Chapter 7 · Integrals

Exercise 7.8 · Q5: $\int_0^{\pi/2}\cos 2x\,dx$

Exercise 7.8Fundamental theorem of calculus · Section 7.8

The question

Evaluate the definite integral

\int_0^{\frac{\pi}{2}}\cos 2x\,dx.

Exam answer

Let $I = \displaystyle\int_0^{\frac{\pi}{2}}\cos 2x\,dx$ ⁠.

\begin{aligned} I &= \left[\frac{\sin 2x}{2}\right]_0^{\frac{\pi}{2}} \\ &= \frac12\sin\pi - \frac12\sin 0 \\ &= 0 - 0 \end{aligned}

$\therefore\ I = 0$ ⁠.

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

The one idea this problem is built on

The second fundamental theorem (Theorem 2, Section 7.8.3) gives $\int_a^b f(x)\,dx = F(b) - F(a)$ with $F(x) = \frac12\sin 2x$ ⁠. The answer turns out to be 0, and that is not a mistake: a definite integral is a signed area. Here $\cos 2x$ is positive on the first half of the interval and negative on the second half, and the two pieces cancel exactly.

Given, to find, and the plan

Given: $f(x) = \cos 2x$ on $\left[0, \frac{\pi}{2}\right]$ ⁠.
To find: The exact value of $\int_0^{\pi/2}\cos 2x\,dx$ ⁠.
Plan: (1) Continuity. (2) Antiderivative by the substitution $u = 2x$ ⁠, checked by differentiating. (3) $F\left(\frac{\pi}{2}\right) - F(0)$ ⁠. (4) Explain the 0 by splitting at $\frac{\pi}{4}$ ⁠, where $\cos 2x$ changes sign.

Basics you need

1Second fundamental theorem

$\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a)$ for $f$ continuous on $[a, b]$ and $F' = f$ ⁠. Lesson: definite integrals and the fundamental theorem

2Integral of $\cos$

$\int\cos u\,du = \sin u + C$ (standard formula (ii), Section 7.2). Basics: trigonometric identities for integrals

3Substitution

If $u = g(x)$ ⁠, then $du = g'(x)\,dx$ (Section 7.3.1). Basics: substitution (the chain rule backwards)

4Signed area

Section 7.8.1 describes $\int_a^b f$ as the area bounded by the curve, the ordinates $x = a$ ⁠, $x = b$ and the x-axis, and its Fig 7.1 assumes $f(x) > 0$ ⁠; this reprint does not discuss regions below the axis. Where the graph is below the x-axis, $F(b) - F(a)$ comes out negative (Step 5 shows this), so in general the integral counts area above the axis as positive and area below it as negative. Basics: area as a sum of thin strips

Solution

Check continuity on $\left[0, \frac{\pi}{2}\right]$

$\cos 2x$ is the composition of the continuous functions $x \mapsto 2x$ and $\cos$ ⁠, so it is continuous for every real $x$ ⁠, and in particular on $\left[0, \frac{\pi}{2}\right]$ ⁠.

Theorem 2 needs a continuous integrand on the closed interval (Remark (iv) after Theorem 2).

Composition of continuous functions: Chapter 5, Theorem 2, Section 5.2.1. Lesson: the fundamental theorem
Antiderivative by the substitution $u = 2x$

Let $u = 2x$ ⁠, so $du = 2\,dx$ and $dx = \frac12\,du$ ⁠:
$\int\cos 2x\,dx = \tfrac12\int\cos u\,du = \tfrac12\sin u + C = \tfrac12\sin 2x + C.$
Take $F(x) = \frac12\sin 2x$ ⁠.

The inner function $2x$ has the constant derivative 2, so the substitution simply contributes the factor $\frac12$ ⁠.

Standard formula (ii), Section 7.2; substitution, Section 7.3.1. Lesson: integration by substitution
Check: differentiate $F$ back

Chain rule (outer $\sin$ ⁠, inner $2x$ ⁠):
$F'(x) = \tfrac12\cos 2x\cdot 2 = \cos 2x.$
The $\frac12$ cancels the inner derivative 2, giving exactly the integrand.

$\frac{d}{dx}\sin u = \cos u\cdot\frac{du}{dx}$ ⁠. Basics: the chain rule
Evaluate $F\left(\frac{\pi}{2}\right) - F(0)$

At $x = \frac{\pi}{2}$ ⁠, $2x = \pi$ ⁠; at $x = 0$ ⁠, $2x = 0$ ⁠:
$\begin{aligned}\int_0^{\frac{\pi}{2}}\cos 2x\,dx &= \tfrac12\sin\pi - \tfrac12\sin 0 \\ &= \tfrac12\cdot 0 - \tfrac12\cdot 0 = 0.\end{aligned}$
Upper minus lower. Both values of $\sin$ are 0, so the integral is 0.

