Integration

Reversing the standard derivatives Paper 3

Every derivative in topic 3.4 can be read backwards. When the inside of the function is linear, $ax+b$, the chain rule multiplied by $a$ when differentiating, so integrating divides by $a$:

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

Check any of these by differentiating the answer. The rule works only for a linear inside: ${\mathrm{e}}^{x^2}$ cannot be integrated this way.

Why the modulus. $\ln x$ is defined only for $x>0$, but $\frac{1}{x}$ exists for negative $x$ too. For $x<0$, $\frac{\mathrm{d}}{\mathrm{d}x}\ln (-x)=\frac{-1}{-x}=\frac{1}{x}$, so $\ln \left|x\right|$ covers both cases. In a definite integral both limits must lie on the same side of the point where $ax+b=0$, since the integrand is not defined there.

Lay definite integrals out with square brackets and exact values: $\ln 1=0$, ${\mathrm{e}}^0=1$, $\cos 0=1$, and use the laws of logarithms to combine terms when a single logarithm is asked for.

Integrals that give inverse tangents Paper 3

Since $\frac{\mathrm{d}}{\mathrm{d}x}{\tan }^{-1}\frac{x}{a}=\frac{1}{a}\times \frac{1}{1+\frac{x^2}{a^2}}=\frac{a}{a^2+x^2}$, dividing by $a$ gives the MF19 result
$$\int \frac{1}{x^2+a^2}\mathrm{d}x=\frac{1}{a}{\tan }^{-1}(\frac{x}{a})+c$$
When $x^2$ has a coefficient, take it out first so that the bracket matches. For example
$$\frac{1}{9+4x^2}=\frac{1}{4}\times \frac{1}{x^2+\frac{9}{4}}$$
so $a=\frac{3}{2}$ and $\int \frac{1}{9+4x^2}\mathrm{d}x=\frac{1}{4}\times \frac{2}{3}{\tan }^{-1}(\frac{2x}{3})+c=\frac{1}{6}{\tan }^{-1}(\frac{2x}{3})+c$. The same method handles $\frac{1}{2+3x^2}$: take out the 3 and use $a=\sqrt{\frac{2}{3}}$.

Definite integrals of this type usually come out as multiples of $\pi$, because ${\tan }^{-1}1=\frac{\pi }{4}$, ${\tan }^{-1}\sqrt{3}=\frac{\pi }{3}$ and ${\tan }^{-1}\frac{1}{\sqrt{3}}=\frac{\pi }{6}$. Work in radians.

Using trigonometric identities Paper 3

There is no direct rule for ${\sin }^{2}x$ or ${\cos }^{2}x$, and the reverse chain rule fails because there is no $\cos x$ factor to absorb. Rewrite them with the double angle formula $\cos 2A=1-2{\sin }^{2}A=2{\cos }^{2}A-1$:
$${\sin }^{2}x=\frac{1}{2}(1-\cos 2x){\cos }^{2}x=\frac{1}{2}(1+\cos 2x)$$
Then $\int {\cos }^{2}x\mathrm{d}x=\frac{1}{2}x+\frac{1}{4}\sin 2x+c$. With a multiple angle, double it: ${\cos }^{2}5x=\frac{1}{2}(1+\cos 10x)$.

Other identities that turn an integrand into something standard:

• ${\tan }^{2}x={\sec }^{2}x-1$, so $\int {\tan }^{2}x\mathrm{d}x=\tan x-x+c$;
• $\sin x\cos x=\frac{1}{2}\sin 2x$;
• $(\sin x+\cos x{)}^2=1+\sin 2x$, using ${\sin }^{2}x+{\cos }^{2}x=1$.

Choose the identity that leaves only first powers of sines, cosines or ${\sec }^2$, each of a linear angle.

Integrals of the form k f'(x) over f(x) Paper 3

Differentiating $\ln |\mathrm{f}(x)|$ gives $\frac{\mathrm{f}\prime (x)}{\mathrm{f}(x)}$, so (as in MF19)
$$\int \frac{\mathrm{f}\prime (x)}{\mathrm{f}(x)}\mathrm{d}x=\ln |\mathrm{f}(x)|+c$$
Look for a numerator that is a multiple of the derivative of the denominator, and adjust the constant. For $\int \frac{x^2}{x^3+5}\mathrm{d}x$ the derivative of $x^3+5$ is $3x^2$, so the integrand is $\frac{1}{3}\times \frac{3x^2}{x^3+5}$ and the integral is $\frac{1}{3}\ln |x^3+5|+c$.

