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Chapter 7 · Integrals

Exercise 7.10 · Q20 (MCQ): ∫−π/2π/2(x3+xcos⁡x+tan⁡5x+1) dx\int_{-\pi/2}^{\pi/2}(x^3+x\cos x+\tan^5x+1)\,dx

Exercise 7.10Properties of definite integrals (P7) · Section 7.10
The question

Choose the correct answer in Exercises 20 and 21.

20. The value of ∫−π/2π/2(x3+xcos⁡x+tan⁡5x+1)dx\displaystyle\int_{-\pi/2}^{\pi/2}\left(x^3 + x\cos x + \tan^5 x + 1\right)dx is

(A) 0    (B) 2    (C) π\pi    (D) 1

Exam answer

Let h(x)=x3+xcos⁡x+tan⁡5xh(x) = x^3 + x\cos x + \tan^5 x⁠. Then h(−x)=−h(x)h(-x) = -h(x)⁠, so hh is odd.

As tan⁡5x\tan^5 x is unbounded at x=±π2x = \pm\frac{\pi}{2}⁠, the integral is understood as the limit over [−b,b][-b, b] as b→π2b \to \frac{\pi}{2}⁠. Using ∫−bbh(x) dx=0\displaystyle\int_{-b}^{b} h(x)\,dx = 0 (hh odd),

∫−bb(h(x)+1) dx=0+2b=2b\int_{-b}^{b}\big(h(x) + 1\big)\,dx = 0 + 2b = 2b

As b→π2b \to \frac{\pi}{2}⁠, 2b→π2b \to \pi⁠.

Answer: (C) π\pi

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. Of the four terms, x3x^3⁠, xcos⁡xx\cos x and tan⁡5x\tan^5 x are odd, so by P7 (ii) they contribute nothing; only the constant 1 is left, and ∫1 dx\int 1\,dx over an interval of length π\pi is π\pi⁠. One extra care point: tan⁡x\tan x is not defined at ±π2\pm\frac{\pi}{2}⁠, and read as an improper integral the given integral diverges. The value π\pi of the answer key is the limit of the integrals over the symmetric intervals [−b,b][-b, b] as b→π2b \to \frac{\pi}{2}⁠.

Given, to find, and the plan
Given
h(x)=x3+xcos⁡x+tan⁡5xh(x) = x^3 + x\cos x + \tan^5 x and the integrand h(x)+1h(x) + 1 on (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})⁠.
To find
The value of the integral, and which option it is.
Plan
(1) Show hh is odd. (2) Look at the endpoints, where tan⁡x\tan x is undefined. (3) Integrate over [−b,b][-b, b] with P7 (ii). (4) Let b→π2b \to \frac{\pi}{2}⁠. (5) Examine every option.

Basics you need

1Odd and even functions

Odd: f(−x)=−f(x)f(-x) = -f(x)⁠; even: f(−x)=f(x)f(-x) = f(x)⁠. xx⁠, sin⁡\sin⁠, tan⁡\tan are odd; cos⁡\cos and constants are even. Odd times even is odd; an odd power of an odd function is odd; a sum of odd functions is odd. Basics: symmetry and mirror images

2Property P7 (Section 7.10)

(ii) ∫−aaf(x) dx=0\int_{-a}^{a} f(x)\,dx = 0 if ff is odd. Lesson: properties of definite integrals

3Where tan⁡x\tan x is defined

tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x} needs cos⁡x≠0\cos x \ne 0⁠; on [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] that fails exactly at the two ends. Basics: trigonometric identities

Solution

  1. Split off the constant; the rest is odd

    Write the integrand as h(x)+1h(x) + 1 with h(x)=x3+xcos⁡x+tan⁡5xh(x) = x^3 + x\cos x + \tan^5 x⁠. For every xx where hh is defined:

    (−x)3=−x3,(−x)cos⁡(−x)=−xcos⁡x,tan⁡5(−x)=(−tan⁡x)5=−tan⁡5x.\begin{aligned}(-x)^3 &= -x^3,\\ (-x)\cos(-x) &= -x\cos x,\\ \tan^5(-x) &= (-\tan x)^5 = -\tan^5 x .\end{aligned}

    Adding: h(−x)=−h(x)h(-x) = -h(x)⁠, so hh is odd. The constant 1 is even, not odd.

