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Chapter 7 · Integrals

Exercise 7.10 · Q21 (MCQ): ∫0π/2log⁡4+3sin⁡x4+3cos⁡x dx\int_0^{\pi/2}\log\frac{4+3\sin x}{4+3\cos x}\,dx

Exercise 7.10Properties of definite integrals (P4) · Section 7.10
The question

Choose the correct answer in Exercises 20 and 21.

21. The value of ∫0π/2log⁡(4+3sin⁡x4+3cos⁡x)dx\displaystyle\int_0^{\pi/2}\log\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right)dx is

(A) 2    (B) 34\frac34    (C) 0    (D) −2-2

Exam answer

Let

I=∫0π/2log⁡4+3sin⁡x4+3cos⁡x dx  …(i)I = \int_0^{\pi/2}\log\frac{4 + 3\sin x}{4 + 3\cos x}\,dx\;\ldots(\text{i})

Using ∫0af(x) dx=∫0af(a−x) dx\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx (sin⁡x\sin x and cos⁡x\cos x swap),

I=∫0π/2log⁡4+3cos⁡x4+3sin⁡x dx  …(ii)I = \int_0^{\pi/2}\log\frac{4 + 3\cos x}{4 + 3\sin x}\,dx\;\ldots(\text{ii})

Adding (i) and (ii), the logarithms cancel:

2I=∫0π/2log⁡1 dx=02I = \int_0^{\pi/2}\log 1\,dx = 0

Answer: (C) 00

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

The one idea this problem is built on

On [0,π2][0, \frac{\pi}{2}]⁠, P4 replaces xx by π2−x\frac{\pi}{2} - x⁠, which swaps sin⁡x\sin x and cos⁡x\cos x⁠. That turns the fraction upside down, and the logarithm of an upside-down fraction is the negative of the original logarithm. So P4 gives I=−II = -I⁠, hence I=0I = 0⁠.

Given, to find, and the plan
Given
f(x)=log⁡4+3sin⁡x4+3cos⁡xf(x) = \log\dfrac{4 + 3\sin x}{4 + 3\cos x} on [0,π2][0, \frac{\pi}{2}]⁠.
To find
I=∫0π/2f(x) dxI = \displaystyle\int_0^{\pi/2} f(x)\,dx⁠, and which option it is.
Plan
(1) Check the logarithm is defined and continuous. (2) P4 with a=π2a = \frac{\pi}{2}⁠: f(π2−x)=−f(x)f(\frac{\pi}{2} - x) = -f(x)⁠. (3) Solve I=−II = -I⁠. (4) Examine every option.

Basics you need

1Property P4 (Section 7.10)

∫0af(x) dx=∫0af(a−x) dx\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx for ff continuous on [0,a][0, a]⁠. Lesson: properties of definite integrals

2Complementary angles

sin⁡(π2−x)=cos⁡x\sin\left(\frac{\pi}{2} - x\right) = \cos x and cos⁡(π2−x)=sin⁡x\cos\left(\frac{\pi}{2} - x\right) = \sin x⁠. Basics: trigonometric identities

3Logarithm of a reciprocal

For q>0q > 0⁠: log⁡1q=−log⁡q\log\frac{1}{q} = -\log q⁠. Basics: logarithms

Solution

  1. The integrand is defined and continuous

    On [0,π2][0, \frac{\pi}{2}]⁠, 0≤sin⁡x≤10 \le \sin x \le 1 and 0≤cos⁡x≤10 \le \cos x \le 1⁠, so

    4≤4+3sin⁡x≤7,4≤4+3cos⁡x≤7.\begin{gathered}4 \le 4 + 3\sin x \le 7, \\ 4 \le 4 + 3\cos x \le 7 .\end{gathered}

    The fraction is a quotient of positive continuous functions, so it is positive and continuous, and its logarithm is defined and continuous on the closed interval.

    A logarithm needs a positive argument; P4 needs a continuous function. Both are guaranteed here. At the ends, f(0)=log⁡47f(0) = \log\frac47 and f(π2)=log⁡74f(\frac{\pi}{2}) = \log\frac74⁠.
    log⁡\log is continuous on (0,∞)(0, \infty)⁠; compositions of continuous functions are continuous (Chapter 5, Section 5.2.1).

