Chapter 7 · Integrals

Miscellaneous · Q20: $\int\frac{2+\sin 2x}{1+\cos 2x}\,e^x\,dx$

Miscellaneous ExerciseDouble-angle identities, then eˣ[f(x) + f′(x)] · Miscellaneous Exercise

The question

Integrate the function:

\frac{2+\sin 2x}{1+\cos 2x}\,e^x

Exam answer

Let $I = \displaystyle\int\frac{2 + \sin 2x}{1 + \cos 2x}\,e^x\,dx$ ⁠.

Using $\sin 2x = 2\sin x\cos x$ and $1 + \cos 2x = 2\cos^2 x$ ⁠,

\begin{aligned} \frac{2 + \sin 2x}{1 + \cos 2x} &= \frac{2 + 2\sin x\cos x}{2\cos^2 x} \\ &= \sec^2 x + \tan x \end{aligned}

\therefore\ I = \int e^x\left(\tan x + \sec^2 x\right)dx

Here $f(x) = \tan x$ and $f'(x) = \sec^2 x$ ⁠. Using $\displaystyle\int e^x\left[f(x) + f'(x)\right]dx$ $= e^x f(x) + C$ ⁠,

$\therefore\ I = e^x\tan x + C$ ⁠.

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

The one idea this problem is built on

When $e^x$ multiplies something, look for the pattern $e^x\left[f(x) + f'(x)\right]$ ⁠, whose integral is simply $e^xf(x)$ (Section 7.6.1). The double-angle identities turn the fraction into $\sec^2x + \tan x$ ⁠, and $\sec^2x$ is the derivative of $\tan x$ ⁠: the pattern with $f(x) = \tan x$ ⁠.

Given, to find, and the plan

Given: $\frac{2+\sin 2x}{1+\cos 2x}\,e^x$ ⁠, defined where $1 + \cos 2x \ne 0$ ⁠, that is where $\cos x \ne 0$ (since $1 + \cos 2x = 2\cos^2x$ ⁠): every $x$ except odd multiples of $\frac{\pi}{2}$ ⁠.
To find: $\displaystyle\int\frac{2+\sin 2x}{1+\cos 2x}\,e^x\,dx$ ⁠.
Plan: (1) Rewrite the fraction with double-angle identities. (2) Split it into two terms. (3) Recognise $e^x[f + f']$ ⁠. (4) Write the answer. (5) Differentiate back and compare with the key.

Basics you need

1Double-angle identities

$1 + \cos 2x = 2\cos^2x$ and $\sin 2x = 2\sin x\cos x$ ⁠. Basics: trigonometric identities

2The $e^x[f + f']$ integral (Section 7.6.1)

\int e^x\left[f(x) + f'(x)\right]dx = e^xf(x) + C

It comes from integration by parts. Lesson: integration by parts

3Derivative of $\tan x$

$\frac{d}{dx}\tan x = \sec^2x$ (Section 7.2, list item (iv)). Basics: integration undoes differentiation

Solution

Rewrite the fraction with double angles
$\frac{2+\sin 2x}{1+\cos 2x} = \frac{2 + 2\sin x\cos x}{2\cos^2x} = \frac{1 + \sin x\cos x}{\cos^2x} .$
Both identities turn $2x$ into $x$ ⁠, and the factor 2 cancels. The denominator becomes a single power $\cos^2x$ ⁠, which is what makes the next split work. We may divide by $\cos^2x$ because $\cos x \ne 0$ on the domain.

Basics: trigonometric identities
Split into two terms
$\frac{1 + \sin x\cos x}{\cos^2x} = \frac{1}{\cos^2x} + \frac{\sin x}{\cos x} = \sec^2x + \tan x .$
So the integrand is $e^x\left(\tan x + \sec^2x\right)$ ⁠.

