Chapter 7 · Integrals

Exercise 7.2 · Q33: $\int\frac{1}{1-\tan x}\,dx$

Exercise 7.2Integration by substitution · Section 7.3.1

The question

Integrate the function (Exercise 7.2, "Integrate the functions in Exercises 1 to 37"):

\frac{1}{1 - \tan x}

Exam answer

Let $I = \displaystyle\int\frac{dx}{1 - \tan x}$ ⁠.

Multiplying the numerator and the denominator by $\cos x$ ⁠,

I = \int\frac{\cos x}{\cos x - \sin x}\,dx

Writing $\cos x = \frac12(\cos x - \sin x)$ $+ \frac12(\cos x + \sin x)$ ⁠,

\begin{aligned} I &= \frac12\int dx \\ &\quad + \frac12\int\frac{\cos x + \sin x}{\cos x - \sin x}\,dx \end{aligned}

Put $\cos x - \sin x = t$ ⁠.

Then $-(\sin x + \cos x)\,dx = dt$ ⁠.

\begin{aligned} I &= \frac x2 - \frac12\int\frac{dt}{t} \\ &= \frac x2 - \frac12\log|t| + C \end{aligned}

$\therefore\ I = \dfrac{x}{2} - \dfrac12\log|\cos x - \sin x|$ $+ C$ ⁠.

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

The one idea this problem is built on

Clear the tangent: $\frac{1}{1 - \tan x} = \frac{\cos x}{\cos x - \sin x}$ ⁠. With $D = \cos x - \sin x$ ⁠, the derivative is $D' = -\sin x - \cos x$ ⁠, and the numerator can be written as

\cos x = \tfrac12 D - \tfrac12 D'.

The $D$ part integrates to $\frac{x}{2}$ ⁠, and the $\frac{D'}{D}$ part to a logarithm. This is the book's Example 6(iii) with a minus sign.

Given, to find, and the plan

Given: $f(x) = \dfrac{1}{1 - \tan x}$ ⁠.
To find: $\int f(x)\,dx$ ⁠, and where the answer is valid.
Plan: (1) Find where $f$ is defined. (2) Rewrite $f$ as $\frac{\cos x}{\cos x - \sin x}$ ⁠. (3) Split the numerator as $\frac12 D - \frac12 D'$ ⁠. (4) Integrate each piece, the second by $t = \cos x - \sin x$ ⁠. (5) Differentiate back.

Basics you need

1 $\tan x = \frac{\sin x}{\cos x}$

Defined where $\cos x \ne 0$ ⁠. Basics: trigonometric facts

2Splitting a fraction

$\frac{a - b}{D} = \frac{a}{D} - \frac{b}{D}$ ⁠. Basics: splitting fractions

3 $\int\frac{g'(x)}{g(x)}\,dx = \log|g(x)| + C$

Substitute $t = g(x)$ and use $\int\frac{dt}{t} = \log|t| + C$ (standard formula (xii), Section 7.2). Basics: substitution Basics: logarithms and 1/x

4Integration undoes differentiation

Every answer can be checked by differentiating it. Basics: integration undoes differentiation

Solution

Where is the integrand defined?
- $\tan x$ must exist: $\cos x \ne 0$ ⁠, that is, $x \ne \frac{\pi}{2} + n\pi$ ⁠.
- The denominator must not vanish: $1 - \tan x \ne 0$ ⁠, that is, $\tan x \ne 1$ ⁠, that is, $x \ne \frac{\pi}{4} + n\pi$ ⁠.
Multiplying by $\cos x \ne 0$ ⁠, the second condition says $\cos x - \sin x \ne 0$ ⁠. So on the domain both $\cos x$ and $\cos x - \sin x$ are non-zero.

We need $\cos x \ne 0$ to multiply through by $\cos x$ in Step 2, and $\cos x - \sin x \ne 0$ for the logarithm in Step 4.

$\tan x = 1$ exactly when $x = \frac{\pi}{4} + n\pi$ ⁠. Basics: trigonometric facts
Rewrite without the tangent

Multiply top and bottom by $\cos x$ (non-zero on the domain):
$\begin{aligned} \frac{1}{1 - \tan x} &= \frac{\cos x}{\cos x - \cos x\tan x} \\ &= \frac{\cos x}{\cos x - \sin x}. \end{aligned}$
Now the denominator $D = \cos x - \sin x$ has a simple derivative, $D' = -\sin x - \cos x$ ⁠, and we can relate the numerator to $D$ and $D'$ ⁠.

$\cos x\tan x = \sin x$ ⁠. Basics: trigonometric facts
Split the numerator

Look for $a, b$ with $a D + b D' = \cos x$ ⁠:
$a(\cos x - \sin x) + b(-\sin x - \cos x) = \cos x.$
Compare coefficients: of $\cos x$ ⁠, $a - b = 1$ ⁠; of $\sin x$ ⁠, $-a - b = 0$ ⁠. So $a = \frac12$ ⁠, $b = -\frac12$ ⁠. Check: $\frac12(\cos x - \sin x) + \frac12(\sin x + \cos x) = \cos x$ ⁠. ✓ Hence
$\begin{aligned} \frac{\cos x}{\cos x - \sin x} &= \frac{\frac12 D - \frac12 D'}{D} \\ &= \tfrac12 + \tfrac12\cdot\frac{\cos x + \sin x}{\cos x - \sin x}. \end{aligned}$
Since $-\frac12 D' = \frac12(\sin x + \cos x)$ ⁠, the second fraction has numerator $\cos x + \sin x$ ⁠. Its sign is plus here, but this numerator is minus the derivative of the denominator, which will bring a minus sign back in Step 4.