  2. P4 with a=π2a = \frac{\pi}{2}⁠: the fraction turns upside down

    Replace xx by π2−x\frac{\pi}{2} - x⁠; sine and cosine swap:

    f ⁣(π2−x)=log⁡4+3cos⁡x4+3sin⁡x=−log⁡4+3sin⁡x4+3cos⁡x=−f(x).\begin{aligned}f\!\left(\tfrac{\pi}{2} - x\right) &= \log\frac{4 + 3\cos x}{4 + 3\sin x} \\ &= -\log\frac{4 + 3\sin x}{4 + 3\cos x} = -f(x) .\end{aligned}

    So P4 gives

    I=∫0π/2f ⁣(π2−x)dx=−∫0π/2f(x) dx=−I.I = \int_0^{\pi/2} f\!\left(\tfrac{\pi}{2} - x\right)dx = -\int_0^{\pi/2} f(x)\,dx = -I .
    P4 holds for every continuous ff (Step 1). The relation f(π2−x)=−f(x)f(\frac{\pi}{2} - x) = -f(x) is also the condition of the second case of P6 (with 2a=π22a = \frac{\pi}{2}⁠), which gives 0 directly. Either way, the graph is unchanged by a half turn about (π4,0)(\frac{\pi}{4}, 0)⁠.
    log⁡pq=−log⁡qp\log\frac{p}{q} = -\log\frac{q}{p} for p,q>0p, q > 0⁠. Basics: logarithms

  3. Solve I=−II = -I

    2I=0,I=0.2I = 0, \qquad I = 0 .
    x y O π/4 π/2 0.5 −0.5 blue: y = log((4 + 3 sin x)/(4 + 3 cos x)) from log(4/7) at x = 0 to log(7/4) at x = π/2, through (π/4, 0) the rose and green regions are congruent: total 0
    The integrand on [0,π2][0, \frac{\pi}{2}] rises from log⁡47\log\frac47 to log⁡74\log\frac74 and crosses the axis at π4\frac{\pi}{4}⁠, where sin⁡x=cos⁡x\sin x = \cos x⁠. The rose region below the axis and the green region above it are swapped by a half turn about (π4,0)(\frac{\pi}{4}, 0)⁠, so they cancel.
    II is a finite number (continuous integrand on a closed interval), so adding II to both sides is legitimate. The integrand is negative on [0,π4)[0, \frac{\pi}{4}) (there sin⁡x<cos⁡x\sin x < \cos x⁠, so the fraction is less than 1) and positive on (π4,π2](\frac{\pi}{4}, \frac{\pi}{2}]⁠: equal and opposite areas.
    Basics: area as a sum of thin strips (signed area)

  4. Examine every option

    (A) 2 and (D) −2-2⁠. Step 2 shows I=−II = -I⁠; a number equal to its own negative must be 0, so neither 22 nor −2-2 is possible. Wrong.

    (B) 34\frac34⁠. This is the ratio of the coefficients 3 and 4 in the fraction, not an integral. It is also not 0, so it fails the same test. Wrong.

    (C) 0. Matches Step 3. Correct.

    The relation I=−II = -I eliminates every nonzero option at once, without any calculation of the integral.
    If I=−II = -I⁠, then 2I=02I = 0⁠, so I=0I = 0⁠.
Result

∫0π/2log⁡(4+3sin⁡x4+3cos⁡x)dx=0,\int_0^{\pi/2}\log\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right)dx = 0, option (C). (Textbook answer: C.)

Check it with numbers

Numerical integration over [0,π2][0, \frac{\pi}{2}] gives 0 (to more than 25 decimal places). Over [0,π4][0, \frac{\pi}{4}] alone it gives −0.216795-0.216795⁠, and over [π4,π2][\frac{\pi}{4}, \frac{\pi}{2}] it gives +0.216795+0.216795⁠.

The relation of Step 2 at x=0.3x = 0.3⁠: f(0.3)≈−0.340094f(0.3) \approx -0.340094 and f(π2−0.3)≈0.340094f(\frac{\pi}{2} - 0.3) \approx 0.340094⁠. The end values: log⁡47≈−0.559616\log\frac47 \approx -0.559616 and log⁡74≈0.559616\log\frac74 \approx 0.559616⁠.

Common mistakes
  1. Splitting the log as log⁡(4+3sin⁡x)\log(4 + 3\sin x) minus log⁡(4+3cos⁡x)\log(4 + 3\cos x) and trying to integrate each term. Neither can be integrated with the methods of this chapter; the symmetry is the way in.
  2. Writing log⁡qp=1log⁡pq\log\frac{q}{p} = \frac{1}{\log\frac{p}{q}}⁠. The log of a reciprocal is the negative, not the reciprocal, of the log.
  3. Guessing 34\frac34 from the numbers in the question.
What to take away
  1. On [0,π2][0, \frac{\pi}{2}]⁠, any integrand of the form log⁡ϕ(sin⁡x)ϕ(cos⁡x)\log\frac{\phi(\sin x)}{\phi(\cos x)} (with ϕ\phi positive and continuous) integrates to 0 by P4.
  2. This is the same mechanism as Q15: P4 turns the integrand into its own negative.

Every fact used, and where it comes from

FactUsed inSource
Continuity of log⁡\log and of compositionsStep 1Chapter 5, Section 5.2.1
Complementary anglesStep 2Class XI trigonometry
log⁡1q=−log⁡q\log\frac1q = -\log qStep 2Laws of logarithms
P4 (and P6, second case)Step 2Section 7.10, properties P4 and P6 (Lesson: properties)
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