Chapter 7 · Integrals

Miscellaneous · Q11: $\int\frac{dx}{\cos(x+a)\cos(x+b)}$

Miscellaneous ExerciseTrigonometric identity trick (multiply by sin(a − b)) · Miscellaneous Exercise on Chapter 7

The question

Integrate the functions in Exercises 1 to 23.

11. $\dfrac{1}{\cos(x+a)\cos(x+b)}$

Exam answer

Let $I = \displaystyle\int\frac{dx}{\cos(x + a)\cos(x + b)}$ ⁠.

Let $k = \sin(a - b)$ ⁠, where $k \ne 0$ ⁠. Writing $a - b = (x + a) - (x + b)$ ⁠,

\begin{aligned} k &= \sin\big[(x + a) - (x + b)\big] \\ &= \sin(x + a)\cos(x + b) \\ &\quad - \cos(x + a)\sin(x + b) \end{aligned}

Multiplying and dividing by $k$ ⁠, then writing $k$ in the expanded form above,

\begin{aligned} I &= \frac1k\int\frac{k\,dx}{\cos(x + a)\cos(x + b)} \\ &= \frac1k\int\tan(x + a)\,dx \\ &\quad - \frac1k\int\tan(x + b)\,dx \\ &= \frac1k\big[-\log|\cos(x + a)| \\ &\qquad + \log|\cos(x + b)|\big] + C \end{aligned}

$\therefore\ I = \dfrac{1}{\sin(a - b)}$ $\log\left|\dfrac{\cos(x + b)}{\cos(x + a)}\right| + C$ (for $\sin(a - b) \ne 0$ ⁠).

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The two angles $x + a$ and $x + b$ differ by the constant $a - b$ ⁠. So the constant $\sin(a - b)$ can be written as $\sin\big[(x+a) - (x+b)\big]$ and expanded with the subtraction formula. Multiplying and dividing by $\sin(a - b)$ then splits the integrand into $\tan(x+a) - \tan(x+b)$ ⁠, whose integrals are standard logarithms. Exercise 7.3 Q22 uses the same trick.

Given, to find, and the plan

Given: $f(x) = \dfrac{1}{\cos(x+a)\cos(x+b)}$ ⁠, $a, b$ constants with $\sin(a - b) \ne 0$ (the remark in the Result covers the other case). Defined where $\cos(x+a) \ne 0$ and $\cos(x+b) \ne 0$ ⁠.
To find: $\displaystyle\int\frac{dx}{\cos(x+a)\cos(x+b)}$ ⁠.
Plan: (1) Why this trick. (2) Multiply and divide by $\sin(a - b)$ and expand. (3) Split into tangents. (4) Integrate. (5) Differentiate back.

Basics you need

1Subtraction formula

$\sin(A - B)$ $= \sin A\cos B - \cos A\sin B$ ⁠. Basics: trigonometric identities

2 $\int\tan u\,du$

$\int\tan u\,du = -\log|\cos u| + C$ $= \log|\sec u| + C$ ⁠, one of the standard trigonometric integrals of Section 7.3.1 (result (i) there). Lesson: integration by substitution

Solution

Why this trick

A product of two cosines in the denominator has no standard integral, and $t = \cos(x+a)$ fails because there is no $\sin(x+a)$ factor. But the angles differ by a constant, so a constant numerator can be "manufactured" from them: $\sin(a - b)$ $= \sin\big[(x+a) - (x+b)\big]$ ⁠.

The expanded form of $\sin\big[(x+a) - (x+b)\big]$ contains exactly the products $\sin(x+a)\cos(x+b)$ and $\cos(x+a)\sin(x+b)$ ⁠. Dividing those by $\cos(x+a)\cos(x+b)$ gives tangents. Multiplying and dividing by the non-zero constant $\sin(a - b)$ does not change the integrand.

Exercise 7.3 Q22 (Exercise 7.3 Q22) uses the same idea.
Multiply and divide by $\sin(a - b)$ ⁠, and expand

Let $k = \sin(a - b) \ne 0$ ⁠. By the subtraction formula with $A = x + a$ ⁠, $B = x + b$ ⁠:
$\begin{aligned} k &= \sin\big[(x+a) - (x+b)\big] \\ &= \sin(x+a)\cos(x+b) \\ &\quad - \cos(x+a)\sin(x+b). \end{aligned}$
$(x + a) - (x + b) = a - b$ ⁠: the $x$ 's cancel, so the left side really is the constant $k$ ⁠, for every $x$ ⁠.

Subtraction formula (Basic 1). Basics: trigonometric identities
Split the integrand into tangents
$\begin{aligned} f(x) &= \frac1k\cdot\frac{k}{\cos(x+a)\cos(x+b)} \\ &= \frac1k\Big[\frac{\sin(x+a)\cos(x+b)}{\cos(x+a)\cos(x+b)} \\ &\qquad - \frac{\cos(x+a)\sin(x+b)}{\cos(x+a)\cos(x+b)}\Big] \\ &= \frac1k\big[\tan(x+a) - \tan(x+b)\big]. \end{aligned}$
In the first fraction $\cos(x+b)$ cancels, leaving $\tan(x+a)$ ⁠; in the second $\cos(x+a)$ cancels, leaving $\tan(x+b)$ ⁠. Both cancellations are valid where $f$ is defined (both cosines non-zero).

