Chapter 7 · Integrals

Exercise 7.6 · Q2: $\int x\sin 3x\,dx$

Exercise 7.6Integration by parts · Section 7.6

The question

Integrate the functions in Exercises 1 to 22.

2. $x\sin 3x$

Exam answer

Let $I = \displaystyle\int x\sin 3x\,dx$ ⁠.

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

\begin{aligned} I &= x\left(-\frac{\cos 3x}{3}\right) \\ &\quad - \int 1\cdot\left(-\frac{\cos 3x}{3}\right)dx \\ &= -\frac x3\cos 3x + \frac13\int\cos 3x\,dx \\ &= -\frac x3\cos 3x + \frac13\cdot\frac{\sin 3x}{3} + C \end{aligned}

$\therefore\ I = -\dfrac x3\cos 3x + \dfrac19\sin 3x + C$ ⁠.

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

The one idea this problem is built on

This is Question 1 with $3x$ in place of $x$ inside the sine. The method is the same: integration by parts with $u = x$ ⁠, so that differentiating $x$ removes it. The only new point is the factor $\frac13$ that appears every time we integrate a sine or cosine of $3x$ ⁠, and it appears twice, giving $\frac19$ in the last term.

Given, to find, and the plan

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

Basics you need

1Integration by parts

Section 7.6, equation (1):

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

with $u$ the first function (differentiated) and $\frac{dv}{dx}$ the second (integrated to give $v$ ⁠). Lesson: integration by parts

2Which factor should be $u$ ⁠?

Remark (iii) after Example 17: a power of $x$ is usually the first function. The memory aid ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential; not in the book) says the same: Algebraic comes before Trigonometric.

3Sine and cosine of $3x$

With $t = 3x$ ⁠, $dt = 3\,dx$ ⁠:

\begin{aligned} \int\sin 3x\,dx &= -\tfrac13\cos 3x + C, \\ \int\cos 3x\,dx &= \tfrac13\sin 3x + C. \end{aligned}

(Substitution, Section 7.3.1, with standard formulae (iii) and (ii).) Basics: substitution

4Product and chain rules

For the check: $(fg)' = f'g + fg'$ and $\frac{d}{dx}\cos 3x = -3\sin 3x$ ⁠. Basics: integration undoes differentiation

Solution

Name $u$ and $\frac{dv}{dx}$
$\begin{aligned} u &= x \quad (\text{first}), \\ \frac{dv}{dx} &= \sin 3x \quad (\text{second}). \end{aligned}$
ILATE: $x$ is Algebraic, $\sin 3x$ is Trigonometric, and A comes before T. Differentiating $x$ gives 1, which removes it; integrating $\sin 3x$ gives another trigonometric function of the same kind, no harder. With the opposite choice the power of $x$ would go up to $x^2$ ⁠.

Remark (iii) after Example 17, Section 7.6. Lesson: integration by parts
Find $\frac{du}{dx}$ and $v$

$\frac{du}{dx} = 1$ ⁠. For $v$ ⁠, put $t = 3x$ ⁠, so $dt = 3\,dx$ and $dx = \frac13\,dt$ ⁠:
$\begin{aligned} v &= \int\sin 3x\,dx = \frac13\int\sin t\,dt \\ &= -\frac13\cos t = -\frac13\cos 3x. \end{aligned}$
The substitution turns $\sin 3x$ into the standard $\sin t$ ⁠; the factor $\frac13$ comes from $dx = \frac13\,dt$ ⁠. (Quick test: $\frac{d}{dx}\left(-\frac13\cos 3x\right)$ $= -\frac13\cdot(-3\sin 3x) = \sin 3x$ ⁠.) No constant is added to $v$ (Remark (ii) after Example 17).

Substitution, Section 7.3.1; standard formula (iii). Basics: substitution
Apply the formula
$\begin{aligned} \int x\sin 3x\,dx &= uv - \int v\,\frac{du}{dx}\,dx \\ &= x\left(-\frac13\cos 3x\right) - \int\left(-\frac13\cos 3x\right)dx \\ &= -\frac x3\cos 3x + \frac13\int\cos 3x\,dx. \end{aligned}$
First function times integral of the second, minus the integral of (derivative of first) times (integral of second). The constant $-\frac13$ comes out of the integral, and minus times minus gives plus.