    P7 (ii) needs the whole function being integrated to be odd. The full integrand h+1h + 1 is neither odd nor even (at x=0x = 0 it equals 1, but an odd function must vanish at 0), so we separate the odd part from the constant.
    cos⁡(−x)=cos⁡x\cos(-x) = \cos x (even), tan⁡(−x)=−tan⁡x\tan(-x) = -\tan x (odd), and (−1)5=−1(-1)^5 = -1⁠. Basics: symmetry

  2. The endpoints: what the integral means

    At x=±π2x = \pm\frac{\pi}{2}⁠, cos⁡x=0\cos x = 0⁠, so tan⁡x\tan x is undefined, and tan⁡5x→±∞\tan^5 x \to \pm\infty as x→±π2x \to \pm\frac{\pi}{2}⁠. The integrand is not continuous on the closed interval, so the integral is not an ordinary one (Remark (iv) after Theorem 2).

    In fact each half is infinite. An antiderivative of tan⁡5x\tan^5 x on (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) is

    T(x)=tan⁡4x4−tan⁡2x2−log⁡cos⁡xT(x) = \frac{\tan^4 x}{4} - \frac{\tan^2 x}{2} - \log\cos x

    (check: T′(x)=tan⁡3xsec⁡2xT'(x) = \tan^3 x\sec^2 x −tan⁡xsec⁡2x+tan⁡x- \tan x\sec^2 x + \tan x⁠, which is tan⁡5x\tan^5 x⁠, using sec⁡2x=1+tan⁡2x\sec^2 x = 1 + \tan^2 x⁠), and T(b)→∞T(b) \to \infty as b→π2−b \to \frac{\pi}{2}^-⁠. So ∫0π/2tan⁡5x dx\int_0^{\pi/2}\tan^5 x\,dx diverges, and so does ∫−π/20tan⁡5x dx\int_{-\pi/2}^{0}\tan^5 x\,dx (to −∞-\infty⁠). Read as an improper integral, the given integral therefore does not exist. What does exist is the limit of the integral over the symmetric intervals [−b,b][-b, b] as b→π2−b \to \frac{\pi}{2}^-⁠, called the principal value. This is the value that P7, the property this exercise is about, leads to, and the answer key's option (C) is this value. Below we compute it.

    The book does not mention this, but a careful student will notice that tan⁡x\tan x does not exist at the limits of integration. Over [−b,b][-b, b] with b<π2b < \frac{\pi}{2} everything is continuous, so P7 applies there without any doubt; the huge positive and negative parts near the two ends cancel exactly for every bb⁠.
    ddxtan⁡x=sec⁡2x\frac{d}{dx}\tan x = \sec^2 x⁠, ddxlog⁡cos⁡x=−tan⁡x\frac{d}{dx}\log\cos x = -\tan x⁠, sec⁡2x=1+tan⁡2x\sec^2 x = 1 + \tan^2 x⁠. Basics: integration undoes differentiation

  3. Integrate over [−b,b][-b, b] with P7 (ii)

    Fix bb with 0<b<π20 < b < \frac{\pi}{2}⁠. On [−b,b][-b, b]⁠, hh is continuous and odd (Step 1), so P7 (ii) with a=ba = b gives ∫−bbh(x) dx=0\int_{-b}^{b} h(x)\,dx = 0⁠. Therefore

    ∫−bb(h(x)+1) dx=0+∫−bb1 dx=b−(−b)=2b.\begin{aligned}\int_{-b}^{b}\big(h(x) + 1\big)\,dx &= 0 + \int_{-b}^{b}1\,dx \\ &= b - (-b) = 2b .\end{aligned}
    x y −π/2 π/2 2 3 −1 −2 −3 blue: odd part y = x³ + x cos x + tan⁵x (cut at y = ±3) on each interval [−b, b] its green and rose parts cancel gold: y = 1, a rectangle of width π and area π
    The odd part h(x)h(x) (blue, drawn while ∣h∣≤3|h| \le 3⁠) has mirror-image regions above and below the axis, which cancel on every symmetric interval [−b,b][-b, b]⁠. The constant 1 contributes the gold rectangle, of width 2b2b⁠, which tends to π\pi⁠.
    Splitting ∫(h+1)\int (h + 1) into ∫h+∫1\int h + \int 1 is the sum property of integrals. The constant part is a rectangle of height 1 and width 2b2b⁠.
    P7 (ii), Section 7.10. Lesson: properties

  4. Let b→π2b \to \frac{\pi}{2}

    lim⁡b→π/2−2b=π.\lim_{b \to \pi/2^-} 2b = \pi .

