Rational, trigonometric and root integrands – Integration (Senior 6 Mathematics) | Ganzaa

Integration

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Senior 6 · Mathematics

Integration

5 lessons

Lesson 2 of 5

Rational, trigonometric and root integrands

35 min

Rewrite first, integrate second

The table of lesson 1 has only a dozen entries, and most integrands you meet are not on it: $\frac{5x + 1}{x^2 + x - 2}$ , $\sin^3 x$ , $\frac{1}{\sqrt{5 + 4x - x^2}}$ . The skill in this lesson is algebra: rewrite the integrand, one family at a time, until every piece is on the table or yields to a substitution. Three families matter at this level: rational functions, powers and products of trigonometric functions, and roots of linear and quadratic expressions.

Definition

Rational function: a quotient $\frac{P(x)}{Q(x)}$ of two polynomials. It is proper if $\deg P \lt \deg Q$ , and improper otherwise.

Partial fractions: writing a proper rational function as a sum of simpler fractions, one group for each factor of the denominator $Q (x)$ .

Example: $\frac{5x + 1}{(x - 1)(x + 2)} = \frac{2}{x - 1} + \frac{3}{x + 2}$ . Add the right side to check.

Integrating a rational function

Step 1 If the fraction is improper, divide first: $\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}$ with $\deg R \lt \deg Q$ . The quotient $S (x)$ is a polynomial and integrates at once.

Step 2 Factorise $Q (x)$ and write one group of terms for each factor:

factor of $Q (x)$	terms in the decomposition
linear, $a x + b$	$\frac{A}{ax + b}$
repeated linear, $(a x + b)^{2}$	$\frac{A}{ax + b} + \frac{B}{(ax + b)^2}$
irreducible quadratic, $x^{2} + p x + q$ with $p^2 - 4q \lt 0$	$\frac{Ax + B}{x^2 + px + q}$

Step 3 Find the constants: multiply through by $Q (x)$ , then substitute the roots of $Q$ (the "cover-up" values) and compare coefficients for the rest.

Step 4 Integrate each term with these results ( $a \ne 0$ ):

\int \frac{dx}{ax + b} = \frac{1}{a}\ln|ax + b| + C, \qquad \int \frac{dx}{(ax + b)^2} = -\frac{1}{a(ax + b)} + C,

\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C, \qquad \int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left|\frac{x - a}{x + a}\right| + C .

The last one is itself a partial-fraction result: $\frac{1}{x^2 - a^2} = \frac{1}{2a}\left(\frac{1}{x - a} - \frac{1}{x + a}\right)$ .

Worked example 1: distinct and repeated linear factors

Find (a) $\int \frac{5x + 1}{(x - 1)(x + 2)}\,dx$ (b) $\int \frac{2x^2 + 3}{x(x - 1)^2}\,dx$ .

Solution

Write $\frac{5x + 1}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}$ , so $5 x + 1 = A (x + 2) + B (x - 1)$ . Put $x = 1$ : $6 = 3 A$ , $A = 2$ . Put $x = - 2$ : $- 9 = - 3 B$ , $B = 3$ . Hence $\int \frac{5x + 1}{(x - 1)(x + 2)}\,dx = 2\ln|x - 1| + 3\ln|x + 2| + C .$
Write $\frac{2x^2 + 3}{x(x - 1)^2} = \frac{A}{x} + \frac{B}{x - 1} + \frac{D}{(x - 1)^2}$ , so $2 x^{2} + 3 = A (x - 1)^{2} + B x (x - 1) + D x$ . Put $x = 0$ : $A = 3$ . Put $x = 1$ : $D = 5$ . Compare the $x^{2}$ coefficients: $2 = A + B$ , so $B = - 1$ . Hence $\int \frac{2x^2 + 3}{x(x - 1)^2}\,dx = 3\ln|x| - \ln|x - 1| - \frac{5}{x - 1} + C .$

Worked example 2: an irreducible quadratic

Find $\int \frac{2x + 5}{x^2 + 2x + 5}\,dx$ .

