Integration

Reverse differentiation, with a linear function inside

Pure 1 integrated $x^n$ by raising the power by one and dividing by the new power. Pure 3 asks the same question with a linear function of $x$ inside the bracket, and the rule survives with one extra factor: $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+c$ for $n\ne -1$. The $a$ in the denominator is the derivative of what is inside the bracket, and leaving it out is the single commonest mistake in this topic.

The same idea produces the six standard results below. Each one is a derivative you already know, read backwards: differentiating $\sin (ax+b)$ gives $a\cos (ax+b)$, so integrating $\cos (ax+b)$ has to produce $\frac{1}{a}\sin (ax+b)$. The factor $\frac{1}{a}$ is what cancels the $a$ the chain rule would bring back.

Two of the results deserve a sentence each. $\int \frac{1}{x}dx=\ln |x|+c$ is the case the power rule cannot reach, because $n=-1$ would give a new power of zero; the modulus matters because $x$ may be negative and the logarithm of a negative number does not exist. And $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c$ is the one result in the list whose answer is a different kind of function from the integrand, so learn it as a shape: a quadratic denominator with no real roots.

• $\int (3x+1{)}^{4}dx=\frac{(3x+1{)}^5}{15}+c$, because $a=3$ and $n+1=5$
• $\int e^{4x-1}dx=\frac{1}{4}{e}^{4x-1}+c$
• $\int \cos (2x+1)dx=\frac{1}{2}\sin (2x+1)+c$, with no sign change
• $\int \sin (2x+1)dx=-\frac{1}{2}\cos (2x+1)+c$, with the sign change
• $\int {\sec }^{2}3xdx=\frac{1}{3}\tan 3x+c$
• $\int \frac{1}{2x+5}dx=\frac{1}{2}\ln |2x+5|+c$
• $\int \frac{1}{x^2+16}dx=\frac{1}{4}{\tan }^{-1}\frac{x}{4}+c$

Logarithmic integrands: the form $k\frac{f^{\prime }(x)}{f(x)}$

If the top of a fraction is a constant multiple of the derivative of the bottom, the integral is a logarithm. The syllabus states the result as $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+c$, and the formula list prints the same line. The test is quick: differentiate the bottom in your head and compare it with the top.

Take $\int \frac{2x}{x^2+3}dx$. The derivative of $x^2+3$ is $2x$, which is exactly the top, so the answer is $\ln (x^2+3)+c$. If the top were $4x$ the answer would be $2\ln (x^2+3)+c$, because $4x=2(2x)$ and a constant factor comes straight out of the integral. If the top were $x$ the answer would be $\frac{1}{2}\ln (x^2+3)+c$, because $x=\frac{1}{2}(2x)$. The factor you pull out is the one you need, and nothing more.

$\tan x$ is the example Cambridge names. Write it as $\frac{\sin x}{\cos x}$ and the top is the derivative of the bottom up to a sign, since $\frac{d}{dx}\cos x=-\sin x$. So $\int \tan xdx=-\int \frac{-\sin x}{\cos x}dx=-\ln |\cos x|+c$, which is also written $\ln |\sec x|+c$. Either form is accepted; the first is easier to reach.

When the top is not a constant multiple of the derivative of the bottom, split it. $\int \frac{3x+2}{x^2+1}dx$ mixes an $x$ term with a constant, so write it as $\frac{3}{2}\int \frac{2x}{x^2+1}dx+2\int \frac{1}{x^2+1}dx=\frac{3}{2}\ln (x^2+1)+2{\tan }^{-1}x+c$. Before splitting, compare the degrees: if the top has the same degree as the bottom or a higher one, divide first, because both the logarithm result and the inverse tangent result need a top of lower degree.

The $\frac{1}{x^2+a^2}$ result, and completing the square

The formula list gives $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c$ for a positive constant $a$. The denominator has to have no $x$ term and no coefficient on $x^2$. When it has either, complete the square first and the result applies again.

$\int \frac{1}{x^2+4x+13}dx$: the denominator is $(x+2{)}^2+9$, so the integral is $\frac{1}{3}{\tan }^{-1}\frac{x+2}{3}+c$. Differentiating that back gives $\frac{1}{3}·\frac{1/3}{1+{\left(\frac{x+2}{3}\right)}^2}$, which simplifies to $\frac{3}{(x+2{)}^2+9}$, and the $\frac{1}{3}$ in front brings it to the integrand. Every step of that check is worth doing once by hand, because the factor of $a$ is easy to misplace.

When the top contains an $x$, split the integral into a logarithm part and an inverse tangent part. For $\int \frac{x+1}{x^2+4x+13}dx$ the derivative of the bottom is $2x+4$, so $x+1=\frac{1}{2}(2x+4)-1$ and the integral is $\frac{1}{2}\ln (x^2+4x+13)-\frac{1}{3}{\tan }^{-1}\frac{x+2}{3}+c$. Two techniques from two different paragraphs appear in one line, which is exactly what a Paper 3 question looks like.

