Chapter 7 · Integrals

Example 17: $\int x\cos x\,dx$

ExamplesIntegration by parts · Section 7.6

The question

Find

\int x\cos x\,dx

Exam answer

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

Taking $x$ as the first function and $\cos x$ as the second, by parts:

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

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

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 two functions, and no basic formula fits it. Integration by parts trades $\int x\cos x\,dx$ for a new integral in which $x$ has been differentiated (to 1) while $\cos x$ has been integrated (to $\sin x$ ⁠). Differentiating $x$ makes it simpler, and integrating $\cos x$ does not make it harder, so the new integral, $\int \sin x\,dx$ ⁠, is easy. Choosing the two functions the other way round would make things worse.

Given, to find, and the plan

Given: The integrand $x\cos x$ ⁠, defined and continuous for every real $x$ ⁠.
To find: $\int x\cos x\,dx$ ⁠.
Plan: (1) Recall the integration by parts formula. (2) Choose the first and second functions. (3) Apply the formula. (4) See what the opposite choice does. (5) See why no constant is needed for the integral of the second function. (6) Differentiate back.

Basics you need

1Integration by parts

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

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

With $u = f(x)$ as the first function and $\frac{dv}{dx} = g(x)$ as the second:

\begin{aligned} \int f(x)\,g(x)\,dx &= f(x)\int g(x)\,dx \\ &\quad - \int\Big[f'(x)\int g(x)\,dx\Big]dx. \end{aligned}

Lesson: integration by parts

2Two basic integrals

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

3The product rule

$(fg)' = f'g + fg'$ ⁠, used for the final check. Basics: integration undoes differentiation

Solution

Why integration by parts?

$x\cos x$ is a product of an algebraic function ( $x$ ⁠) and a trigonometric one ( $\cos x$ ⁠). It is not in the list of standard formulae, it is not a sum we can split, and no simple substitution turns it into a standard form. Integration by parts is the method designed for exactly this kind of product.

The formula comes from reading the product rule backwards: since $(uv)' = uv' + vu'$ ⁠, the integral of one of the two products, $uv'$ ⁠, can be traded for the integral of the other, $vu'$ ⁠. It helps when the second integral is easier than the first.

Section 7.6: "the integral of the product of two functions = (first function) × (integral of the second function) − integral of [(differential coefficient of the first function) × (integral of the second function)]". Lesson: integration by parts
Choose the first and second functions

Take
$f(x) = x \ \ (\text{first}), \qquad g(x) = \cos x \ \ (\text{second}).$
Then $f'(x) = 1$ and $\int g(x)\,dx = \sin x$ ⁠.

The first function gets differentiated, the second gets integrated. Differentiating $x$ turns it into the constant 1, which removes the troublesome factor; integrating $\cos x$ just gives $\sin x$ ⁠, no harder. This is the book's Remark (iii): a power of $x$ or a polynomial is usually taken as the first function.

$\frac{d}{dx}x = 1$ ⁠; $\int\cos x\,dx = \sin x + C$ (formula (ii)). Basics: powers of x
Apply the formula
$\begin{aligned} \int x\cos x\,dx &= x\int\cos x\,dx - \int\Big[\frac{d}{dx}(x)\int\cos x\,dx\Big]dx \\ &= x\sin x - \int 1\cdot\sin x\,dx \\ &= x\sin x - (-\cos x) + C \\ &= x\sin x + \cos x + C. \end{aligned}$
After one use of the formula the remaining integral is $\int\sin x\,dx$ ⁠, a standard one. The minus in front of it and the minus in $\int\sin x\,dx = -\cos x$ multiply to give $+\cos x$ ⁠.

$\int\sin x\,dx = -\cos x + C$ (formula (iii)). Basics: trigonometric functions
What the opposite choice would do

Suppose instead $f(x) = \cos x$ (first) and $g(x) = x$ (second). Then $f'(x) = -\sin x$ and $\int x\,dx = \frac{x^2}{2}$ ⁠, so
$\begin{aligned} \int x\cos x\,dx &= \cos x\cdot\frac{x^2}{2} - \int(-\sin x)\frac{x^2}{2}\,dx \\ &= \frac{x^2}{2}\cos x + \int\frac{x^2}{2}\sin x\,dx. \end{aligned}$
This is true, but the new integral contains $x^2$ instead of $x$ ⁠: it is harder than the one we started with.

Integrating $x$ raises its power, so the algebraic factor grows instead of disappearing. The formula is always correct; the choice only decides whether it is useful. That is why the order of the two functions matters.