Solution The discriminant of $x^{2} + 2 x + 5$ is $4 - 20 \lt 0$ , so it does not factorise. Split the numerator into a multiple of the denominator's derivative, $2 x + 2$ , plus a constant: $2 x + 5 = (2 x + 2) + 3$ . Then complete the square, $x^{2} + 2 x + 5 = (x + 1)^{2} + 4$ :

\int \frac{2x + 2}{x^2 + 2x + 5}\,dx + 3\int \frac{dx}{(x + 1)^2 + 2^2} = \ln(x^2 + 2x + 5) + \frac{3}{2}\tan^{-1}\frac{x + 1}{2} + C .

The first part is the $\frac{f'}{f}$ rule of lesson 1; the second is the $\tan^{-1}$ result with $u = x + 1$ and $a = 2$ .

Warning

Divide before you decompose. Partial fractions are for proper fractions only. For $\frac{x^3}{x^2 - 1}$ , first write $x^{3} = x (x^{2} - 1) + x$ , so the integrand is $x + \frac{x}{x^2 - 1}$ . Setting $\frac{x^3}{x^2 - 1} = \frac{A}{x - 1} + \frac{B}{x + 1}$ straight away leads to equations with no solution.

Trigonometric integrands

Four tools cover nearly every case at this level.

Odd power of $\sin x$ or $\cos x$ : split off one factor and change the rest with $\sin^2 x + \cos^2 x = 1$ . For $\sin^{2k+1}x$ , write $\sin^{2k}x\sin x = (1 - \cos^2 x)^k\sin x$ and put $u = \cos x$ .
Even powers: lower the power with the double-angle identities (the linearisation of Unit 1): $\sin^2 x = \frac{1 - \cos 2x}{2}, \qquad \cos^2 x = \frac{1 + \cos 2x}{2} .$
Products of sines and cosines of different angles: turn the product into a sum (the factor formulae of S5): $\sin A\cos B = \tfrac{1}{2}\left[\sin(A + B) + \sin(A - B)\right], \quad \cos A\cos B = \tfrac{1}{2}\left[\cos(A + B) + \cos(A - B)\right],$ $\sin A\sin B = \tfrac{1}{2}\left[\cos(A - B) - \cos(A + B)\right] .$
A rational function of $\sin x$ and $\cos x$ : the substitution $t = \tan\frac{x}{2}$ turns it into a rational function of $t$ , because $\sin x = \frac{2t}{1 + t^2}, \qquad \cos x = \frac{1 - t^2}{1 + t^2}, \qquad dx = \frac{2\,dt}{1 + t^2} .$

Two more results complete the table of lesson 1. Since $\tan x = \frac{\sin x}{\cos x}$ is a $-\frac{f'}{f}$ form, $\int \tan x\,dx = -\ln|\cos x| + C = \ln|\sec x| + C$ . Multiplying $\sec x$ above and below by $\sec x + \tan x$ gives $\frac{\sec^2 x + \sec x\tan x}{\sec x + \tan x}$ , again an $\frac{f'}{f}$ form, so $\int \sec x\,dx = \ln|\sec x + \tan x| + C$ .

Worked example 3: powers and products

Find (a) $\int \sin^2 x\,dx$ (b) $\int \sin^3 x\,dx$ (c) $\int \sin 5x\cos 3x\,dx$ .

Solution

$\int \frac{1 - \cos 2x}{2}\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$ .
$\int (1 - \cos^2 x)\sin x\,dx$ . Let $u = \cos x$ , $du = -\sin x\,dx$ : $-\int (1 - u^2)\,du = -u + \frac{u^3}{3} + C = -\cos x + \frac{\cos^3 x}{3} + C .$
$\sin 5x\cos 3x = \frac{1}{2}(\sin 8x + \sin 2x)$ , so the integral is $-\frac{\cos 8x}{16} - \frac{\cos 2x}{4} + C$ .