Section 7.6, equation (1); constant multiple rule (Section 7.2.1). Lesson: integration by parts
Finish the remaining integral

By the same substitution, $\int\cos 3x\,dx = \frac13\sin 3x$ ⁠. So
$\begin{aligned} \int x\sin 3x\,dx &= -\frac x3\cos 3x + \frac13\cdot\frac13\sin 3x + C \\ &= -\frac x3\cos 3x + \frac19\sin 3x + C. \end{aligned}$
Each integration of a sine or cosine of $3x$ brings a factor $\frac13$ ⁠. The last term was integrated twice (once for $v$ ⁠, once now), so it carries $\frac13\cdot\frac13 = \frac19$ ⁠.

Substitution $t = 3x$ ⁠; standard formula (ii). Basics: substitution
Check: differentiate the answer

Product rule on the first term, chain rule on both:
$\begin{aligned} &\frac{d}{dx}\Big(-\frac x3\cos 3x + \frac19\sin 3x\Big) \\ &= -\frac13\cos 3x - \frac x3(-3\sin 3x) + \frac39\cos 3x \\ &= -\frac13\cos 3x + x\sin 3x + \frac13\cos 3x \\ &= x\sin 3x. \end{aligned}$
The two cosine terms cancel and exactly the integrand remains, for every real $x$ ⁠.

Product rule and chain rule (Chapter 5). Basics: integration undoes differentiation

Result

\int x\sin 3x\,dx = -\frac{x}{3}\cos 3x + \frac19\sin 3x + C,

valid for every real $x$ ⁠. (Textbook answer: $-\frac{x}{3}\cos 3x + \frac19\sin 3x + C$ ⁠, the same.)

Check it with numbers

With $F(x) = -\frac x3\cos 3x + \frac19\sin 3x$ ⁠, on $\left[0, \frac\pi3\right]$ ⁠, where $\cos\pi = -1$ and $\sin\pi = 0$ ⁠:

\begin{aligned} F\left(\tfrac\pi3\right) - F(0) &= -\frac{\pi}{9}(-1) + 0 - 0 \\ &= \frac{\pi}{9} = 0.3490658504\ldots \end{aligned}

Numerical integration of $x\sin 3x$ over $\left[0, \frac\pi3\right]$ gives $0.3490658504\ldots$ ⁠, the same.

Common mistakes

Forgetting the $\frac13$ ⁠. $\int\sin 3x\,dx = -\frac13\cos 3x$ ⁠, not $-\cos 3x$ ⁠.
Only one factor $\frac13$ in the last term. The last term needs $\frac19$ ⁠, because it has been integrated twice.
Multiplying by 3 instead of dividing. Differentiating $\cos 3x$ multiplies by 3; integrating divides by 3.

What to take away

The same steps give, for any $a \ne 0$ ⁠, $\int x\sin ax\,dx = -\frac xa\cos ax + \frac{1}{a^2}\sin ax + C.$ Question 1 is the case $a = 1$ ⁠.
When integrating $\sin ax$ or $\cos ax$ ⁠, divide by $a$ every time.
Always test $v$ by differentiating it before using it in the formula.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts	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
No constant needed in $v$	Step 2	Remark (ii) after Example 17, Section 7.6
Substitution $t = 3x$	Steps 2, 4	Section 7.3.1 (Lesson: substitution)
$\int\sin t\,dt = -\cos t + C$ ⁠, $\int\cos t\,dt = \sin t + C$	Steps 2, 4	Standard formulae (iii), (ii), Section 7.2
Constant multiple rule	Step 3	Property (IV), Section 7.2.1
Product and chain rules	Step 5	Chapter 5, Sections 5.3 and 5.3.1

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Exercise 7.6 · Q2: ∫xsin⁡3x dx\int x\sin 3x\,dx∫xsin3xdx

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 · Q2: $\int x\sin 3x\,dx$

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

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