$\int x\,dx = \frac{x^2}{2} + C$ (power rule, formula (i)). Lesson: integration by parts
Why no constant in $\int\cos x\,dx$

In Step 3 we wrote $\int\cos x\,dx = \sin x$ ⁠, without a constant. If we used $\sin x + k$ for any constant $k$ ⁠:
$\begin{aligned} \int x\cos x\,dx &= x(\sin x + k) - \int(\sin x + k)\,dx \\ &= x\sin x + kx + \cos x - kx + C \\ &= x\sin x + \cos x + C. \end{aligned}$
The $k$ terms cancel, so adding a constant there changes nothing.

This is the book's Remark (ii). The extra $kx$ from the first part is removed by $-\int k\,dx = -kx$ from the second part. Only one constant, $C$ ⁠, is needed, at the very end.

$\int k\,dx = kx + C$ ⁠. Lesson: integration by parts
Check: differentiate the answer

By the product rule, $\frac{d}{dx}(x\sin x) = 1\cdot\sin x + x\cos x$ ⁠. So
$\begin{aligned} \frac{d}{dx}\big(x\sin x + \cos x\big) &= \sin x + x\cos x - \sin x \\ &= x\cos x. \end{aligned}$
The derivative is the integrand, for every real $x$ ⁠. Integration by parts is the product rule run backwards, so it is natural that the product rule confirms the answer.

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

Result

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

valid for every real $x$ ⁠. This is the book's answer.

Check it with numbers

At a point. At $x = 1$ ⁠: the integrand is $\cos 1 = 0.5403023\ldots$ ⁠; the derivative of our answer is $\sin 1 + \cos 1 - \sin 1$ ⁠, the same $0.5403023\ldots$

Over an interval. On $\left[0, \frac\pi2\right]$ ⁠:

\begin{aligned} F\left(\tfrac\pi2\right) - F(0) &= \left(\tfrac\pi2 + 0\right) - (0 + 1) \\ &= \tfrac\pi2 - 1 \approx 0.5707963268 \end{aligned}

Numerical integration of $x\cos x$ over $\left[0, \frac\pi2\right]$ gives $0.5707963268$ to ten decimal places, the same.

Common mistakes

Choosing $\cos x$ as the first function. The new integral gets harder (Step 4).
Sign of the last term. $-\int\sin x\,dx = +\cos x$ ⁠, not $-\cos x$ ⁠.
Writing $\int x\cos x\,dx = \int x\,dx\cdot\int\cos x\,dx$ ⁠. The integral of a product is not the product of the integrals: $\frac{x^2}{2}\sin x$ differentiates to something else.

What to take away

By parts: first function differentiated, second integrated. Choose the first so that differentiating simplifies it.
No constant is needed when integrating the second function; add one $C$ at the end.
Integration by parts does not finish every product. The book's Remark (i) gives $\int\sqrt{x}\sin x\,dx$ ⁠. Strictly, $\sqrt{x}\sin x$ does have antiderivatives on $x \ge 0$ ⁠: it is continuous there, and by Theorem 1 of Section 7.8.2 (stated in the book without proof) its area function is one. What is true is that no antiderivative can be written with the functions of this chapter (a result beyond this book), so no choice of first and second function finishes it.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts formula	Steps 1, 3, 4, 5	Section 7.6, equation (1) and the rule after it (Lesson: integration by parts)
Choice of first function (polynomial first)	Step 2	Remark (iii) after Example 17, Section 7.6
No constant needed for $\int g\,dx$	Step 5	Remark (ii) after Example 17, Section 7.6
$\int\cos x\,dx = \sin x + C$ ⁠; $\int\sin x\,dx = -\cos x + C$ ⁠; $\int x^n dx$	Steps 3, 4	Standard formulae (ii), (iii), (i), Section 7.2
Product rule	Steps 1, 6	Chapter 5, Section 5.3
A continuous function has an antiderivative (its area function)	Take-away 3	First fundamental theorem (Theorem 1), Section 7.8.2, stated without proof
$\sqrt{x}\sin x$ has no antiderivative built from the functions of this chapter	Take-away 3	Beyond this book (stated, not proved)

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Example 17: ∫xcos⁡x dx\int x\cos x\,dx∫xcosxdx

Basics you need

Solution

Why integration by parts?

Choose the first and second functions

Apply the formula

What the opposite choice would do

Why no constant in ∫cos⁡x dx\int\cos x\,dx∫cosxdx

Check: differentiate the answer

Every fact used, and where it comes from

Example 17: $\int x\cos x\,dx$

Why no constant in $\int\cos x\,dx$