$\sin 0 = \sin\pi = 0$ ⁠. Basics: trigonometric values
Why 0? Split at $\frac{\pi}{4}$

$\cos 2x = 0$ when $2x = \frac{\pi}{2}$ ⁠, that is at $x = \frac{\pi}{4}$ ⁠. For $0 \le x < \frac{\pi}{4}$ ⁠, $2x$ is in $\left[0, \frac{\pi}{2}\right)$ and $\cos 2x > 0$ ⁠; for $\frac{\pi}{4} < x \le \frac{\pi}{2}$ ⁠, $2x$ is in $\left(\frac{\pi}{2}, \pi\right]$ and $\cos 2x < 0$ ⁠. Using the same $F$ on each half:
$\begin{aligned}\int_0^{\frac{\pi}{4}}\cos 2x\,dx &= \tfrac12\sin\tfrac{\pi}{2} - \tfrac12\sin 0 = \tfrac12, \\ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos 2x\,dx &= \tfrac12\sin\pi - \tfrac12\sin\tfrac{\pi}{2} = -\tfrac12.\end{aligned}$
The sum is $\frac12 + \left(-\frac12\right) = 0$ ⁠.

$y = \cos 2x$ on $\left[0, \frac{\pi}{2}\right]$ ⁠. The blue region above the axis has area $\frac12$ and counts $+\frac12$ ⁠; the rose region below the axis has the same area and counts $-\frac12$ ⁠. The signed total is 0.

The two pieces are mirror images: $\cos\left(2\left(\frac{\pi}{2} - x\right)\right) = \cos(\pi - 2x) = -\cos 2x$ ⁠, so the graph on the second half is the first half turned upside down about the point $\left(\frac{\pi}{4}, 0\right)$ ⁠. Equal areas, opposite signs. The geometric area enclosed by the curve and the axis is 1, but the integral is 0.

$\cos(\pi - \theta) = -\cos\theta$ ⁠. Basics: symmetry and mirror images

Result

\int_0^{\frac{\pi}{2}}\cos 2x\,dx = 0.

(Textbook answer: 0, the same.)

Check it with numbers

Numerical integration (thin strips, no antiderivative used) gives $0.5000000$ on $\left[0, \frac{\pi}{4}\right]$ ⁠, $-0.5000000$ on $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$ ⁠, and $0.0000000$ on the whole interval $\left[0, \frac{\pi}{2}\right]$ ⁠.

Common mistakes

Thinking 0 must be wrong. A definite integral is a signed area; positive and negative parts can cancel.
Using $\sin\frac{\pi}{2}$ at the upper limit. The angle is $2x$ ⁠, so at $x = \frac{\pi}{2}$ it is $\pi$ ⁠, and $\sin\pi = 0$ ⁠.
Calling 0 "the area". If a question asks for the area between the curve and the axis, add the absolute values: $\frac12 + \frac12 = 1$ ⁠.

What to take away

$\int\cos kx\,dx = \frac1k\sin kx + C$ for a constant $k \ne 0$ ⁠.
The definite integral is a signed area: parts below the axis subtract.
When an integral comes out 0, look for a sign change and a symmetry that explain it.

Every fact used, and where it comes from

Fact	Used in	Source
$\int_a^b f = F(b) - F(a)$ for continuous $f$	Steps 1, 4, 5	Theorem 2, Section 7.8.3 (Lesson)
Composition of continuous functions is continuous	Step 1	Chapter 5, Theorem 2, Section 5.2.1
$\int\cos u\,du = \sin u + C$	Step 2	Standard formula (ii), Section 7.2
Substitution $u = 2x$	Step 2	Section 7.3.1 (Lesson)
Chain rule	Step 3	Chapter 5, Section 5.3.1 (Basics)
Signed area: Theorem 2 on each half of the interval	Step 5	Theorem 2, Section 7.8.3; area meaning, Section 7.8.1 (Basics)
$\cos(\pi - \theta) = -\cos\theta$	Step 5	Trigonometric identities (Basics)

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 →

Exercise 7.8 · Q5: ∫0π/2cos⁡2x dx\int_0^{\pi/2}\cos 2x\,dx∫0π/2​cos2xdx

Basics you need

Solution

Check continuity on [0,π2]\left[0, \frac{\pi}{2}\right][0,2π​]

Antiderivative by the substitution u=2xu = 2xu=2x

Check: differentiate FFF back

Evaluate F(π2)−F(0)F\left(\frac{\pi}{2}\right) - F(0)F(2π​)−F(0)

Why 0? Split at π4\frac{\pi}{4}4π​

Every fact used, and where it comes from

Exercise 7.8 · Q5: $\int_0^{\pi/2}\cos 2x\,dx$

Check continuity on $\left[0, \frac{\pi}{2}\right]$

Antiderivative by the substitution $u = 2x$

Check: differentiate $F$ back

Evaluate $F\left(\frac{\pi}{2}\right) - F(0)$

Why 0? Split at $\frac{\pi}{4}$