Dividing each term of the numerator by $\cos^2x$ ⁠: the first gives $\sec^2x$ ⁠; in the second one factor $\cos x$ cancels, leaving $\tan x$ ⁠.
Choose the method: recognise $e^x[f + f']$

With $f(x) = \tan x$ we have $f'(x) = \sec^2x$ ⁠, so
$e^x\left(\tan x + \sec^2x\right) = e^x\left[f(x) + f'(x)\right] .$
Whenever an integrand is $e^x$ times a sum, check whether one term is the derivative of the other. Here it is, so the whole integral is done in one line by Section 7.6.1, with no need to integrate $e^x\sec^2x$ or $e^x\tan x$ separately (by parts, each of these only leads back to the other).

Section 7.6.1. Lesson: integration by parts
Write the answer
$\int\frac{2+\sin 2x}{1+\cos 2x}\,e^x\,dx = e^x\tan x + C .$
The rule $\int e^x[f + f']\,dx = e^xf + C$ with $f(x) = \tan x$ ⁠.
Check: differentiate the answer
$\begin{aligned}\frac{d}{dx}\left(e^x\tan x\right) &= e^x\tan x + e^x\sec^2x\\ &= e^x\left(\tan x + \sec^2x\right)\\ &= \frac{2+\sin 2x}{1+\cos 2x}\,e^x\ \checkmark\end{aligned}$
Product rule, then Steps 1 and 2 read backwards. The answer is valid on each interval between consecutive odd multiples of $\frac{\pi}{2}$ ⁠, where $\cos x \ne 0$ ⁠.

Product rule (Chapter 5). Basics: integration undoes differentiation
Compare with the textbook's answer

The key gives $e^x\tan x + C$ ⁠, exactly our answer.

The key does not mention the domain; we add it: the answer holds where $\cos x \ne 0$ ⁠.

Result

Where $\cos x \ne 0$ ⁠:

\int\frac{2+\sin 2x}{1+\cos 2x}\,e^x\,dx = e^x\tan x + C

(Textbook answer: $e^x\tan x + C$ ⁠.)

Check it with numbers

The rewrite at $x = \frac{\pi}{4}$ ⁠: $\frac{2 + 1}{1 + 0} = 3$ ⁠, and $\sec^2\frac{\pi}{4} + \tan\frac{\pi}{4} = 2 + 1 = 3$ ⁠.

Area from 0 to $\frac{\pi}{4}$ (the integrand is continuous there): $F(\frac{\pi}{4}) - F(0) = e^{\pi/4}\cdot 1 - 0$ ⁠, that is $e^{\pi/4} \approx 2.1932800507$ ⁠. Numerical integration of the integrand from 0 to $\frac{\pi}{4}$ also gives $2.1932800507$ ⁠.

Common mistakes

Using $1 + \cos 2x = 2\sin^2x$ ⁠. That is $1 - \cos 2x$ ⁠; here it is $2\cos^2x$ ⁠.
Integrating $e^x$ and the fraction separately. The integral of a product is not the product of the integrals.
Choosing $f = \sec^2x$ ⁠. Its derivative is $2\sec^2x\tan x$ ⁠, not $\tan x$ ⁠; the term whose derivative is the other term is $\tan x$ ⁠.

What to take away

With $e^x$ times a sum, test whether the sum is $f + f'$ ⁠; if so, the answer is $e^xf + C$ ⁠.
Simplify trigonometric fractions with double-angle identities first; the pattern often only appears afterwards.

Every fact used, and where it comes from

Fact	Used in	Source
$1 + \cos 2x = 2\cos^2x$ ⁠, $\sin 2x = 2\sin x\cos x$	Step 1	Class XI trigonometry (Basics)
$\frac{d}{dx}\tan x = \sec^2x$	Steps 3, 5	Section 7.2, list item (iv)
$\int e^x[f + f']\,dx = e^xf + C$	Steps 3, 4	Section 7.6.1 (Lesson: integration by parts)
Product rule	Step 5	Chapter 5

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Miscellaneous · Q20: ∫2+sin⁡2x1+cos⁡2x ex dx\int\frac{2+\sin 2x}{1+\cos 2x}\,e^x\,dx∫1+cos2x2+sin2x​exdx

Basics you need

Solution

Rewrite the fraction with double angles

Split into two terms

Choose the method: recognise ex[f+f′]e^x[f + f']ex[f+f′]