Chapter 7 · Integrals

Exercise 7.8 · Q17: $\int_0^{\pi/4}(2\sec^2 x + x^3 + 2)\,dx$

Exercise 7.8Fundamental theorem of calculus · Section 7.8

The question

Evaluate the definite integral

\int_0^{\frac{\pi}{4}}\left(2\sec^2 x + x^3 + 2\right)dx.

Exam answer

Let $I = \displaystyle\int_0^{\frac{\pi}{4}}\left(2\sec^2 x + x^3 + 2\right)dx$ ⁠.

\begin{aligned} I &= \left[2\tan x + \frac{x^4}{4} + 2x\right]_0^{\frac{\pi}{4}} \\ &= \left(2\tan\frac{\pi}{4} + \frac{\pi^4}{4\cdot 256} + \frac{2\pi}{4}\right) \\ &\quad - (2\tan 0 + 0 + 0) \end{aligned}

$\therefore\ I = \dfrac{\pi^4}{1024} + \dfrac{\pi}{2} + 2$ ⁠.

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

The one idea this problem is built on

Each of the three terms has a standard antiderivative: $\sec^2 x$ gives $\tan x$ ⁠, $x^3$ gives $\frac{x^4}{4}$ ⁠, and the constant 2 gives $2x$ ⁠. The only care needed is that $\sec^2 x$ is continuous on $\left[0, \frac{\pi}{4}\right]$ (it is, since $\cos x \ne 0$ there), and then Theorem 2 (Section 7.8.3) gives $F\left(\frac{\pi}{4}\right) - F(0)$ ⁠.

Given, to find, and the plan

Given: The integrand $f(x) = 2\sec^2 x + x^3 + 2$ on $\left[0, \frac{\pi}{4}\right]$ ⁠.
To find: The exact value of the integral, and a decimal.
Plan: (1) Check continuity, especially of $\sec^2 x$ ⁠. (2) Integrate term by term to get $F$ ⁠. (3) Differentiate $F$ back. (4) Compute $F\left(\frac{\pi}{4}\right) - F(0)$ ⁠.

Basics you need

1Second fundamental theorem

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

2 $\sec^2 x$ and $\tan x$

$\frac{d}{dx}\tan x = \sec^2 x$ ⁠, so $\int\sec^2 x\,dx = \tan x + C$ ⁠; $\sec x = \frac{1}{\cos x}$ is defined only where $\cos x \ne 0$ ⁠. Basics: trigonometric identities for integrals

3Power rule

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for $n \ne -1$ ⁠. Basics: powers of x and the power rule

Solution

Check continuity on $\left[0, \frac{\pi}{4}\right]$

On $\left[0, \frac{\pi}{4}\right]$ ⁠, $\cos x$ decreases from 1 to $\frac{1}{\sqrt2}$ ⁠, so $\cos x \ge \frac{1}{\sqrt2} > 0$ ⁠. Hence $\sec^2 x = \frac{1}{\cos^2 x}$ is defined and continuous there, with values between 1 and 2. The terms $x^3$ and 2 are polynomials. So the integrand is continuous on the whole interval.

$\sec^2 x$ blows up where $\cos x = 0$ ⁠, at $x = \frac{\pi}{2} + n\pi$ ⁠. The nearest such point, $\frac{\pi}{2}$ ⁠, lies outside $\left[0, \frac{\pi}{4}\right]$ ⁠, so Theorem 2 can be used.

$\cos\frac{\pi}{4} = \frac{1}{\sqrt2}$ ⁠, so $\sec^2\frac{\pi}{4} = 2$ ⁠. Basics: trigonometric identities
Integrate term by term
$\begin{aligned}\int(2\sec^2 x + x^3 + 2)\,dx &= 2\tan x + \frac{x^4}{4} + 2x + C.\end{aligned}$
Take $F(x) = 2\tan x + \frac{x^4}{4} + 2x$ ⁠.

The integral of a sum is the sum of the integrals, and constants come outside (Properties (III), (IV), Section 7.2.1). Then $\int\sec^2 x\,dx = \tan x$ ⁠, $\int x^3 dx = \frac{x^4}{4}$ and $\int 2\,dx = 2x$ ⁠.

Standard formulae (iv) and (i), Section 7.2. Basics: the power rule
Check: differentiate $F$ back
$F'(x) = 2\sec^2 x + \frac{4x^3}{4} + 2 = 2\sec^2 x + x^3 + 2.\ \checkmark$
Differentiating returns the integrand on every interval where $\cos x \ne 0$ ⁠, in particular on $\left[0, \frac{\pi}{4}\right]$ ⁠.

