Chapter 7 · Integrals

Exercise 7.3 · Q1: $\int\sin^2(2x+5)\,dx$

Exercise 7.3Integration using trigonometric identities · Section 7.3.2

The question

Find the integral of the function

\sin^2(2x+5).

Exam answer

Let $I = \displaystyle\int \sin^2(2x + 5)\,dx$ ⁠.

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

\begin{aligned} I &= \frac12\int\big(1 - \cos(4x + 10)\big)\,dx \\ &= \frac12\left(x - \frac{\sin(4x + 10)}{4}\right) + C \end{aligned}

$\therefore\ I = \dfrac{x}{2} - \dfrac18\sin(4x + 10) + C$ ⁠.

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 the square of a sine. But a double-angle identity rewrites a square of a sine as a constant minus a cosine with no square: $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ ⁠. After that "power-lowering" step, both pieces are standard integrals.

Given, to find, and the plan

Given: The function $\sin^2(2x+5)$ ⁠, defined for every real $x$ ⁠.
To find: $\displaystyle\int \sin^2(2x+5)\,dx$ ⁠, the family of all antiderivatives.
Plan: (1) Lower the power with $\sin^2\theta = \frac{1-\cos 2\theta}{2}$ ⁠, taking $\theta = 2x+5$ ⁠. (2) Split into two integrals. (3) Integrate the constant and the cosine (substitution $t = 4x+10$ ⁠). (4) Assemble the answer. (5) Differentiate it back to check.

Basics you need

1The power-lowering identity for $\sin^2$

From $\cos 2\theta = 1 - 2\sin^2\theta$ (a form of the double-angle formula) we get $\sin^2\theta = \frac{1-\cos 2\theta}{2}$ ⁠, true for every real $\theta$ ⁠. Basics: trigonometric identities for integrals

2Cosine of a linear expression

$\int \cos(ax+b)\,dx = \frac{1}{a}\sin(ax+b) + C$ for $a \ne 0$ ⁠. It comes from $\int\cos t\,dt = \sin t + C$ with the substitution $t = ax+b$ ⁠, $dt = a\,dx$ (Section 7.3.1; Example 5(i) does the same for $\sin mx$ ⁠). Basics: substitution

3Integrals split over sums and constants

$\int [f + g]\,dx = \int f\,dx + \int g\,dx$ and $\int k f\,dx = k\int f\,dx$ (Properties (III) and (IV), Section 7.2.1). Lesson: integration as anti-differentiation

4Checking an integral

$F(x) + C$ is right exactly when $F'(x)$ equals the integrand (Property (I), Section 7.2.1). Basics: integration undoes differentiation

Solution

Lower the power with the double-angle identity

Start from $\cos 2\theta = 1 - 2\sin^2\theta$ and solve for $\sin^2\theta$ ⁠:
$\sin^2\theta = \frac{1 - \cos 2\theta}{2}.$
Take $\theta = 2x + 5$ ⁠. Then $2\theta = 2(2x+5) = 4x + 10$ ⁠, so
$\sin^2(2x+5) = \frac{1 - \cos(4x+10)}{2}.$
The identity holds for every real angle $\theta$ ⁠, so we may put any expression in place of $\theta$ ⁠, including $2x+5$ ⁠. The point of the rewrite: $\sin^2$ has no formula, but a constant and a plain cosine do. The price is that the angle doubles, so be careful to double the whole angle: $4x + 10$ ⁠, not $4x + 5$ ⁠.

The book uses the twin identity $\cos^2 x = \frac{1+\cos 2x}{2}$ in Example 7(i), Section 7.3.2. For $\sin^2$ the sign in front of $\cos 2\theta$ is minus. Basics: trigonometric identities
Split the integral
$\begin{aligned} \int \sin^2(2x+5)\,dx &= \int \frac{1 - \cos(4x+10)}{2}\,dx \\ &= \frac12\int dx - \frac12\int \cos(4x+10)\,dx. \end{aligned}$
The constant $\frac12$ comes out of the integral and the integral of a difference is the difference of the integrals. Each piece is now something we know how to integrate.

Properties (III) and (IV), Section 7.2.1 (sum rule and constant multiple rule). Lesson: integration as anti-differentiation
Integrate each piece

The first piece is $\int dx = x$ ⁠.

For the second, substitute $t = 4x + 10$ ⁠. Then $dt = 4\,dx$ ⁠, so $dx = \frac{dt}{4}$ ⁠, and
$\begin{aligned} \int \cos(4x+10)\,dx &= \frac14\int \cos t\,dt \\ &= \frac14\sin t = \frac14\sin(4x+10). \end{aligned}$
(We add one constant $C$ at the end instead of one per piece.)

$\int\cos t\,dt = \sin t$ is a standard formula, but our cosine has the angle $4x+10$ ⁠, not $x$ ⁠. The substitution makes the angle the new variable. Since $t$ changes 4 times as fast as $x$ ⁠, we must divide by 4. Forgetting that factor is the commonest slip in this exercise.