Worked example 4: the substitution $t = \tan\frac{x}{2}$

Find $\int \frac{dx}{5 + 3\cos x}$ for $-\pi \lt x \lt \pi$ .

Solution With $t = \tan\frac{x}{2}$ ,

5 + 3\cos x = \frac{5(1 + t^2) + 3(1 - t^2)}{1 + t^2} = \frac{8 + 2t^2}{1 + t^2},

\int \frac{1 + t^2}{8 + 2t^2}\cdot\frac{2\,dt}{1 + t^2} = \int \frac{dt}{4 + t^2} = \frac{1}{2}\tan^{-1}\frac{t}{2} + C = \frac{1}{2}\tan^{-1}\left(\frac{1}{2}\tan\frac{x}{2}\right) + C .

Roots of linear and quadratic expressions

A root of a linear expression, $\sqrt{ax + b}$ : put $u = \sqrt{ax + b}$ . Then $u^{2} = a x + b$ , $x = \frac{u^2 - b}{a}$ and $dx = \frac{2u}{a}\,du$ , and the root disappears.

A root of a quadratic: complete the square, so that the expression under the root becomes one of three standard shapes, then use the matching result or substitution:

shape	substitution	standard result
$\sqrt{a^2 - x^2}$	$x = a\sin\theta$	$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C$
$\sqrt{x^2 + a^2}$	$x = a\tan\theta$	$\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left\|x + \sqrt{x^2 + a^2}\right\| + C$
$\sqrt{x^2 - a^2}$	$x = a\sec\theta$	$\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left\|x + \sqrt{x^2 - a^2}\right\| + C$

Each substitution works because of an identity: $a^2 - a^2\sin^2\theta = a^2\cos^2\theta$ , $a^2\tan^2\theta + a^2 = a^2\sec^2\theta$ , and $a^2\sec^2\theta - a^2 = a^2\tan^2\theta$ . After integrating in $\theta$ , a right triangle takes you back to $x$ .

Draw the triangle that matches your substitution, and read every other ratio off it. Left:

\cos\theta = \frac{\sqrt{a^2 - x^2}}{a}

. Right:

\sin\frac{x}{2} = \frac{t}{\sqrt{1 + t^2}}

and

\cos\frac{x}{2} = \frac{1}{\sqrt{1 + t^2}}

Worked example 5: roots

Find (a) $\int \frac{dx}{x\sqrt{x + 1}}$ (b) $\int \frac{dx}{\sqrt{5 + 4x - x^2}}$ (c) $\int \sqrt{9 - x^2}\,dx$ .

Solution

Let $u = \sqrt{x + 1}$ : $x = u^{2} - 1$ , $dx = 2u\,du$ . $\int \frac{2u\,du}{(u^2 - 1)u} = \int \frac{2\,du}{u^2 - 1} = \ln\left|\frac{u - 1}{u + 1}\right| + C = \ln\left|\frac{\sqrt{x + 1} - 1}{\sqrt{x + 1} + 1}\right| + C .$
Complete the square: $5 + 4 x - x^{2} = 9 - (x^{2} - 4 x + 4) = 9 - (x - 2)^{2}$ . With $a = 3$ and $u = x - 2$ : $\sin^{-1}\frac{x - 2}{3} + C$ , valid for $-1 \lt x \lt 5$ .
Let $x = 3\sin\theta$ , $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ , so $dx = 3\cos\theta\,d\theta$ and $\sqrt{9 - x^2} = 3\cos\theta$ : $\int 9\cos^2\theta\,d\theta = \frac{9}{2}\left(\theta + \sin\theta\cos\theta\right) + C .$ From the triangle, $\sin\theta = \frac{x}{3}$ and $\cos\theta = \frac{\sqrt{9 - x^2}}{3}$ , so $\int \sqrt{9 - x^2}\,dx = \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\frac{x}{3} + C .$

Warning

Go back to $x$ , and state where it holds. An answer left in $\theta$ or $t$ is unfinished. Also note the domain: $\sqrt{9 - x^2}$ needs $|x| \le 3$ , and $\tan\frac{x}{2}$ is undefined at $x = \pm\pi$ .

