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Chapter 7 · Integrals

Exercise 7.5 · Q17: ∫cos⁡x dx(1−sin⁡x)(2−sin⁡x)\int\frac{\cos x\,dx}{(1 - \sin x)(2 - \sin x)}

Exercise 7.5Substitution and partial fractions · Section 7.5
The question

Integrate the function

cos⁡x(1−sin⁡x)(2−sin⁡x)\frac{\cos x}{(1 - \sin x)(2 - \sin x)}

[Hint: Put sin⁡x=t\sin x = t⁠.]

Exam answer

Let I=∫cos⁡x dx(1−sin⁡x)(2−sin⁡x)I = \displaystyle\int\frac{\cos x\,dx}{(1 - \sin x)(2 - \sin x)}⁠.

Put sin⁡x=t\sin x = t⁠. Then cos⁡x dx=dt\cos x\,dx = dt⁠.

I=∫dt(1−t)(2−t)I = \int\frac{dt}{(1 - t)(2 - t)}

Let

1(1−t)(2−t)=A1−t+B2−t\frac{1}{(1 - t)(2 - t)} = \frac{A}{1 - t} + \frac{B}{2 - t} ⇒ 1=A(2−t)+B(1−t)\Rightarrow\ 1 = A(2 - t) + B(1 - t)

Putting t=1t = 1⁠: A=1A = 1⁠. Putting t=2t = 2⁠: B=−1B = -1⁠.

∴ I=∫dt1−t−∫dt2−t=−log⁡∣1−t∣+log⁡∣2−t∣+C=log⁡∣2−t1−t∣+C\begin{aligned} \therefore\ I &= \int\frac{dt}{1 - t} - \int\frac{dt}{2 - t} \\ &= -\log|1 - t| \\ &\quad + \log|2 - t| + C \\ &= \log\left|\frac{2 - t}{1 - t}\right| + C \end{aligned}

∴ I=log⁡∣2−sin⁡x1−sin⁡x∣+C\therefore\ I = \log\left|\dfrac{2 - \sin x}{1 - \sin x}\right| + C⁠.

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

The one idea this problem is built on

The numerator cos⁡x\cos x is the derivative of sin⁡x\sin x⁠, and the denominator is built only from sin⁡x\sin x⁠. So t=sin⁡xt = \sin x turns the integral into ∫dt(1−t)(2−t)\int\frac{dt}{(1 - t)(2 - t)}⁠, a rational function with two different linear factors, which splits by Table 7.2 (row 1).

Given, to find, and the plan
Given
The integrand cos⁡x(1−sin⁡x)(2−sin⁡x)\frac{\cos x}{(1 - \sin x)(2 - \sin x)}⁠. Since sin⁡x≤1<2\sin x \le 1 < 2⁠, the factor 2−sin⁡x2 - \sin x is never 0; the integrand is defined except where sin⁡x=1\sin x = 1⁠, that is, x≠2kπ+π2x \ne 2k\pi + \frac{\pi}{2} (kk an integer).
To find
∫cos⁡x dx(1−sin⁡x)(2−sin⁡x)\int\frac{\cos x\,dx}{(1 - \sin x)(2 - \sin x)}⁠.
Plan
(1) Put t=sin⁡xt = \sin x⁠. (2) Split 1(1−t)(2−t)\frac{1}{(1 - t)(2 - t)}⁠. (3) Integrate, watching signs. (4) Return to xx and simplify the moduli. (5) Differentiate back.

Basics you need

1Substitution

With t=sin⁡xt = \sin x⁠, dt=cos⁡x dxdt = \cos x\,dx (since ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x⁠). Basics: substitution

2Table 7.2, row 1

1(t−a)(t−b)=At−a+Bt−b\frac{1}{(t - a)(t - b)} = \frac{A}{t - a} + \frac{B}{t - b}⁠, a≠ba \ne b⁠. Lesson: partial fractions

3∫dtk−t\int\frac{dt}{k - t}

Put u=k−tu = k - t⁠, du=−dtdu = -dt⁠: ∫dtk−t=−log⁡∣k−t∣+C\int\frac{dt}{k - t} = -\log|k - t| + C⁠. Basics: logarithms and 1/x