Standard formula $\int\cos x\,dx = \sin x + C$ (list in Section 7.2, from $\frac{d}{dx}\sin x = \cos x$ ⁠); change of variable, Section 7.3.1. Basics: substitution
Assemble the answer
$\begin{aligned} \int \sin^2(2x+5)\,dx &= \frac12\,x - \frac12\cdot\frac14\sin(4x+10) + C \\ &= \frac{x}{2} - \frac18\sin(4x+10) + C. \end{aligned}$
The constants from the two pieces combine into one arbitrary constant $C$ ⁠. The term $\frac{x}{2}$ has a nice meaning: $\sin^2(2x+5)$ swings between 0 and 1 and its average value is $\frac12$ ⁠, so its integral grows like $\frac{x}{2}$ plus a small wobble (see the figure).
Check: differentiate the answer back

Let $F(x) = \frac{x}{2} - \frac18\sin(4x+10)$ ⁠. By the chain rule (outer: $\sin u$ ⁠; inner: $u = 4x+10$ ⁠, $u' = 4$ ⁠):
$\begin{aligned} F'(x) &= \frac12 - \frac18\cdot 4\cos(4x+10) \\ &= \frac{1 - \cos(4x+10)}{2} \\ &= \sin^2(2x+5). \end{aligned}$
The last line is Step 1 read backwards. $F'$ is the integrand, so the answer is correct, and it is valid for every real $x$ ⁠.

The factor 4 from the chain rule cancels the $\frac14$ we introduced in Step 3. Seeing that cancellation is a quick sign that the "divide by the coefficient of $x$ " step was done right.

Chain rule, Theorem 4, Section 5.3.1. Basics: integration undoes differentiation

y = \sin^2(2x+5)

0 \le x \le \pi

(two full periods, each of length

\frac{\pi}{2}

⁠). The curve swings between 0 and 1, balanced around the dashed line

y = \frac12

⁠. That average value is why the answer starts with

\frac{x}{2}

⁠; the term

-\frac18\sin(4x+10)

is the small periodic wobble.

Result

\int \sin^2(2x+5)\,dx = \frac{x}{2} - \frac18\sin(4x+10) + C,

valid for all real $x$ ⁠. (Textbook answer: $\dfrac{x}{2} - \dfrac18\sin(4x+10) + C$ ⁠.)

Check it with numbers

Take $x = 1$ and $h = 0.001$ ⁠, with $F(x) = \frac{x}{2} - \frac18\sin(4x+10)$ ⁠:

$F(1.001) \approx 0.376606702717$ and $F(0.999) \approx 0.375743439571$ ⁠.
Slope estimate: $\dfrac{F(1.001) - F(0.999)}{0.002}$ $\approx 0.4316316$ ⁠.
Integrand: $\sin^2 7 \approx 0.4316314$ ⁠.

They agree to five significant figures (the tiny gap is the error of the difference quotient).

Common mistakes

Using the power rule on $\sin^2$ ⁠. $\int\sin^2 u\,du$ is not $\frac{\sin^3 u}{3}$ ⁠: the power rule needs the derivative $\cos u$ as a factor, and it is not there.
Doubling only part of the angle. $2(2x+5) = 4x+10$ ⁠, not $4x+5$ ⁠.
Forgetting to divide by 4. $\int\cos(4x+10)\,dx = \frac14\sin(4x+10)$ ⁠, not $\sin(4x+10)$ ⁠.
Mixing up the signs. $\sin^2\theta = \frac{1-\cos 2\theta}{2}$ but $\cos^2\theta = \frac{1+\cos 2\theta}{2}$ ⁠.

What to take away

Even powers of sine or cosine: lower the power with the double-angle identities. Questions 10 and 11 repeat this twice for fourth powers.
$\int\cos(ax+b)\,dx = \frac1a\sin(ax+b) + C$ and $\int\sin(ax+b)\,dx = -\frac1a\cos(ax+b) + C$ ⁠.
An identity in $\theta$ holds with any expression in place of $\theta$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\cos 2\theta = 1 - 2\sin^2\theta$ ⁠, so $\sin^2\theta = \frac{1-\cos 2\theta}{2}$	Steps 1, 5	Double-angle formula (Class XI trigonometry); twin form used in Example 7(i), Section 7.3.2
Sum and constant-multiple rules	Step 2	Properties (III), (IV), Section 7.2.1
$\int\cos x\,dx = \sin x + C$	Step 3	Standard formulae, Section 7.2
Substitution $t = 4x+10$	Step 3	Section 7.3.1 (Lesson: integration by substitution)
Chain rule	Step 5	Theorem 4, Section 5.3.1
Differentiation undoes integration	Step 5	Property (I), Section 7.2.1

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 7.3 · Q1: ∫sin⁡2(2x+5) dx\int\sin^2(2x+5)\,dx∫sin2(2x+5)dx

Basics you need

Solution

Lower the power with the double-angle identity

Split the integral

Integrate each piece

Assemble the answer

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.3 · Q1: $\int\sin^2(2x+5)\,dx$