    So the value of the integral (in the sense of Step 2) is π\pi⁠.

    If the term tan⁡5x\tan^5 x were absent, the integrand x3+xcos⁡x+1x^3 + x\cos x + 1 would be continuous on the closed interval, and P7 (ii) would give π\pi directly. The term tan⁡5x\tan^5 x changes nothing in the symmetric limit, because it cancels exactly on every [−b,b][-b, b]⁠.
    2b2b depends continuously on bb⁠; at b=π2b = \frac{\pi}{2} it is π\pi⁠.

  5. Examine every option

    (A) 0. This is what you get by treating the whole integrand as odd and applying P7 (ii) to all of it. But the constant 1 is even (Step 1), and its integral is π\pi⁠, not 0. Wrong.

    (B) 2. This is ∫−π/2π/2cos⁡x dx\int_{-\pi/2}^{\pi/2}\cos x\,dx⁠. It comes from confusing xcos⁡xx\cos x (odd) with cos⁡x\cos x (even); the term xcos⁡xx\cos x contributes 0. Wrong.

    (C) π\pi⁠. Matches Steps 3 and 4. Correct.

    (D) 1. This is the integrand's constant term, not its integral; integrating 1 over an interval of length π\pi gives π\pi⁠. Wrong.

    An MCQ is safest when each wrong option is matched to the slip that produces it; then we know our answer is not one of those slips.
    ∫−π/2π/2cos⁡x dx=sin⁡π2−sin⁡(−π2)=2\int_{-\pi/2}^{\pi/2}\cos x\,dx = \sin\frac{\pi}{2} - \sin\left(-\frac{\pi}{2}\right) = 2⁠.
Result

∫−π/2π/2(x3+xcos⁡x+tan⁡5x+1)dx=π≈3.141593,\begin{aligned}&\int_{-\pi/2}^{\pi/2}\left(x^3 + x\cos x + \tan^5 x + 1\right)dx \\ &= \pi \approx 3.141593,\end{aligned} understood as the limit over [−b,b][-b, b] as b→π2b \to \frac{\pi}{2}⁠. This is option (C). (Textbook answer: C.)

Check it with numbers

Numerical integration over [−b,b][-b, b]⁠: for b=1b = 1 the value is 2.0000002.000000⁠, for b=1.5b = 1.5 it is 3.0000003.000000⁠, and for b=1.57b = 1.57 it is 3.1400003.140000⁠; each equals 2b2b⁠, and 2b2b approaches π\pi⁠.

The divergence of each half: ∫01.5tan⁡5x dx≈9788.56\int_0^{1.5}\tan^5 x\,dx \approx 9788.56 already, and it grows without bound as the upper limit approaches π2\frac{\pi}{2}⁠. Without tan⁡5x\tan^5 x⁠, ∫0π/2(x3+xcos⁡x) dx≈2.092813\int_0^{\pi/2}(x^3 + x\cos x)\,dx \approx 2.092813⁠, and the left half gives −2.092813-2.092813⁠.

Common mistakes
  1. Calling the whole integrand odd and answering 0. A constant is even.
  2. Thinking xcos⁡xx\cos x is even because cos⁡x\cos x is. Odd times even is odd.
  3. Integrating tan⁡5x\tan^5 x separately over each half. Each half diverges; only the symmetric combination has a value.
What to take away
  1. Over [−a,a][-a, a]⁠, split the integrand into its odd and even parts: the odd part contributes 0 (P7 (ii)), and only the even part needs work.
  2. Always check that the integrand is defined at the limits; when it is not, say what the integral means. Compare Q13 and Example 31, where the odd integrands are continuous on the whole interval.

Every fact used, and where it comes from

FactUsed inSource
sin⁡\sin⁠, tan⁡\tan odd; cos⁡\cos evenStep 1Class XI trigonometry
A definite integral needs ff continuous on [a,b][a, b]Step 2Remark (iv) after Theorem 2, Section 7.8.3
Derivatives of tan⁡x\tan x and log⁡cos⁡x\log\cos xStep 2Chapter 5, Sections 5.3 and 5.4
P7 (ii): ∫−aaf(x) dx=0\int_{-a}^{a} f(x)\,dx = 0 for odd ffStep 3Section 7.10, property P7 (Lesson: properties)
Integral of a sum; ∫−bb1 dx=2b\int_{-b}^{b}1\,dx = 2bStep 3Section 7.2.1, Property (III); Theorem 2, Section 7.8.3
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