Trigonometric integrals: use an identity first

You cannot integrate ${\sin }^{2}x$ by raising a power, because $\sin x$ is a function and not a power of $x$. The way in is a double-angle formula, which turns a square into something linear in $\cos 2x$: ${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$ and ${\cos }^{2}x=\frac{1}{2}(1+\cos 2x)$. Both come from $\cos 2x=1-2{\sin }^{2}x=2{\cos }^{2}x-1$, which is printed in the trigonometry section of the formula list.

So $\int {\sin }^{2}xdx=\frac{1}{2}\int (1-\cos 2x)dx=\frac{1}{2}x-\frac{1}{4}\sin 2x+c$. Note the $\frac{1}{4}$: integrating $\cos 2x$ brings its own $\frac{1}{2}$, and the $\frac{1}{2}$ already outside multiplies it. The same working on ${\cos }^2(2x)$ uses ${\cos }^2(2x)=\frac{1}{2}(1+\cos 4x)$ and gives $\frac{1}{2}x+\frac{1}{8}\sin 4x+c$. The angle doubles, so the $a$ in the standard result doubles with it.

Two other identities earn their place on this page. $\sin 2x=2\sin x\cos x$ handles a product such as $\int \sin x\cos xdx$, which is $\frac{1}{2}{\sin }^{2}x+c$, or $-\frac{1}{4}\cos 2x+c$ if you use the double angle instead: the two answers differ by a constant, so both are right. And ${\sec }^{2}x=1+{\tan }^{2}x$ handles $\int {\tan }^{2}xdx=\int ({\sec }^{2}x-1)dx=\tan x-x+c$.

A definite integral with a squared trigonometric function is finished in the usual way once the square has gone. ${\int }_0^{\pi /2}{\sin }^{2}xdx={[\frac{1}{2}x-\frac{1}{4}\sin 2x]}_0^{\pi /2}=\frac{\pi }{4}-0=\frac{\pi }{4}$, because $\sin \pi =0$.

Partial fractions, then integrate

A rational function whose denominator factorises becomes a sum of simpler fractions, and each piece integrates to a logarithm, or to a power when the factor is repeated. Cambridge restricts the types to the three in topic 3.1: three distinct linear factors $(ax+b)(cx+d)(ex+f)$, a repeated linear factor $(ax+b)(cx+d{)}^2$, and a linear factor times an irreducible quadratic $(ax+b)(c{x}^2+d)$. The numerator must be of lower degree than the denominator.

$\int \frac{3x+2}{(x-1)(x+2)}dx$: write $\frac{3x+2}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}$, so $3x+2=A(x+2)+B(x-1)$. Putting $x=1$ gives $5=3A$, so $A=\frac{5}{3}$; putting $x=-2$ gives $-4=-3B$, so $B=\frac{4}{3}$. The integral is $\frac{5}{3}\ln |x-1|+\frac{4}{3}\ln |x+2|+c$.

A repeated factor needs a term for each power. $\frac{1}{x(x+1{)}^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}$, so $1=A(x+1{)}^2+Bx(x+1)+Cx$. From $x=0$, $A=1$; from $x=-1$, $C=-1$; comparing $x^2$ terms, $A+B=0$, so $B=-1$. The integral is $\ln |x|-\ln |x+1|+\frac{1}{x+1}+c$. The repeated factor produced the $\frac{1}{x+1}$ term, not a third logarithm.

Sometimes the split is not needed at all. $\int \frac{2x+3}{(x+1{)}^2}dx$ is quicker once you notice that $2x+3=2(x+1)+1$, so the integral is $2\ln |x+1|-\frac{1}{x+1}+c$. Look for that before you start writing out partial fractions.

An irreducible quadratic in the denominator brings an inverse tangent with it. $\frac{x+4}{(x+1)(x^2+1)}=\frac{3/2}{x+1}+\frac{-\frac{3}{2}x+\frac{5}{2}}{x^2+1}$, so the integral is $\frac{3}{2}\ln |x+1|-\frac{3}{4}\ln (x^2+1)+\frac{5}{2}{\tan }^{-1}x+c$. The quadratic factor always takes a numerator of the form $Bx+C$, never a single constant.

Integration by parts

The formula list prints $\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dx}dx$. It is the product rule read backwards, and it is the tool for an integrand that is a product of two different kinds of function: an algebraic factor times an exponential, a trigonometric function or a logarithm.

