Chapter 7 · Integrals

Exercise 7.6 · Q1: $\int x\sin x\,dx$

Exercise 7.6Integration by parts · Section 7.6

The question

Integrate the functions in Exercises 1 to 22.

1. $x\sin x$

Exam answer

Let $I = \displaystyle\int x\sin x\,dx$ ⁠.

Taking $x$ as the first function, by parts:

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

$\therefore\ I = -x\cos x + \sin x + C$ ⁠.

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

The one idea this problem is built on

The integrand is a product of an algebraic factor $x$ and a trigonometric factor $\sin x$ ⁠, and no standard formula fits it. Integration by parts lets us differentiate one factor and integrate the other. Differentiating $x$ turns it into 1, so the integral that is left contains only a trigonometric function, which is a standard one.

Given, to find, and the plan

Given: The function $x\sin x$ ⁠, defined and continuous for every real $x$ ⁠.
To find: $\int x\sin x\,dx$ ⁠, the family of all its antiderivatives.
Plan: (1) Name $u$ and $\frac{dv}{dx}$ and explain the choice. (2) Find $\frac{du}{dx}$ and $v$ ⁠. (3) Apply the integration by parts formula. (4) Finish the remaining integral. (5) Differentiate the answer back.

Basics you need

1Integration by parts

Integrating the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$ gives Section 7.6, equation (1):

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx.

Here $u$ is the first function (it gets differentiated) and $\frac{dv}{dx}$ is the second function (it gets integrated, giving $v$ ⁠). Lesson: integration by parts

2Which factor should be $u$ ⁠?

The book's Remark (iii) after Example 17: a power of $x$ or a polynomial is usually taken as the first function, unless the other factor is an inverse trigonometric or logarithmic function. A common memory aid (not in the book) is the order ILATE: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. The factor whose type comes earlier is taken as $u$ ⁠.

3Two standard integrals

$\int\sin x\,dx = -\cos x + C$ and $\int\cos x\,dx = \sin x + C$ (standard formulae (iii) and (ii), Section 7.2). Basics: trigonometric functions

4The product rule

$(fg)' = f'g + fg'$ ⁠, used to differentiate the answer back. Basics: integration undoes differentiation

Solution

Name $u$ and $\frac{dv}{dx}$

Write the integrand as a product of two factors and take
$\begin{aligned} u &= x \quad (\text{first}), \\ \frac{dv}{dx} &= \sin x \quad (\text{second}). \end{aligned}$
In ILATE order, $x$ is Algebraic (A) and $\sin x$ is Trigonometric (T); A comes first, so $u = x$ ⁠. The reason behind the rule: differentiating $x$ gives the constant 1, which removes the troublesome factor, while integrating $\sin x$ only gives $-\cos x$ ⁠, no harder. The opposite choice ( $u = \sin x$ ⁠, $\frac{dv}{dx} = x$ ⁠) would integrate $x$ to $\frac{x^2}{2}$ and leave $\int\frac{x^2}{2}\cos x\,dx$ ⁠, which is harder than the question.

Remark (iii) after Example 17, Section 7.6. Lesson: integration by parts
Find $\frac{du}{dx}$ and $v$
$\frac{du}{dx} = 1, \qquad v = \int\sin x\,dx = -\cos x.$
The formula needs the derivative of $u$ and an integral of $\frac{dv}{dx}$ ⁠. No constant is added to $v$ ⁠: the book's Remark (ii) after Example 17 shows that any constant added here cancels out of the final answer.

$\frac{d}{dx}x = 1$ ⁠; standard formula (iii). Basics: trigonometric functions
Apply the formula
$\begin{aligned} \int x\sin x\,dx &= uv - \int v\,\frac{du}{dx}\,dx \\ &= x(-\cos x) - \int(-\cos x)\cdot 1\,dx \\ &= -x\cos x + \int\cos x\,dx. \end{aligned}$
First function times the integral of the second, minus the integral of (derivative of the first) times (integral of the second). The two minus signs in $-\int(-\cos x)\,dx$ multiply to a plus.

Section 7.6, equation (1). Lesson: integration by parts
Finish the remaining integral
$\int x\sin x\,dx = -x\cos x + \sin x + C.$
$\int\cos x\,dx = \sin x$ is standard formula (ii). One constant $C$ is added at the very end, for the whole family of antiderivatives.

Standard formula (ii), Section 7.2. Basics: trigonometric functions
Check: differentiate the answer

By the product rule, $\frac{d}{dx}(x\cos x) = \cos x - x\sin x$ ⁠. So
$\begin{aligned} &\frac{d}{dx}\big(-x\cos x + \sin x\big) \\ &= -(\cos x - x\sin x) + \cos x \\ &= -\cos x + x\sin x + \cos x \\ &= x\sin x. \end{aligned}$
The derivative is exactly the integrand for every real $x$ ⁠, so the answer is right. Integration by parts is the product rule run backwards, so the product rule is the natural check.

Product rule; $\frac{d}{dx}\cos x = -\sin x$ ⁠. Basics: integration undoes differentiation

Result

\int x\sin x\,dx = -x\cos x + \sin x + C,

valid for every real $x$ ⁠. (Textbook answer: $-x\cos x + \sin x + C$ ⁠, the same.)

Check it with numbers

With $F(x) = -x\cos x + \sin x$ ⁠, on the interval $\left[0, \frac\pi2\right]$ ⁠:

\begin{aligned} F\left(\tfrac\pi2\right) - F(0) &= \left(-\tfrac\pi2\cdot 0 + 1\right) - (0 + 0) \\ &= 1. \end{aligned}

Numerical integration of $x\sin x$ over $\left[0, \frac\pi2\right]$ (adding up very thin strips) gives $1.0000000000$ ⁠, the same.

Common mistakes

Sign of $v$ ⁠. $\int\sin x\,dx = -\cos x$ ⁠, not $+\cos x$ ⁠. Losing this minus changes the answer to $x\cos x - \sin x$ ⁠, whose derivative is $-x\sin x$ ⁠.
Choosing $u = \sin x$ ⁠. The new integral then contains $x^2$ and is harder (Step 1).
Multiplying integrals. $\int x\sin x\,dx$ is not $\int x\,dx\cdot\int\sin x\,dx$ ⁠.

What to take away

For $x$ times a sine or cosine, take $u = x$ ⁠: one use of the formula removes the $x$ ⁠.
Write $u$ ⁠, $\frac{dv}{dx}$ ⁠, $\frac{du}{dx}$ and $v$ down before using the formula; most errors are sign errors in $v$ ⁠.
One constant $C$ at the end is enough.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts, $\int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx$	Step 3	Section 7.6, equation (1) (Lesson: integration by parts)
Choice of the first function (ILATE)	Step 1	Remark (iii) after Example 17, Section 7.6; same choice as Example 17
No constant needed in $v$	Step 2	Remark (ii) after Example 17, Section 7.6
$\int\sin x\,dx = -\cos x + C$ ⁠, $\int\cos x\,dx = \sin x + C$	Steps 2, 4	Standard formulae (iii), (ii), Section 7.2
Product rule	Step 5	Chapter 5, Section 5.3

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Exercise 7.6 · Q1: ∫xsin⁡x dx\int x\sin x\,dx∫xsinxdx

Basics you need

Solution

Name uuu and dvdx\frac{dv}{dx}dxdv​

Find dudx\frac{du}{dx}dxdu​ and vvv

Apply the formula

Finish the remaining integral

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.6 · Q1: $\int x\sin x\,dx$

Name $u$ and $\frac{dv}{dx}$

Find $\frac{du}{dx}$ and $v$