Property (I), Section 7.2.1. Basics: integration undoes differentiation
Evaluate $F\left(\frac{\pi}{4}\right) - F(0)$

At the upper limit: $\tan\frac{\pi}{4} = 1$ and $\left(\frac{\pi}{4}\right)^4 = \frac{\pi^4}{256}$ ⁠. At the lower limit everything is 0 ( $\tan 0 = 0$ ⁠). So
$\begin{aligned}I &= \left[2\cdot 1 + \frac{\pi^4}{4\cdot 256} + \frac{2\pi}{4}\right] - [0] \\ &= \frac{\pi^4}{1024} + \frac{\pi}{2} + 2.\end{aligned}$
In decimals: $\frac{\pi^4}{1024} \approx 0.0951261$ and $\frac{\pi}{2} \approx 1.5707963$ ⁠, so $I \approx 3.6659224$ ⁠.

The shaded region under $y = 2\sec^2 x + x^3 + 2$ from 0 to $\frac{\pi}{4}$ has area about 3.666. The curve rises from 4 to about 6.48, and the width is $\frac{\pi}{4} \approx 0.785$ ⁠, so the area lies between $4 \times 0.785$ and $6.48 \times 0.785$ ⁠.

The $x^4$ term is small (about 0.095) because $\frac{\pi}{4} < 1$ ⁠, so its fourth power is smaller still. The bound from the figure, between about 3.14 and 5.09, contains 3.666.

$4\cdot 256 = 1024$ ⁠; $\frac{2\pi}{4} = \frac{\pi}{2}$ ⁠. Basics: area as a sum of thin strips

Result

\int_0^{\frac{\pi}{4}}(2\sec^2 x + x^3 + 2)\,dx = \frac{\pi^4}{1024} + \frac{\pi}{2} + 2

$\approx 3.6659224$ ⁠. (Textbook answer: $\frac{\pi^4}{1024} + \frac{\pi}{2} + 2$ ⁠.)

Check it with numbers

A numerical integration of $2\sec^2 x + x^3 + 2$ over $\left[0, \frac{\pi}{4}\right]$ gives about 3.6659224, the same as the exact value to 7 decimal places.

Separately: $\int_0^{\pi/4}2\sec^2 x\,dx = 2$ ⁠, $\int_0^{\pi/4}x^3dx \approx 0.0951261$ ⁠, $\int_0^{\pi/4}2\,dx \approx 1.5707963$ ⁠.

Common mistakes

Integrating $\sec^2 x$ as $\frac{\sec^3 x}{3}$ ⁠. The power rule does not apply to a function of $x$ without its derivative; $\sec^2 x$ is the derivative of $\tan x$ ⁠.
Writing $\left(\frac{\pi}{4}\right)^4 = \frac{\pi^4}{16}$ ⁠. $4^4 = 256$ ⁠, and the extra $\frac14$ from $\frac{x^4}{4}$ makes 1024.
Ignoring continuity. Over an interval containing $\frac{\pi}{2}$ the same calculation would be meaningless, since $\sec^2 x$ is undefined there.

What to take away

For a sum, integrate each term with its own standard formula.
Before using Theorem 2 with $\tan$ ⁠, $\sec$ or a quotient, check that no point of the interval makes a denominator 0.
Rough area bounds (smallest height × width, largest height × width) catch big slips.

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	Theorem 2 and Remark (iv), Section 7.8.3 (Lesson)
$\int\sec^2 x\,dx = \tan x + C$	Step 2	Standard formula (iv), Section 7.2 (Basics)
$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$	Step 2	Standard formula (i), Section 7.2 (Basics)
Sum and constant-multiple rules	Step 2	Properties (III), (IV), Section 7.2.1
$\tan\frac{\pi}{4} = 1$ ⁠, $\cos\frac{\pi}{4} = \frac{1}{\sqrt2}$	Steps 1, 4	Trigonometric values (Basics)
Definite integral as area	Figure	Section 7.8.1 (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 · Q17: ∫0π/4(2sec⁡2x+x3+2) dx\int_0^{\pi/4}(2\sec^2 x + x^3 + 2)\,dx∫0π/4​(2sec2x+x3+2)dx

Basics you need

Solution

Check continuity on [0,π4]\left[0, \frac{\pi}{4}\right][0,4π​]

Integrate term by term

Check: differentiate FFF back

Evaluate F(π4)−F(0)F\left(\frac{\pi}{4}\right) - F(0)F(4π​)−F(0)

Every fact used, and where it comes from

Exercise 7.8 · Q17: $\int_0^{\pi/4}(2\sec^2 x + x^3 + 2)\,dx$

Check continuity on $\left[0, \frac{\pi}{4}\right]$

Check: differentiate $F$ back

Evaluate $F\left(\frac{\pi}{4}\right) - F(0)$