=3,\qquad -2A+B=0 \Rightarrow A=1,\ B=2$
Integrate the simpler form:
$dx=ln⁡∣x+1∣+2ln⁡∣x−2∣+c\int\frac{3x}{(x+1)(x-2)}\,dx=\ln|x+1|+2\ln|x-2|+c$

Common Mistake

Vertical asymptotes

For indefinite integrals, use modulus signs in $ln⁡∣x−a∣\ln|x-a|$ . For definite integrals, check the interval does not cross a point where the denominator is zero.

6. Areas, including parametric curves

A definite integral gives signed area. If the curve is above the $x$ -axis, the area is positive. If finding the area between two curves, integrate “upper curve minus lower curve”.

Definite integrals count area above the x-axis as positive and area below it as negative.

Definition

Parametric equations

Parametric equations describe $x$ and $y$ using a third variable, often $t$ . For area under a parametric curve, use $dtA=\int y\,dx=\int y(t)\frac{dx}{dt}\,dt$ .

For a parametric curve, the area element is the height y(t) times the horizontal change dx.

Example

Area under a parametric curve

A curve is given by $x=2sin⁡tx=2\sin t$ , $y=3sin⁡2ty=3\sin2t$ , where $0≤t≤π20\leq t\leq \frac{\pi}{2}$ . Since $x$ increases from 0 to 2, use the area formula directly.
Differentiate $x$ with respect to $t$ :
$dxdt=2cos⁡t\frac{dx}{dt}=2\cos t$
Substitute into $dtA=\int y\frac{dx}{dt}\,dt$ :
$dt\begin{aligned} A&=\int_0^{\frac{\pi}{2}}3\sin2t(2\cos t)\,dt \\ &=\int_0^{\frac{\pi}{2}}12\sin t\cos^2t\,dt \end{aligned}$
Use $u=cos⁡tu=\cos t$ , so $dtdu=-\sin t\,dt$ , with limits $u = 1$ to $u = 0$ :
$du=4A=\int_1^0 -12u^2\,du=\int_0^1 12u^2\,du=4$

Common Mistake

Negative parametric area

If $dxdt\frac{dx}{dt}$ is negative over the interval, $dt\int y\frac{dx}{dt}\,dt$ may give a negative signed area. Reverse the limits or take the positive area when the question asks for area.

7. Differential equations

Definition

Differential equation

A differential equation contains a derivative, such as $dydx\frac{dy}{dx}$ . It is separable if you can rearrange it so all the $y$ terms are with $d y$ and all the $x$ terms are with $d x$ .

When a rate is directly proportional to the amount present, you often get an exponential model such as $P=Ae^{kt}$ .

Example

Solving a separable differential equation

Solve $dydx=(y+2)2\frac{dy}{dx}=(y+2)^2$ , given that $y = - 1$ when $x = 0$ .
Separate the variables:
$dy=dx\frac{1}{(y+2)^2}\,dy=dx$
Integrate both sides:
$−1y+2=x+c-\frac{1}{y+2}=x+c$
Use the initial condition:
$\Rightarrow c=-1$
Rearrange into the form $y = f (x)$ :
$−1y+2=x−1⇒1y+2=1−x⇒y=11−x−2-\frac{1}{y+2}=x-1 \Rightarrow \frac{1}{y+2}=1-x \Rightarrow y=\frac{1}{1-x}-2$

8. The trapezium rule

The trapezium rule estimates an area by splitting it into trapezia. The vertical values used are called ordinates.

The trapezium rule replaces strips under a curve with trapezia whose parallel sides are the ordinates.

For width $h$ and ordinates $y0,y1,…,yny_0,y_1,\dots,y_n$ ,

dx≈h2(y0+yn+2(y1+y2+⋯+yn−1))\int_a^b y\,dx\approx \frac{h}{2}\left(y_0+y_n+2(y_1+y_2+\cdots+y_{n-1})\right)

Example

Using five ordinates

Estimate $dx\int_0^2 e^{x/2}\,dx$ using ordinates at $x = 0, 0.5, 1, 1.5, 2$ . The corresponding values are 1.000, 1.284, 1.649, 2.117 and 2.718.
There are four strips, so the strip width is 0.5.
Substitute into the trapezium rule:
$Estimate=0.52(1.000+2.718+2(1.284+1.649+2.117))=3.4545\begin{aligned} \text{Estimate} &=\frac{0.5}{2}\left(1.000+2.718+2(1.284+1.649+2.117)\right) \\ &=3.4545 \end{aligned}$
To three significant figures, the estimate is 3.45.

Tip

Overestimate or underestimate

Trapezia use straight chords. For a concave-up curve, the chords lie above the curve, so the estimate is an overestimate; for a concave-down curve, it is an underestimate.

Chord positions explain why the trapezium rule overestimates concave-up curves and underestimates concave-down curves.

Exam technique

In the exam

For every indefinite integral, include $+ c$ unless the question is definite.
Differentiate your answer quickly to check missing scale factors.
In substitution, change the limits or change back to the original variable, but do not mix methods.
For areas, decide whether you need signed area, positive area, or upper curve minus lower curve.
Keep exact answers exact: simplify logs using laws of logarithms only at the end.

Self review

Check yourself

Why does $dx\int(2x+1)^5\,dx$ need a factor involving the derivative of $2 x + 1$ ?
In the substitution $u=1+cos⁡xu=1+\cos x$ , what happens to the limits $x = 0$ and $x=π2x=\frac{\pi}{2}$ ?
Which technique would you try first for $dx\int x\ln x\,dx$ , and why?

Integration

What you'll learn

1. The big idea: integration reverses differentiation

2. Trigonometric integrals and identities

3. Integration by substitution

4. Integration by parts

5. Algebraic integration and partial fractions

6. Areas, including parametric curves

7. Differential equations

8. The trapezium rule

Recap questions

You've reached the end

Integration

What you'll learn

1. The big idea: integration reverses differentiation

2. Trigonometric integrals and identities

3. Integration by substitution

4. Integration by parts

5. Algebraic integration and partial fractions

6. Areas, including parametric curves

7. Differential equations

8. The trapezium rule

Recap questions

You've reached the end

Integration

What you'll learn

1. The big idea: integration reverses differentiation

2. Trigonometric integrals and identities

3. Integration by substitution

4. Integration by parts

5. Algebraic integration and partial fractions

6. Areas, including parametric curves

7. Differential equations

8. The trapezium rule

Recap questions

You've reached the end

Integration Revision Guide

Integration

What you'll learn

1. The big idea: integration reverses differentiation

2. Trigonometric integrals and identities

3. Integration by substitution

4. Integration by parts

5. Algebraic integration and partial fractions

6. Areas, including parametric curves

7. Differential equations

8. The trapezium rule

Recap questions

You've reached the end

Integration Revision Guide