Chapter 7 · Integrals

Example 29: $\int_{-\pi/4}^{\pi/4}\sin^2x\,dx$

ExamplesProperties of definite integrals · Section 7.10

The question

Evaluate $\displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x\; dx$ ⁠.

Exam answer

Let $I = \displaystyle\int_{-\pi/4}^{\pi/4}\sin^2 x\,dx$ ⁠.

Let $f(x) = \sin^2 x$ ⁠. Then $f(-x) = \sin^2(-x)$ $= \sin^2 x = f(x)$ ⁠, so $f$ is even.

Using $\displaystyle\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$ for even $f$ ⁠,

I = 2\int_0^{\pi/4}\sin^2 x\,dx

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

\begin{aligned} I &= 2\int_0^{\pi/4}\frac{1 - \cos 2x}{2}\,dx \\ &= \Big[x - \frac12\sin 2x\Big]_0^{\pi/4} \\ &= \left(\frac{\pi}{4} - \frac12\sin\frac{\pi}{2}\right) - 0 \end{aligned}

$\therefore\ I = \dfrac{\pi}{4} - \dfrac12$ ⁠.

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

The one idea this problem is built on

The interval is symmetric about 0 and $\sin^2 x$ is an even function: its graph is a mirror image across the y-axis. Property P7 (i) says the area on the left equals the area on the right, so $\int_{-a}^{a} f = 2\int_0^a f$ ⁠. The remaining integral is done with the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$ ⁠, which replaces a square by something we can integrate directly.

Given, to find, and the plan

Given: $f(x) = \sin^2 x$ on $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$ ⁠.
To find: $\int_{-\pi/4}^{\pi/4}\sin^2 x\,dx$ ⁠.
Plan: Show $f$ is even; apply P7 (i); rewrite $\sin^2 x$ by the double-angle identity; integrate and evaluate; check without using symmetry.

Basics you need

1Even functions

$f$ is even when $f(-x) = f(x)$ for every $x$ ⁠; its graph is symmetric about the y-axis. Basics: symmetry and mirror images

2Property P7 (i)

$\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$ if $f$ is even (Section 7.10). Lesson: properties of definite integrals

3Double-angle identity

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

4 $\int\cos 2x\,dx$

$\int\cos 2x\,dx = \frac12\sin 2x + C$ ⁠, because $\frac{d}{dx}\sin 2x = 2\cos 2x$ ⁠. Basics: substitution

Solution

$\sin^2 x$ is an even function

For every real $x$ ⁠, $\sin(-x) = -\sin x$ ⁠, so
$f(-x) = \sin^2(-x) = (-\sin x)^2 = \sin^2 x = f(x).$
Also $f$ is continuous everywhere, in particular on $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$ ⁠.

P7 (i) has one condition, $f(-x) = f(x)$ ⁠, and it must hold for all $x$ in the interval. Squaring removes the sign that the odd function $\sin$ introduces.

$\sin$ is odd: $\sin(-x) = -\sin x$ ⁠. An odd function squared is even. Basics: symmetry
Apply P7 (i) with $a = \frac{\pi}{4}$
$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2 x\,dx = 2\int_0^{\frac{\pi}{4}}\sin^2 x\,dx.$
The interval $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$ has the form $[-a, a]$ with $a = \frac{\pi}{4}$ ⁠, and $f$ is even (Step 1), so both conditions of P7 (i) hold. In the figure the gold half (left) is the mirror image of the blue half (right).

P7 (i), Section 7.10. Its proof splits at 0 by P2 and substitutes $t = -x$ in the left part. Lesson: properties of definite integrals
Replace $\sin^2 x$ using the double angle
$2\int_0^{\frac{\pi}{4}}\sin^2 x\,dx = 2\int_0^{\frac{\pi}{4}}\frac{1 - \cos 2x}{2}\,dx = \int_0^{\frac{\pi}{4}}(1 - \cos 2x)\,dx.$
There is no standard formula for $\int\sin^2 x\,dx$ ⁠, but there is one for $\int 1\,dx$ and for $\int\cos 2x\,dx$ ⁠. The identity trades the square for a cosine of the double angle, and the 2 in front cancels the $\frac12$ ⁠.

From $\cos 2x = 1 - 2\sin^2 x$ ⁠: $2\sin^2 x = 1 - \cos 2x$ (Section 7.3.2 uses the same idea for $\cos^2 x$ ⁠). Basics: trigonometric identities
Integrate and evaluate

An antiderivative of $1 - \cos 2x$ is $F(x) = x - \frac12\sin 2x$ ⁠. Check: $F'(x) = 1 - \frac12\cdot 2\cos 2x = 1 - \cos 2x$ ⁠. Then
$\begin{aligned}\int_0^{\frac{\pi}{4}}(1 - \cos 2x)\,dx &= \left[x - \tfrac12\sin 2x\right]_0^{\frac{\pi}{4}} \\ &= \left(\tfrac{\pi}{4} - \tfrac12\sin\tfrac{\pi}{2}\right) - 0 \\ &= \frac{\pi}{4} - \frac12.\end{aligned}$
In decimals $\frac{\pi}{4} - \frac12 \approx 0.2853982$ ⁠.