Tip

Before choosing a method, ask in this order: Is it on the table? Is it $\frac{f'}{f}$ or $f(g)\,g'$ ? Is it a rational function (divide, then decompose)? Is it trigonometric (odd power, even power, product)? Is there a root (substitute, or complete the square)? The first "yes" usually gives the method.

Dr. Trefor Bazett: Integration by Partial Fractions, Big Idea and First Example. Watch for the moment the integrand is split into simpler fractions that each integrate to a logarithm, as in Worked example 1.

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Practice

Questions 7 and 8 are harder.

Find $\int \frac{7}{(x - 2)(x + 5)}\,dx$ .
Find $\int \frac{x^3}{x^2 - 1}\,dx$ .
Find $\int \frac{3x + 1}{(x + 1)^2}\,dx$ .
Find $\int \frac{dx}{x^2 + 4x + 13}$ .
Find (a) $\int \cos^2 3x\,dx$ (b) $\int \cos^3 x\,dx$ (c) $\int \tan^2 x\,dx$ (d) $\int \cos 4x\cos 2x\,dx$ .
Use $t = \tan\frac{x}{2}$ to find $\int \frac{dx}{1 + \sin x}$ .
Find (a) $\int \frac{dx}{\sqrt{4 - x^2}}$ (b) $\int \frac{dx}{\sqrt{x^2 - 2x + 10}}$ (c) $\int \frac{dx}{\sqrt{3 - 2x - x^2}}$ .
Use $x = 2\tan\theta$ to find $\int \frac{dx}{x^2\sqrt{x^2 + 4}}$ for $x \gt 0$ .

Show answersHide answers

$\frac{7}{(x - 2)(x + 5)} = \frac{1}{x - 2} - \frac{1}{x + 5}$ : $\ln\left|\frac{x - 2}{x + 5}\right| + C$
$x + \frac{x}{x^2 - 1}$ : $\frac{x^2}{2} + \frac{1}{2}\ln|x^2 - 1| + C$
$\frac{3}{x + 1} - \frac{2}{(x + 1)^2}$ : $3\ln|x + 1| + \frac{2}{x + 1} + C$
$(x + 2)^{2} + 9$ : $\frac{1}{3}\tan^{-1}\frac{x + 2}{3} + C$
(a) $\frac{x}{2} + \frac{\sin 6x}{12} + C$ (b) $\sin x - \frac{\sin^3 x}{3} + C$ (c) $\tan x - x + C$ , since $\tan^2 x = \sec^2 x - 1$ (d) $\frac{\sin 6x}{12} + \frac{\sin 2x}{4} + C$
$1 + \sin x = \frac{(1 + t)^2}{1 + t^2}$ , so the integral is $\int \frac{2\,dt}{(1 + t)^2} = -\frac{2}{1 + \tan\frac{x}{2}} + C$ .
(a) $\sin^{-1}\frac{x}{2} + C$ (b) $(x - 1)^{2} + 9$ : $\ln\left|x - 1 + \sqrt{x^2 - 2x + 10}\right| + C$ (c) $4 - (x + 1)^{2}$ : $\sin^{-1}\frac{x + 1}{2} + C$
The integral becomes $\frac{1}{4}\int \frac{\cos\theta}{\sin^2\theta}\,d\theta = -\frac{1}{4\sin\theta} + C$ . Since $\sin\theta = \frac{x}{\sqrt{x^2 + 4}}$ , the answer is $-\frac{\sqrt{x^2 + 4}}{4x} + C$ .

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples, the practice items and the quiz questions. Every antiderivative was checked by numerical differentiation.
Video: Integration by Partial Fractions | Big Idea + First Example, Dr. Trefor Bazett, embedded from YouTube.

1Rewrite first, integrate second

2Integrating a rational function

3Trigonometric integrands

4Roots of linear and quadratic expressions

Rewrite first, integrate second

Integrating a rational function

Trigonometric integrands

Roots of linear and quadratic expressions