Chapter 7 · Integrals

Exercise 7.3 · Q12: $\int\frac{\sin^2x}{1+\cos x}\,dx$

Exercise 7.3Integration using trigonometric identities · Section 7.3.2

The question

Find the integral of the function

\frac{\sin^2 x}{1 + \cos x}.

Exam answer

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

Using $\sin^2 x = 1 - \cos^2 x$ ⁠,

\begin{aligned} I &= \int\frac{(1 - \cos x)(1 + \cos x)}{1 + \cos x}\,dx \\ &= \int(1 - \cos x)\,dx \end{aligned}

$\therefore\ I = x - \sin x + C$ ⁠.

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

The one idea this problem is built on

Turn the numerator into cosines: $\sin^2 x = 1 - \cos^2 x$ ⁠, a difference of two squares, $(1 - \cos x)(1 + \cos x)$ ⁠. The factor $1 + \cos x$ cancels with the denominator, leaving the easy integrand $1 - \cos x$ ⁠.

Given, to find, and the plan

Given: $\dfrac{\sin^2 x}{1+\cos x}$ ⁠, defined where $\cos x \ne -1$ ⁠, that is, $x \ne (2k+1)\pi$ for integers $k$ ⁠.
To find: $\displaystyle\int \frac{\sin^2 x}{1+\cos x}\,dx$ ⁠.
Plan: (1) Rewrite $\sin^2 x$ and factorise. (2) Cancel $1 + \cos x$ ⁠, noting where this is allowed. (3) Integrate. (4) Differentiate back to check and state where the answer is valid.

Basics you need

1The Pythagorean identity

$\sin^2 x + \cos^2 x = 1$ ⁠, so $\sin^2 x = 1 - \cos^2 x$ ⁠. Basics: trigonometric identities for integrals

2Difference of two squares

$a^2 - b^2 = (a - b)(a + b)$ ⁠. Basics: splitting fractions

3Standard integrals

$\int dx = x + C$ and $\int\cos x\,dx = \sin x + C$ (standard formulae, Section 7.2). Lesson: integration as anti-differentiation

4Checking an integral

$F(x) + C$ is right exactly when $F'(x)$ is the integrand (Property (I), Section 7.2.1). Basics: integration undoes differentiation

Solution

Rewrite the numerator and factorise
$\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x).$
The denominator is $1 + \cos x$ ⁠, so we want the numerator in terms of $\cos x$ too. The Pythagorean identity does that, and the result is a difference of two squares ( $1^2 - \cos^2 x$ ⁠) with $1 + \cos x$ as one of its factors.

$\sin^2 x = 1 - \cos^2 x$ ⁠; $a^2 - b^2 = (a-b)(a+b)$ with $a = 1$ ⁠, $b = \cos x$ ⁠. Basics: trigonometric identities
Cancel the common factor

Wherever $1 + \cos x \ne 0$ ⁠, that is, wherever the integrand is defined:
$\frac{\sin^2 x}{1 + \cos x} = \frac{(1 - \cos x)(1 + \cos x)}{1 + \cos x} = 1 - \cos x.$
We may only cancel a factor that is not zero. Here that is automatic: the points where $1 + \cos x = 0$ (the odd multiples of $\pi$ ⁠) are exactly the points excluded from the integrand's domain. So on its domain the integrand equals $1 - \cos x$ ⁠.
Integrate
$\begin{aligned} \int \frac{\sin^2 x}{1 + \cos x}\,dx &= \int (1 - \cos x)\,dx \\ &= x - \sin x + C. \end{aligned}$
Both terms are standard: $\int 1\,dx = x$ and $\int\cos x\,dx = \sin x$ ⁠.

Standard formulae, Section 7.2; Properties (III), (IV), Section 7.2.1. Lesson: integration as anti-differentiation
Check: differentiate the answer back, and say where it is valid

With $F(x) = x - \sin x$ ⁠:
$\begin{aligned} F'(x) &= 1 - \cos x = \frac{(1 - \cos x)(1 + \cos x)}{1 + \cos x} \\ &= \frac{\sin^2 x}{1 + \cos x} \end{aligned}$
for every $x$ with $\cos x \ne -1$ ⁠. So the answer is valid on each interval $\big((2k-1)\pi,\ (2k+1)\pi\big)$ ⁠.

$x - \sin x$ itself is defined for all $x$ ⁠, but the integrand is not defined at the odd multiples of $\pi$ ⁠, so the statement " $F' =$ integrand" only makes sense away from those points. (Near them the integrand tends to $1 - (-1) = 2$ ⁠, so these are only "holes"; but the question's function is still undefined there.)

$\frac{d}{dx}\sin x = \cos x$ ⁠; Property (I), Section 7.2.1. Basics: integration undoes differentiation

Result

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

valid wherever $\cos x \ne -1$ ⁠, i.e. $x \ne (2k+1)\pi$ ⁠. (Textbook answer: $x - \sin x + C$ ⁠.)

Check it with numbers

The simplification at $x = 1$ ⁠: $\dfrac{\sin^2 1}{1 + \cos 1}$ $\approx \dfrac{0.7080734}{1.5403023}$ $\approx 0.4596977$ ⁠, and $1 - \cos 1 \approx 0.4596977$ ⁠.

The derivative at $x = 1$ ⁠, $h = 0.001$ ⁠: $F(1.001) \approx 0.1589891337$ ⁠, $F(0.999) \approx 0.1580697381$ ⁠, slope $\approx 0.4596978$ ⁠, matching the integrand $0.4596977$ to six significant figures.

Common mistakes

Splitting the fraction as $\frac{\sin^2 x}{1} + \frac{\sin^2 x}{\cos x}$ ⁠. A sum in the denominator cannot be split.
Taking the long road. Half-angles ( $1 + \cos x = 2\cos^2\frac{x}{2}$ and $\sin^2 x = 4\sin^2\frac{x}{2}\cos^2\frac{x}{2}$ ⁠) also work (it gives $2\sin^2\frac{x}{2} = 1 - \cos x$ ⁠), but it is a longer road to the same place.
Cancelling without a word. Say that $1 + \cos x \ne 0$ on the domain; that is what makes the cancellation legal.

What to take away

Before reaching for half-angles, look for a factor to cancel: $\sin^2 x = (1 - \cos x)(1 + \cos x)$ ⁠.
Cancellation is legal where the cancelled factor is nonzero; state the domain.
Similarly $\frac{\sin^2 x}{1 - \cos x} = 1 + \cos x$ wherever $\cos x \ne 1$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin^2 x + \cos^2 x = 1$	Step 1	Pythagorean identity (Class XI trigonometry)
$a^2 - b^2 = (a - b)(a + b)$	Steps 1, 4	Algebra
$\int dx = x + C$ ⁠; $\int\cos x\,dx = \sin x + C$	Step 3	Standard formulae, Section 7.2
Sum and constant-multiple rules	Step 3	Properties (III), (IV), Section 7.2.1
Differentiation undoes integration	Step 4	Property (I), Section 7.2.1

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Exercise 7.3 · Q12: ∫sin⁡2x1+cos⁡x dx\int\frac{\sin^2x}{1+\cos x}\,dx∫1+cosxsin2x​dx

Basics you need

Solution

Rewrite the numerator and factorise

Cancel the common factor

Integrate

Check: differentiate the answer back, and say where it is valid

Every fact used, and where it comes from

Exercise 7.3 · Q12: $\int\frac{\sin^2x}{1+\cos x}\,dx$