At the upper limit, $2x = \frac{\pi}{2}$ and $\sin\frac{\pi}{2} = 1$ ⁠. At the lower limit both terms are 0.

Theorem 2, Section 7.8.3: F(upper) minus F(lower). Lesson: the fundamental theorem
The picture, and a check without symmetry

$y = \sin^2 x$ is symmetric about the y-axis. The shaded region from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$ consists of two mirror-image halves, each of area $\frac{\pi - 2}{8} \approx 0.143$ ⁠.

Without P7, use the antiderivative $\frac{x}{2} - \frac{\sin 2x}{4}$ of $\sin^2 x$ over the whole interval:
$\begin{aligned}&\left(\tfrac{\pi}{8} - \tfrac14\right) - \left(-\tfrac{\pi}{8} + \tfrac14\right) \\ &= \tfrac{\pi}{4} - \tfrac12.\end{aligned}$
Both routes must agree. The symmetry route is shorter here, but the direct route confirms that the factor 2 from P7 was handled correctly. (At $x = -\frac{\pi}{4}$ ⁠, $\sin 2x = \sin\left(-\frac{\pi}{2}\right) = -1$ ⁠.)

$\frac{d}{dx}\left(\frac{x}{2} - \frac{\sin 2x}{4}\right) = \frac12 - \frac{\cos 2x}{2} = \sin^2 x$ ⁠. Basics: area as a sum of thin strips

Result

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2 x\,dx = \frac{\pi}{4} - \frac12 \approx 0.2853982,$ the same as the book's answer.

Check it with numbers

A numerical integration of $\sin^2 x$ over $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$ gives 0.2853982, and $\frac{\pi}{4} - \frac12$ is $0.7853982 - 0.5 = 0.2853982$ ⁠. The value is sensible: $\sin^2 x$ rises from 0 to $\frac12$ on each half, and the interval has length $\frac{\pi}{2} \approx 1.571$ ⁠, so the area must be well below $\frac12\times 1.571 \approx 0.785$ ⁠.

Common mistakes

Forgetting the factor 2 from P7 (i) (or forgetting that it cancels the $\frac12$ of the identity).
$\int\cos 2x\,dx = \sin 2x$ ⁠. The chain rule needs a $\frac12$ ⁠: $\frac12\sin 2x$ ⁠.
Using P7 (ii) instead. $\sin^2 x$ is even, not odd; the answer is not 0 (it is the area of a region above the axis).
Mixing up the identity: $\sin^2 x = \frac{1 - \cos 2x}{2}$ ⁠, while $\cos^2 x = \frac{1 + \cos 2x}{2}$ ⁠.

What to take away

On a symmetric interval $[-a, a]$ ⁠, first test whether the integrand is even or odd: P7 halves the work or ends it.
Squares of sine and cosine are integrated through the double-angle identities.
Always check a symmetry shortcut once by a direct computation when you are learning it.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin(-x) = -\sin x$ ⁠, so $\sin^2$ is even	Step 1	Trigonometry (Basics)
P7 (i): $\int_{-a}^a f = 2\int_0^a f$ for even $f$	Step 2	Section 7.10 (Lesson)
$\sin^2 x = \frac{1 - \cos 2x}{2}$	Step 3	Double-angle formula (Basics); Section 7.3.2
$\int\cos 2x\,dx = \frac12\sin 2x + C$	Step 4	Standard formula (ii), Section 7.2, with substitution
$\int_a^b f = F(b) - F(a)$	Steps 4 and 5	Theorem 2, Section 7.8.3 (Lesson)

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Example 29: ∫−π/4π/4sin⁡2x dx\int_{-\pi/4}^{\pi/4}\sin^2x\,dx∫−π/4π/4​sin2xdx

Basics you need

Solution

sin⁡2x\sin^2 xsin2x is an even function

Apply P7 (i) with a=π4a = \frac{\pi}{4}a=4π​

Replace sin⁡2x\sin^2 xsin2x using the double angle

Integrate and evaluate

The picture, and a check without symmetry

Every fact used, and where it comes from

Example 29: $\int_{-\pi/4}^{\pi/4}\sin^2x\,dx$

$\sin^2 x$ is an even function

Apply P7 (i) with $a = \frac{\pi}{4}$

Replace $\sin^2 x$ using the double angle