Same method as Example 6(iii), Section 7.3.1. Basics: splitting fractions
Integrate each piece

First piece: $\int\frac12\,dx = \frac{x}{2}$ ⁠.

Second piece: put $t = \cos x - \sin x$ ⁠, so $dt = (-\sin x - \cos x)\,dx$ ⁠, that is, $(\cos x + \sin x)\,dx = -dt$ ⁠:
$\int\frac{\cos x + \sin x}{\cos x - \sin x}\,dx = \int\frac{-dt}{t} = -\log|t|.$
Together (Properties (III) and (IV), Section 7.2.1):
$\int\frac{dx}{1 - \tan x} = \frac{x}{2} - \tfrac12\log|\cos x - \sin x| + C.$
The substitution works because the numerator is the derivative of $t$ up to the factor $-1$ (Section 7.3.1). The modulus is needed: $\cos x - \sin x$ is positive near $0$ but negative near $x = 1$ ⁠, for example.

Standard formula (xii), Section 7.2. Lesson: integration by substitution
Check: differentiate the answer back

Let $F(x) = \frac{x}{2} - \frac12\log|\cos x - \sin x|$ ⁠. With $\frac{d}{dx}\log|g| = \frac{g'}{g}$ ⁠:
$\begin{aligned} F'(x) &= \tfrac12 - \tfrac12\cdot\frac{-\sin x - \cos x}{\cos x - \sin x} \\ &= \frac{(\cos x - \sin x) + (\sin x + \cos x)}{2(\cos x - \sin x)} \\ &= \frac{2\cos x}{2(\cos x - \sin x)} = \frac{\cos x}{\cos x - \sin x}. \end{aligned}$
Dividing top and bottom by $\cos x$ gives $\frac{1}{1 - \tan x}$ ⁠, the integrand. ✓

This is the test of an antiderivative (Property (I), Section 7.2.1), and it confirms the sign handled in Steps 3 and 4.

Basics: integration undoes differentiation

Result

$\int\frac{dx}{1 - \tan x} = \frac{x}{2} - \tfrac12\log|\cos x - \sin x| + C,$ valid wherever $\cos x \ne 0$ and $\tan x \ne 1$ ⁠. (Textbook answer: $\frac{x}{2} - \frac12\log|\cos x - \sin x| + C$ ⁠.)

Check it with numbers

Take $x = 0.5$ and $h = 0.001$ ⁠:

$F(0.501) = 0.7126617$ and $F(0.499) = 0.7082534$ ⁠,
slope $\approx \frac{0.7126617 - 0.7082534}{0.002} = 2.2041$ ⁠,
integrand: $\frac{1}{1 - \tan 0.5} = \frac{1}{1 - 0.546302} = 2.2041$ ⁠. ✓

Common mistakes

Wrong derivative of the denominator. $\frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x$ ⁠, not $\sin x - \cos x$ ⁠.
Sign of the log term. The numerator $\cos x + \sin x$ is minus $D'$ ⁠, so the log term comes out with a minus sign.
Copying Q32's answer. Q32 has $\log|\cos x + \sin x|$ ⁠; here it is $\log|\cos x - \sin x|$ ⁠.

What to take away

For a ratio of linear combinations of $\sin x$ and $\cos x$ ⁠, write the numerator as $\lambda D + \mu D'$ and compare coefficients.
Then $\int\frac{\lambda D + \mu D'}{D}\,dx = \lambda x + \mu\log|D| + C$ ⁠.
Check by differentiating: it catches sign slips in $D'$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\tan x = \frac{\sin x}{\cos x}$ ⁠; $\tan x = 1$ at $x = \frac{\pi}{4} + n\pi$	Steps 1, 2	Trigonometric functions (Class XI, Chapter 3)
Numerator written as $\frac12 D - \frac12 D'$	Step 3	Comparing coefficients; same method as Example 6(iii), Section 7.3.1
$\int(f + g) = \int f + \int g$ ⁠, $\int kf = k\int f$	Step 4	Properties (III), (IV), Section 7.2.1
Substitution $t = \cos x - \sin x$	Step 4	Section 7.3.1 (Lesson: substitution)
$\int\frac{dt}{t} = \log\|t\| + C$	Step 4	Standard formula (xii), Section 7.2
$F' = f$ checks the answer	Step 5	Property (I), Section 7.2.1

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 7.2 · Q33: ∫11−tan⁡x dx\int\frac{1}{1-\tan x}\,dx∫1−tanx1​dx

Basics you need

Solution

Where is the integrand defined?

Rewrite without the tangent

Split the numerator

Integrate each piece

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.2 · Q33: $\int\frac{1}{1-\tan x}\,dx$