The whole skill is choosing $u$. Take for $u$ the factor that becomes simpler when it is differentiated. In practice that means the logarithm first, because $\ln x$ turns into $\frac{1}{x}$; then the power of $x$, because it loses a degree; and the exponential or trigonometric factor goes into $\frac{dv}{dx}$, because integrating it never makes things worse. The syllabus names four examples: $x\sin 2x$, $x^2{e}^{-x}$, $\ln x$ and $x{\tan }^{-1}x$.

$\int x\sin 2xdx$: take $u=x$ and $\frac{dv}{dx}=\sin 2x$, so $\frac{du}{dx}=1$ and $v=-\frac{1}{2}\cos 2x$. Then $\int x\sin 2xdx=-\frac{1}{2}x\cos 2x+\frac{1}{2}\int \cos 2xdx=-\frac{1}{2}x\cos 2x+\frac{1}{4}\sin 2x+c$. The new integral was simpler than the original, which is the sign that the choice was right.

$\int x^2{e}^{-x}dx$ needs parts twice. First with $u=x^2$ and $\frac{dv}{dx}=e^{-x}$, which gives $-x^2{e}^{-x}+2\int x{e}^{-x}dx$. Then parts again on $\int x{e}^{-x}dx$ with $u=x$: it is $-x{e}^{-x}+\int e^{-x}dx=-x{e}^{-x}-e^{-x}$. Collecting the two results, $\int x^2{e}^{-x}dx=-e^{-x}(x^2+2x+2)+c$, and differentiating that back is a quick check that the double application was carried through correctly.

$\int \ln xdx$ is the case with only one factor, and it still works: take $u=\ln x$ and $\frac{dv}{dx}=1$, so $v=x$ and the integral is $x\ln x-\int x·\frac{1}{x}dx=x\ln x-x+c$. $\int x{\tan }^{-1}xdx$ is the fourth named example. With $u={\tan }^{-1}x$ and $\frac{dv}{dx}=x$ you reach $\frac{1}{2}{x}^2{\tan }^{-1}x-\frac{1}{2}\int \frac{x^2}{1+x^2}dx$, and the leftover integral is finished by writing $\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}$, giving $\frac{1}{2}(x^2+1){\tan }^{-1}x-\frac{1}{2}x+c$.

For a definite integral, put the limits on the $uv$ term straight away: ${\int }_0^{1}x{e}^{2x}dx={\left[\frac{1}{2}x{e}^{2x}\right]}_0^1-\frac{1}{2}{\int }_0^1{e}^{2x}dx=\frac{1}{2}{e}^2-\frac{1}{4}(e^2-1)=\frac{1}{4}(e^2+1)$. Carrying the limits along to the end is how the arithmetic goes wrong.

Substitution: a given change of variable

When a question says 'use the substitution $u=\sin x$', the method mark is for using it. Write $u=\sin x$, differentiate to get $\frac{du}{dx}=\cos x$, and then $du=\cos xdx$. That last line is the one that changes the integral into an integral in $u$. The syllabus's own example is $\int {\sin }^{2}x\cos xdx$: the $\cos xdx$ is exactly $du$, so the integral becomes $\int u^{2}du=\frac{1}{3}{u}^3+c=\frac{1}{3}{\sin }^{3}x+c$.

A substitution also handles a root of a linear expression multiplied by a power of $x$. With $u=3x+1$, $x=\frac{u-1}{3}$ and $dx=\frac{1}{3}du$, so an integrand such as $x^2(3x+1{)}^{1/2}$ becomes $\frac{1}{27}(u-1{)}^2{u}^{1/2}$, which is a sum of powers of $u$ and integrates term by term.

For a definite integral, convert the limits at the same time as the variable and then never write $x$ again. ${\int }_0^{1}x(x^2+1{)}^{3}dx$ with $u=x^2+1$: $du=2xdx$, so $xdx=\frac{1}{2}du$, and the limits become $u=1$ at $x=0$ and $u=2$ at $x=1$. The integral is $\frac{1}{2}{\int }_1^2{u}^{3}du=\frac{1}{2}{\left[\frac{u^4}{4}\right]}_1^2=\frac{1}{2}(4-\frac{1}{4})=\frac{15}{8}$.

A substitution is also the quick way to an integral that looks like a standard result but is not quite one. $\int \frac{2x}{x^2+5}dx$ is the logarithm result with $f(x)=x^2+5$, so it is $\ln (x^2+5)+c$; a student who writes $u=x^2+5$ reaches the same line and can see why the constant multiple is what it is.

Choosing the technique

In the exam the question usually names the method, and when it does not, the shape of the integrand names it for you. This table is the one to have in your head when you turn the page and read the question for the first time.

Definite integrals work exactly as they did in Pure 1: evaluate the antiderivative at the upper limit and subtract its value at the lower limit. What is new is only the integrand. Unless the question says otherwise, a non-exact answer is given to three significant figures, and the rounding is done once, at the end, from a full-accuracy calculator value.