$\tan u = \frac{\sin u}{\cos u}$ ⁠.
Integrate

With $u = x + a$ (or $x + b$ ⁠), $du = dx$ ⁠, and $\int\tan u\,du = -\log|\cos u|$ ⁠:
$\begin{aligned} \int f(x)\,dx &= \frac1k\big[-\log|\cos(x+a)| \\ &\qquad + \log|\cos(x+b)|\big] + C \\ &= \frac{1}{\sin(a-b)}\log\left|\frac{\cos(x+b)}{\cos(x+a)}\right| + C. \end{aligned}$
The difference of two logarithms is the logarithm of the quotient (applied to the positive numbers $|\cos(x+b)|$ and $|\cos(x+a)|$ ⁠). This is exactly the textbook's answer.

$\int\tan u\,du$ ⁠, Section 7.3.1 (Basic 2); $\log p - \log q = \log\frac pq$ ⁠. Basics: logarithms
Check: differentiate the answer

Since $\frac{d}{dx}\log|\cos(x+b)| = -\tan(x+b)$ and $\frac{d}{dx}\log|\cos(x+a)| = -\tan(x+a)$ ⁠:
$\begin{aligned} F' &= \frac1k\big[\tan(x+a) - \tan(x+b)\big] \\ &= \frac1k\cdot\frac{\sin(x+a)\cos(x+b) - \cos(x+a)\sin(x+b)}{\cos(x+a)\cos(x+b)} \\ &= \frac1k\cdot\frac{\sin(a-b)}{\cos(x+a)\cos(x+b)} \\ &= \frac{1}{\cos(x+a)\cos(x+b)}. \ \checkmark \end{aligned}$
Putting the two tangents over the common denominator rebuilds $\sin(a - b) = k$ in the numerator (Step 2 in reverse), which cancels the $\frac1k$ ⁠. The answer is valid on every interval where both cosines are non-zero.

Chain rule, Chapter 5. Basics: integration undoes differentiation

Result

For $\sin(a - b) \ne 0$ ⁠, wherever $\cos(x+a)\cos(x+b) \ne 0$ ⁠:

\begin{aligned} &\int\frac{dx}{\cos(x+a)\cos(x+b)} \\ &= \frac{1}{\sin(a-b)}\log\left|\frac{\cos(x+b)}{\cos(x+a)}\right| + C. \end{aligned}

(Textbook answer: $\frac{1}{\sin(a-b)}\log\left|\frac{\cos(x+b)}{\cos(x+a)}\right| + C$ ⁠, the same.)

Remark: if $\sin(a - b) = 0$ ⁠, then $a - b = n\pi$ and $\cos(x+a) = (-1)^n\cos(x+b)$ ⁠, so the integrand is $(-1)^n\sec^2(x+b)$ and the integral is $(-1)^n\tan(x+b) + C$ ⁠.

Check it with numbers

Take $a = \frac\pi3$ ⁠, $b = 0$ ⁠, so $\sin(a - b) = \frac{\sqrt3}{2}$ ⁠, and integrate from 0 to $\frac{\pi}{12}$ (both cosines stay positive). With $\cos\frac{5\pi}{12} = \sin\frac{\pi}{12}$ and $\cot\frac{\pi}{12} = 2 + \sqrt3$ ⁠:

\begin{aligned} F\big(\tfrac{\pi}{12}\big) - F(0) &= \tfrac{2}{\sqrt3}\big[\log(2 + \sqrt3) - \log 2\big] \\ &\approx 0.7203146. \end{aligned}

Numerical integration over the same interval also gives $0.7203146$ ⁠.

Common mistakes

Using $\sin(b - a)$ instead of $\sin(a - b)$ : the expansion must start with $\sin(x+a)\cos(x+b)$ for the signs in Step 3 to come out as $\tan(x+a) - \tan(x+b)$ ⁠.
Inverting the fraction inside the log. $\int\tan(x+a)\,dx = -\log|\cos(x+a)|$ ⁠, so $\cos(x+a)$ ends up in the denominator.
Ignoring the case $\sin(a - b) = 0$ , where the trick divides by zero.

What to take away

For $\frac{1}{\cos(x+a)\cos(x+b)}$ ⁠, $\frac{1}{\sin(x+a)\sin(x+b)}$ or $\frac{1}{\sin(x+a)\cos(x+b)}$ ⁠: multiply and divide by the sine or cosine of the constant difference of the angles.
$\int\tan u\,du = -\log|\cos u| + C$ and $\int\cot u\,du = \log|\sin u| + C$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin(A - B)$ $= \sin A\cos B - \cos A\sin B$	Steps 2, 5	Trigonometry (Class XI) (Basics)
$\tan u = \frac{\sin u}{\cos u}$	Step 3	Trigonometry
$\int\tan x\,dx = \log\|\sec x\| + C$	Step 4	Section 7.3.1, result (i)
Substitution $u = x + a$	Step 4	Section 7.3.1
$\log p - \log q = \log\frac pq$	Step 4	Algebra (Basics)
Chain rule	Step 5	Chapter 5
All decimals	Check	Recomputed in the site's verification script

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Miscellaneous · Q11: ∫dxcos⁡(x+a)cos⁡(x+b)\int\frac{dx}{\cos(x+a)\cos(x+b)}∫cos(x+a)cos(x+b)dx​

Basics you need

Solution

Why this trick

Multiply and divide by sin⁡(a−b)\sin(a - b)sin(a−b)⁠, and expand

Split the integrand into tangents

Integrate

Check: differentiate the answer

Every fact used, and where it comes from

Miscellaneous · Q11: $\int\frac{dx}{\cos(x+a)\cos(x+b)}$

Multiply and divide by $\sin(a - b)$ ⁠, and expand