Chapter 7 · Integrals

Exercise 7.6 · Q18: $\int e^x\frac{1+\sin x}{1+\cos x}\,dx$

Exercise 7.6Integral of the type eˣ[f + f′] · Section 7.6.1

The question

Integrate the functions in Exercises 1 to 22.

18.

e^x\left(\frac{1 + \sin x}{1 + \cos x}\right)

Exam answer

Let $I = \displaystyle\int e^x\left(\frac{1 + \sin x}{1 + \cos x}\right)dx$ ⁠.

Using $1 + \cos x = 2\cos^2\frac x2$ and $\sin x = 2\sin\frac x2\cos\frac x2$ ⁠,

\begin{aligned} &\frac{1 + \sin x}{1 + \cos x} \\ &= \frac{1 + 2\sin\frac x2\cos\frac x2}{2\cos^2\frac x2} \\ &= \frac12\sec^2\frac x2 + \tan\frac x2 \end{aligned}

Let $f(x) = \tan\frac x2$ ⁠. Then $f'(x) = \frac12\sec^2\frac x2$ ⁠.

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

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

$\therefore\ I = e^x\tan\dfrac x2 + C$ ⁠.

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

The one idea this problem is built on

Rewrite the fraction with half angles. Since $1 + \cos x = 2\cos^2\frac x2$ and $\sin x = 2\sin\frac x2\cos\frac x2$ ⁠,

\frac{1 + \sin x}{1 + \cos x} = \frac12\sec^2\frac x2 + \tan\frac x2,

and $\frac12\sec^2\frac x2$ is the derivative of $\tan\frac x2$ ⁠. So the bracket is $f + f'$ with $f = \tan\frac x2$ ⁠, and Section 7.6.1 gives the answer $e^x\tan\frac x2 + C$ ⁠.

Given, to find, and the plan

Given: $e^x\,\frac{1+\sin x}{1+\cos x}$ ⁠, defined where $\cos x \ne -1$ ⁠, that is $x \ne (2n+1)\pi$ for integers $n$ ⁠.
To find: $\int e^x\,\frac{1+\sin x}{1+\cos x}\,dx$ ⁠.
Plan: (1) Rewrite the fraction with half-angle identities. (2) Recognise $f = \tan\frac x2$ and apply the rule. (3) Differentiate back.

Basics you need

1Half-angle identities

$1 + \cos x = 2\cos^2\frac x2$ (from $\cos 2\theta = 2\cos^2\theta - 1$ with $\theta = \frac x2$ ⁠) and $\sin x = 2\sin\frac x2\cos\frac x2$ ⁠. Basics: trigonometric identities

2The rule $\int e^x[f + f']\,dx = e^x f + C$

Section 7.6.1. Lesson: integration by parts

3Derivative of $\tan\frac x2$

$\frac{d}{dx}\tan\frac x2 = \frac12\sec^2\frac x2$ (chain rule, inner function $\frac x2$ ⁠). Basics: the chain rule backwards

Solution

Rewrite the fraction with half angles

Using $1 + \cos x = 2\cos^2\frac x2$ and $\sin x = 2\sin\frac x2\cos\frac x2$ ⁠:
$\begin{aligned} \frac{1 + \sin x}{1 + \cos x} &= \frac{1}{2\cos^2\frac x2} + \frac{2\sin\frac x2\cos\frac x2}{2\cos^2\frac x2} \\ &= \frac12\sec^2\frac x2 + \tan\frac x2. \end{aligned}$
Split the fraction into $\frac{1}{1+\cos x} + \frac{\sin x}{1+\cos x}$ ⁠, then rewrite each part with half angles. In the second part one factor $\cos\frac x2$ cancels, which is allowed: $\cos\frac x2 = 0$ exactly when $\cos^2\frac x2 = 0$ ⁠, that is when $1 + \cos x = 0$ ⁠, and those points are excluded from the domain.