The classic case is $\tan x=\frac{\sin x}{\cos x}$. The derivative of $\cos x$ is $-\sin x$, so
$$\int \tan x\mathrm{d}x=-\ln \left|\cos x\right|+c=\ln \left|\sec x\right|+c$$
Only a constant can be adjusted. $\int \frac{1}{x^2+1}\mathrm{d}x$ is not of this type: its numerator is not a multiple of $2x$, and the answer is ${\tan }^{-1}x$.

Partial fractions Paper 3

A rational function whose denominator factorises is split into partial fractions (topic 3.1), and each piece is then a standard integral. The denominators are no more complicated than these, with the numerator of lower degree:

• $(ax+b)(cx+d)(ex+f)$: $\frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{ex+f}$, each giving a logarithm;
• $(ax+b)(cx+d{)}^2$: $\frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{(cx+d{)}^2}$, where the last term integrates to $-\frac{C}{c(cx+d)}$, not a logarithm;
• $(ax+b)(c{x}^2+d)$: $\frac{A}{ax+b}+\frac{Bx+C}{c{x}^2+d}$. Split the second fraction: $\frac{Bx}{c{x}^2+d}$ is a $\frac{k\mathrm{f}\prime (x)}{\mathrm{f}(x)}$ logarithm, and $\frac{C}{c{x}^2+d}$ gives an inverse tangent.

Find the constants by substituting the roots of the factors into the identity, then compare coefficients for any that remain. Remember the factor $\frac{1}{a}$ from each $\frac{1}{ax+b}$.

Integration by parts Paper 3

Integrating the product rule $\frac{\mathrm{d}}{\mathrm{d}x}(uv)=u\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}u}{\mathrm{d}x}$ and rearranging gives the MF19 formula
$$\int u\frac{\mathrm{d}v}{\mathrm{d}x}\mathrm{d}x=uv-\int v\frac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x$$
Use it when the integrand is a product of two different kinds of function, such as $x{\mathrm{e}}^{2x}$ or $x\cos x$. It trades one integral for another, so choose $u$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ to make the new integral easier:

• $u$ is usually the power of $x$, because differentiating it lowers the power;
• $\frac{\mathrm{d}v}{\mathrm{d}x}$ is the factor you can integrate, such as ${\mathrm{e}}^{kx}$, $\sin kx$ or $\cos kx$;
• except $\ln x$ and ${\tan }^{-1}x$, which must be $u$, since you can differentiate them but have no rule to integrate them. To integrate $\ln x$ on its own, write it as $1\times \ln x$ with $\frac{\mathrm{d}v}{\mathrm{d}x}=1$.

With $x^2$ as $u$, one application leaves $\int x(\dots )\mathrm{d}x$, so apply the formula a second time. Watch the signs: the minus in the formula applies to the whole of the second integral. For a definite integral, ${\int }_a^{b}u\frac{\mathrm{d}v}{\mathrm{d}x}\mathrm{d}x=[uv{]}_a^b-{\int }_a^{b}v\frac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x$.

Integration by substitution Paper 3

A substitution $u=\mathrm{g}(x)$ reverses the chain rule. The question gives the substitution; your job is to rewrite the whole integral in $u$:

1. differentiate: find $\frac{\mathrm{d}u}{\mathrm{d}x}$ and so replace $\mathrm{d}x$ by $\frac{\mathrm{d}u}{\frac{\mathrm{d}u}{\mathrm{d}x}}$;
2. replace every remaining $x$ by an expression in $u$;
3. for a definite integral, change the limits to values of $u$; for an indefinite one, substitute back to $x$ at the end.

The syllabus illustration is $\int {\sin }^{2}2x\cos x\mathrm{d}x$ with $u=\sin x$. Here $\frac{\mathrm{d}u}{\mathrm{d}x}=\cos x$, so $\cos x\mathrm{d}x$ becomes $\mathrm{d}u$. Also ${\sin }^{2}2x=4{\sin }^{2}x{\cos }^{2}x=4u^2(1-u^2)$. The integral becomes $\int (4u^2-4u^4)\mathrm{d}u=\frac{4}{3}{u}^3-\frac{4}{5}{u}^5+c=\frac{4}{3}{\sin }^{3}x-\frac{4}{5}{\sin }^{5}x+c$.

Changing the limits saves substituting back and is where most marks go: once the integral is in $u$, the old $x$ limits are wrong. A Show that part usually asks you to reach a stated integral in $u$, so show the $\mathrm{d}x$ replacement and both new limits.