Chapter 7 · Integrals

Exercise 7.8 · Q12: $\int_0^{\pi/2}\cos^2 x\,dx$

Exercise 7.8Fundamental theorem of calculus · Section 7.8

The question

Evaluate the definite integral

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

Exam answer

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

Using $\cos^2 x = \frac{1 + \cos 2x}{2},$

\begin{aligned} I &= \frac12\int_0^{\frac{\pi}{2}}(1 + \cos 2x)\,dx \\ &= \frac12\left[x + \frac{\sin 2x}{2}\right]_0^{\frac{\pi}{2}} \\ &= \frac12\left[\left(\frac{\pi}{2} + \frac{\sin\pi}{2}\right) - 0\right] \end{aligned}

$\therefore\ I = \dfrac{\pi}{4}$ ⁠.

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

The one idea this problem is built on

There is no standard formula for $\cos^2 x$ directly, but the double-angle identity $\cos 2x = 2\cos^2 x - 1$ rewrites it as $\frac12 + \frac12\cos 2x$ ⁠, and each of those has a known antiderivative. Then the second fundamental theorem (Theorem 2, Section 7.8.3) gives the value as $F\left(\frac{\pi}{2}\right) - F(0)$ ⁠.

Given, to find, and the plan

Given: The integrand $f(x) = \cos^2 x$ on the interval $\left[0, \frac{\pi}{2}\right]$ ⁠.
To find: The exact value of $\int_0^{\pi/2}\cos^2 x\,dx$ ⁠, and a decimal.
Plan: (1) Check continuity on $\left[0, \frac{\pi}{2}\right]$ ⁠. (2) Lower the power with the double-angle identity. (3) Integrate term by term to get $F$ ⁠. (4) Differentiate $F$ back. (5) Compute $F\left(\frac{\pi}{2}\right) - F(0)$ ⁠.

Basics you need

1Second fundamental theorem

If $f$ is continuous on $[a, b]$ and $F' = f$ ⁠, then $\int_a^b f(x)\,dx = F(b) - F(a)$ ⁠. No "+ C" is needed: it cancels in the subtraction. Lesson: definite integrals and the fundamental theorem

2Double-angle identity

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

3Integral of $\cos kx$

$\int\cos kx\,dx = \frac{\sin kx}{k} + C$ for a constant $k \ne 0$ ⁠, because $\frac{d}{dx}\sin kx = k\cos kx$ (chain rule). Basics: substitution (the chain rule backwards)

Solution

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

$\cos x$ is continuous for every real $x$ ⁠, and $\cos^2 x = \cos x\cdot\cos x$ is a product of continuous functions, so it is continuous on all of $\mathbb{R}$ ⁠, in particular on $\left[0, \frac{\pi}{2}\right]$ ⁠.

Theorem 2 may only be used when the integrand is continuous on the whole closed interval (Remark (iv), Section 7.8.3). Here there are no denominators, roots or logarithms, so nothing can go wrong.

A product of continuous functions is continuous (Chapter 5). Lesson: the fundamental theorem
Lower the power with the double-angle identity

Start from $\cos 2x = 2\cos^2 x - 1$ ⁠. Add 1 to both sides and divide by 2:
$\cos^2 x = \frac{1 + \cos 2x}{2} = \frac12 + \frac12\cos 2x.$
The square of a cosine has no formula in the book's list, but a constant and $\cos 2x$ do. The identity trades the square for a doubled angle, which is exactly what Example 7(i), Section 7.3.2, does.

$\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1$ ⁠, using $\sin^2 x = 1 - \cos^2 x$ ⁠. Basics: trigonometric identities
Integrate term by term
$\begin{aligned}\int\cos^2 x\,dx &= \frac12\int dx + \frac12\int\cos 2x\,dx \\ &= \frac{x}{2} + \frac12\cdot\frac{\sin 2x}{2} + C \\ &= \frac{x}{2} + \frac{\sin 2x}{4} + C.\end{aligned}$
Take $F(x) = \frac{x}{2} + \frac{\sin 2x}{4}$ ⁠.

The integral of a sum is the sum of the integrals and constants come outside (Properties (III) and (IV), Section 7.2.1). For $\int\cos 2x\,dx$ ⁠, put $u = 2x$ ⁠, $du = 2\,dx$ ⁠: it becomes $\frac12\int\cos u\,du = \frac12\sin u = \frac12\sin 2x$ ⁠.

Standard formula (ii), Section 7.2: $\int\cos x\,dx = \sin x + C$ ⁠. Lesson: integration by substitution
Check: differentiate $F$ back
$\begin{aligned}F'(x) &= \frac12 + \frac{2\cos 2x}{4} = \frac{1 + \cos 2x}{2} \\ &= \frac{1 + (2\cos^2 x - 1)}{2} = \cos^2 x.\ \checkmark\end{aligned}$
Differentiating returns the integrand, so $F$ really is an antiderivative of $\cos^2 x$ for every real $x$ ⁠. The factor 2 from the chain rule cancels the 4 down to 2.

