Integration

Integrals of the new functions Paper 2

Each derivative in topic 2.4, read backwards, gives an integral. In Paper 1 you integrated $(ax+b{)}^n$ by increasing the power, then dividing by the new power and by $a$. The same idea, divide by $a$, the coefficient of $x$ inside the bracket, works for every function of $ax+b$:

• $\int {\mathrm{e}}^{ax+b}\mathrm{d}x=\frac{1}{a}{\mathrm{e}}^{ax+b}+c$
• $\int \frac{1}{ax+b}\mathrm{d}x=\frac{1}{a}\ln |ax+b|+c$
• $\int \sin (ax+b)\mathrm{d}x=-\frac{1}{a}\cos (ax+b)+c$
• $\int \cos (ax+b)\mathrm{d}x=\frac{1}{a}\sin (ax+b)+c$
• $\int {\sec }^2(ax+b)\mathrm{d}x=\frac{1}{a}\tan (ax+b)+c$

Why divide by $a$: differentiating $\frac{1}{a}\sin (ax+b)$ multiplies by $a$ (the chain rule), which gives back $\cos (ax+b)$. Always check an integral by differentiating it.

The logarithm and the modulus. $\int \frac{1}{x}\mathrm{d}x=\ln \left|x\right|+c$. The power rule fails at $x^{-1}$, because it would divide by zero, and the logarithm fills the gap. The modulus is there because $\frac{1}{x}$ exists for negative $x$ too, and $\frac{\mathrm{d}}{\mathrm{d}x}\ln (-x)=\frac{-1}{-x}=\frac{1}{x}$ as well. When the bracket is positive throughout a question, you may write $\ln (ax+b)$.

Signs. $\sin$ integrates to $-\cos$, and $\cos$ to $+\sin$: the opposite way round from differentiation.

What is not in this paper. The general method of integration by substitution is not required. Every integral here is one of these standard forms, perhaps after some algebra or an identity: expand brackets, split fractions over a single term, or use the laws of indices, such as $\frac{1}{{\mathrm{e}}^{2x}}={\mathrm{e}}^{-2x}$.

In the answer boxes on this site, type $\ln |2x+1|$ as ln(2x+1), ${\sin }^{2}x$ as sin(x)^2, and include + c when asked.

Definite integrals and exact values Paper 2

Definite integrals work exactly as in Paper 1: integrate, then work out $\mathrm{F}(b)-\mathrm{F}(a)$. The new functions give exact answers involving $\mathrm{e}$, $\ln$, $\pi$ and surds, and you should simplify them with the laws of logarithms and indices: $\ln 9-\ln 3=\ln 3$, ${\mathrm{e}}^{\ln 2}=2$, ${\mathrm{e}}^{2\ln 3}=9$ and $\ln 1=0$.

Limits for trigonometric integrals are in radians, and you need the exact values of $\sin$, $\cos$ and $\tan$ at $\frac{1}{6}\pi$, $\frac{1}{4}\pi$ and $\frac{1}{3}\pi$. If you check an answer on your calculator, set it to radians.

Integrating with trigonometric identities Paper 2

Many trigonometric integrands are not in the table, but an identity from topic 2.3 turns them into ones that are. The ones you need:

• ${\cos }^{2}x=\frac{1}{2}(1+\cos 2x)$ and ${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$, from $\cos 2x=2{\cos }^{2}x-1=1-2{\sin }^{2}x$;
• $\sin x\cos x=\frac{1}{2}\sin 2x$;
• ${\tan }^{2}x={\sec }^{2}x-1$, from ${\sec }^{2}x=1+{\tan }^{2}x$.

Each replaces a square, which you cannot integrate directly, with a first power of a double angle, which you can. For the syllabus example ${\cos }^{2}2x$: ${\cos }^{2}2x=\frac{1}{2}(1+\cos 4x)$, so $\int {\cos }^{2}2x\mathrm{d}x=\frac{1}{2}x+\frac{1}{8}\sin 4x+c$. The double angle of $2x$ is $4x$.

The syllabus also names $\sin 2x$ itself, which you can integrate directly, $\int \sin 2x\mathrm{d}x=-\frac{1}{2}\cos 2x+c$, or recognise inside a product: $\int 6\sin x\cos x\mathrm{d}x=\int 3\sin 2x\mathrm{d}x=-\frac{3}{2}\cos 2x+c$.

A common wrong approach is $\int {\sin }^{2}x\mathrm{d}x=\frac{1}{3}{\sin }^{3}x$, treating $\sin x$ like $x$. Differentiate it and you get ${\sin }^{2}x\cos x$, which is not the integrand: the check exposes the error at once.

The trapezium rule Paper 2

Some definite integrals cannot be found exactly with the methods of this paper, for example ${\int }_0^1{\mathrm{e}}^{x^2}\mathrm{d}x$. The trapezium rule estimates the area under the curve by splitting it into $n$ vertical strips of equal width $h$ and treating each strip as a trapezium: the curve across the top of the strip is replaced by a straight chord.

With $h=\frac{b-a}{n}$ and the heights (ordinates) $y_0,y_1,\dots ,y_n$ at $x=a,a+h,\dots ,b$:
$${\int }_a^{b}y\mathrm{d}x\approx \frac{1}{2}h\{y_0+y_n+2(y_1+y_2+\dots +y_{n-1})\}$$
Each trapezium has area $\frac{1}{2}h(y_k+y_{k+1})$. Adding them, the two end ordinates appear once and every middle ordinate twice, because it is shared by two strips. The formula is not in the formula list, so learn it.

Practical points:

• $n$ strips need $n+1$ ordinates: four intervals means $h=\frac{b-a}{4}$ and five $y$-values.
• Work in a table, keep at least 4 significant figures in each ordinate, and round only the final answer.
• Trigonometric functions use radians.

Over-estimate or under-estimate? Paper 2

Sketch the curve over the interval and look at the chords.

• If the curve bends upwards (it is convex, like $y={\mathrm{e}}^x$ or $y=x^2$), each chord lies above the curve, so every trapezium is slightly too big: the rule gives an over-estimate.
• If the curve bends downwards (it is concave, like $y=\ln x$ or $y=\sqrt{x}$), each chord lies below the curve: an under-estimate.
• If the curve changes the way it bends within the interval, the errors partly cancel and a sketch alone may not decide.

A reason in the exam should name the shape and say where the tops of the trapezia are, for example: 'the curve bends downwards, so the tops of the trapezia lie below the curve'. More strips give a better estimate, because the chords hug the curve more closely, but for a curve that bends one way only, the error never changes sign.

For a curve that bends upwards, each straight top lies above the curve, so the trapezium rule over-estimates.