$\cos 2\theta = 2\cos^2\theta - 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$ with $\theta = \frac x2$ ⁠. Basics: trigonometric identities
Recognise $f + f'$ and apply the rule

Take $f(x) = \tan\frac x2$ ⁠. By the chain rule,
$f'(x) = \sec^2\frac x2\cdot\frac12 = \frac12\sec^2\frac x2.$
So the integrand is $e^x[f(x) + f'(x)]$ ⁠, and by Section 7.6.1
$\int e^x\,\frac{1+\sin x}{1+\cos x}\,dx = e^x\tan\frac x2 + C.$
Of the two terms, $\frac12\sec^2\frac x2$ is the derivative of $\tan\frac x2$ ⁠, so $f = \tan\frac x2$ ⁠. It is differentiable wherever $\cos\frac x2 \ne 0$ ⁠, which is the same set as the domain of the integrand, so the answer holds on each interval $\big((2n-1)\pi, (2n+1)\pi\big)$ ⁠.

$\frac{d}{dx}\tan u = \sec^2 u\cdot\frac{du}{dx}$ ⁠; $\int e^x[f + f']\,dx = e^x f + C$ ⁠. Lesson: integration by parts
Check: differentiate the answer
$\begin{aligned} \frac{d}{dx}\left[e^x\tan\frac x2\right] &= e^x\tan\frac x2 + e^x\cdot\frac12\sec^2\frac x2 \\ &= e^x\left(\tan\frac x2 + \frac{1}{2\cos^2\frac x2}\right) \\ &= e^x\,\frac{2\sin\frac x2\cos\frac x2 + 1}{2\cos^2\frac x2} \\ &= e^x\,\frac{1 + \sin x}{1 + \cos x}. \end{aligned}$
Product rule, then put everything over the common bottom $2\cos^2\frac x2$ and read the half-angle identities of Step 1 backwards: the top is $\sin x + 1$ and the bottom is $1 + \cos x$ ⁠.

Product rule; half-angle identities. Basics: integration undoes differentiation

Result

\int e^x\left(\frac{1+\sin x}{1+\cos x}\right)dx = e^x\tan\frac x2 + C,

valid on any interval on which $\cos x \ne -1$ (no odd multiple of $\pi$ inside). (Textbook answer: $e^x\tan\frac x2 + C$ ⁠, the same.)

Check it with numbers

On $\left[0, \frac{\pi}{2}\right]$ ⁠, where $\cos x \ne -1$ ⁠:

\begin{aligned} F\left(\tfrac{\pi}{2}\right) - F(0) &= e^{\pi/2}\tan\frac{\pi}{4} - e^0\tan 0 \\ &= e^{\pi/2} = 4.8104773810\ldots \end{aligned}

Numerical integration of $e^x\,\frac{1+\sin x}{1+\cos x}$ over $\left[0, \frac{\pi}{2}\right]$ gives $4.8104773810\ldots$ ⁠, the same.

Common mistakes

Trying to integrate by parts directly with the fraction as it stands; the structure only appears after the half-angle rewrite.
Losing the $\frac12$ from the chain rule: $\frac{d}{dx}\tan\frac x2 = \frac12\sec^2\frac x2$ ⁠, which matches $\frac{1}{2\cos^2\frac x2}$ exactly.
Writing $1 + \cos x = 2\sin^2\frac x2$ ⁠. That is $1 - \cos x$ ⁠.

What to take away

$1 + \cos x$ and $1 - \cos x$ in a bottom call for half angles: $2\cos^2\frac x2$ and $2\sin^2\frac x2$ ⁠.
After rewriting, look for $f + f'$ ⁠; here $f = \tan\frac x2$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$1 + \cos x = 2\cos^2\frac x2$ ⁠; $\sin x = 2\sin\frac x2\cos\frac x2$	Steps 1, 3	Class XI, Chapter 3 (trigonometric functions of multiple angles)
$\int e^x[f + f']\,dx = e^x f + C$	Step 2	Section 7.6.1 (Lesson: integration by parts)
$\frac{d}{dx}\tan x = \sec^2 x$ ⁠; chain rule	Step 2	Standard formula (iv), Section 7.2; Chapter 5, Section 5.3.1
Product rule	Step 3	Chapter 5, Section 5.3

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Exercise 7.6 · Q18: ∫ex1+sin⁡x1+cos⁡x dx\int e^x\frac{1+\sin x}{1+\cos x}\,dx∫ex1+cosx1+sinx​dx

Basics you need

Solution

Rewrite the fraction with half angles

Recognise f+f′f + f'f+f′ and apply the rule

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.6 · Q18: $\int e^x\frac{1+\sin x}{1+\cos x}\,dx$

Recognise $f + f'$ and apply the rule