Property (I), Section 7.2.1: differentiation undoes integration. Basics: integration undoes differentiation
Evaluate $F\left(\frac{\pi}{2}\right) - F(0)$

At the upper limit, $2x = \pi$ and $\sin\pi = 0$ ⁠; at the lower limit, $\sin 0 = 0$ ⁠:
$\begin{aligned}I &= \left[\frac{\pi}{4} + \frac{\sin\pi}{4}\right] - \left[0 + \frac{\sin 0}{4}\right] \\ &= \left[\frac{\pi}{4} + 0\right] - [0 + 0] \\ &= \frac{\pi}{4} \approx 0.7853982.\end{aligned}$

The shaded region under $y = \cos^2 x$ from 0 to $\frac{\pi}{2}$ has area $\frac{\pi}{4}$ ⁠. It equals the area of a rectangle of height $\frac12$ on the same base: the part of the curve above the dashed line exactly fills the gap below it.

Theorem 2: value of the antiderivative at the upper limit minus its value at the lower limit. The sine term vanishes at both ends, so only $\frac{x}{2}$ contributes. The figure explains why: $\cos^2 x = \frac12 + \frac12\cos 2x$ ⁠, and $\cos 2x$ has equal areas above and below the axis on $\left[0, \frac{\pi}{2}\right]$ ⁠, so the average height is $\frac12$ and the area is $\frac12\cdot\frac{\pi}{2}$ ⁠.

$\sin\pi = 0$ ⁠, $\sin 0 = 0$ ⁠. Basics: area as a sum of thin strips

Result

\int_0^{\frac{\pi}{2}}\cos^2 x\,dx = \frac{\pi}{4} \approx 0.7853982.

(Textbook answer: $\frac{\pi}{4}$ ⁠.)

Check it with numbers

A numerical integration of $\cos^2 x$ over $\left[0, \frac{\pi}{2}\right]$ (no antiderivative used) gives about 0.7853982, and $\frac{\pi}{4} \approx 0.7853982$ ⁠. They agree.

Spot check of the identity at $x = 0.5$ ⁠: $\cos^2 0.5 \approx 0.7701512$ and $\frac{1 + \cos 1}{2} \approx 0.7701512$ ⁠.

Common mistakes

Writing $\int\cos^2 x\,dx = \frac{\cos^3 x}{3}$ ⁠. The power rule needs the derivative of the inside as a factor; differentiating $\frac{\cos^3 x}{3}$ gives $-\cos^2 x\sin x$ ⁠, not $\cos^2 x$ ⁠.
Using $\cos^2 x = \frac{1 - \cos 2x}{2}$ ⁠. That is $\sin^2 x$ ⁠. The sign is "+" for cosine.
Forgetting the $\frac12$ in $\int\cos 2x\,dx = \frac{\sin 2x}{2}$ ⁠. It would give $\frac{x}{2} + \frac{\sin 2x}{2}$ ⁠, whose derivative is $\frac12 + \cos 2x$ ⁠.

What to take away

Even powers of sine and cosine are integrated by lowering the power with a double-angle identity.
Over a quarter period, $\cos^2 x$ and $\sin^2 x$ both have average value $\frac12$ ⁠, so each integral over $\left[0, \frac{\pi}{2}\right]$ is $\frac{\pi}{4}$ ⁠.
Always differentiate the antiderivative back before substituting the limits.

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, 5	Theorem 2 and Remark (iv), Section 7.8.3 (Lesson)
$\cos^2 x = \frac{1 + \cos 2x}{2}$	Step 2	Example 7(i), Section 7.3.2 (Basics)
Sum and constant-multiple rules	Step 3	Properties (III), (IV), Section 7.2.1
$\int\cos x\,dx = \sin x + C$	Step 3	Standard formula (ii), Section 7.2
Substitution $u = 2x$	Step 3	Section 7.3.1 (Lesson)
Differentiating an integral returns the integrand	Step 4	Property (I), Section 7.2.1 (Basics)
Definite integral as area	Figure	Section 7.8.1 (Basics)

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Exercise 7.8 · Q12: ∫0π/2cos⁡2x dx\int_0^{\pi/2}\cos^2 x\,dx∫0π/2​cos2xdx

Basics you need

Solution

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

Lower the power with the double-angle identity

Integrate term by term

Check: differentiate FFF back

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

Every fact used, and where it comes from

Exercise 7.8 · Q12: $\int_0^{\pi/2}\cos^2 x\,dx$

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

Check: differentiate $F$ back

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