4Range of sine

−1≤sin⁡x≤1-1 \le \sin x \le 1 for every real xx⁠. Basics: trigonometric identities

Solution

  1. Substitute t=sin⁡xt = \sin x

    Then dt=cos⁡x dxdt = \cos x\,dx⁠, and

    ∫cos⁡x dx(1−sin⁡x)(2−sin⁡x)=∫dt(1−t)(2−t),\int\frac{\cos x\,dx}{(1 - \sin x)(2 - \sin x)} = \int\frac{dt}{(1 - t)(2 - t)},

    with −1≤t<1-1 \le t < 1 on the domain.

    The product cos⁡x dx\cos x\,dx is exactly dtdt⁠, and every other appearance of xx is through sin⁡x=t\sin x = t⁠. So the whole integrand is rewritten in tt⁠, as the hint suggests. Since sin⁡x≠1\sin x \ne 1 on the domain, t≠1t \ne 1⁠.
    Integration by substitution, Section 7.3.1. Lesson: substitution

  2. Split into partial fractions

    1(1−t)(2−t)=A1−t+B2−t,1=A(2−t)+B(1−t).\begin{gathered} \frac{1}{(1 - t)(2 - t)} = \frac{A}{1 - t} + \frac{B}{2 - t}, \\ 1 = A(2 - t) + B(1 - t). \end{gathered}

    Put t=1t = 1⁠: 1=A(1)1 = A(1)⁠, so A=1A = 1⁠. Put t=2t = 2⁠: 1=B(−1)1 = B(-1)⁠, so B=−1B = -1⁠. Hence

    1(1−t)(2−t)=11−t−12−t.\frac{1}{(1 - t)(2 - t)} = \frac{1}{1 - t} - \frac{1}{2 - t}.
    The factors 1−t=−(t−1)1 - t = -(t - 1) and 2−t=−(t−2)2 - t = -(t - 2) are different linear factors, so row 1 of Table 7.2 applies. The cleared identity holds for all tt⁠, so we may substitute the roots t=1,2t = 1, 2⁠. Check at t=0t = 0⁠: 12=1−12\frac12 = 1 - \frac12⁠. ✓
    Table 7.2, row 1. Basics: splitting fractions

  3. Integrate, watching the signs

    ∫dt1−t=−log⁡∣1−t∣+C1,∫dt2−t=−log⁡∣2−t∣+C2.\begin{aligned} \int\frac{dt}{1 - t} &= -\log|1 - t| + C_1, \\ \int\frac{dt}{2 - t} &= -\log|2 - t| + C_2. \end{aligned}

    So

    ∫dt(1−t)(2−t)=−log⁡∣1−t∣+log⁡∣2−t∣+C=log⁡∣2−t1−t∣+C.\begin{aligned} \int\frac{dt}{(1 - t)(2 - t)} &= -\log|1 - t| + \log|2 - t| + C \\ &= \log\left|\frac{2 - t}{1 - t}\right| + C. \end{aligned}
    Each bracket has derivative −1-1 with respect to tt⁠, which gives a minus sign in front of each logarithm. Subtracting the second integral turns its minus into a plus. Then log⁡∣p∣−log⁡∣q∣=log⁡∣pq∣\log|p| - \log|q| = \log\left|\frac pq\right|⁠.
    Standard formula (xii), Section 7.2, with substitution. Basics: logarithms and 1/x

  4. Return to xx

    ∫cos⁡x dx(1−sin⁡x)(2−sin⁡x)=log⁡∣2−sin⁡x1−sin⁡x∣+C.\int\frac{\cos x\,dx}{(1 - \sin x)(2 - \sin x)} = \log\left|\frac{2 - \sin x}{1 - \sin x}\right| + C.

    On the domain, 1−sin⁡x>01 - \sin x > 0 and 2−sin⁡x≥1>02 - \sin x \ge 1 > 0⁠, so the modulus may be dropped: the answer is also log⁡2−sin⁡x1−sin⁡x+C\log\frac{2 - \sin x}{1 - \sin x} + C⁠.

    sin⁡x≤1\sin x \le 1 always, and sin⁡x≠1\sin x \ne 1 on the domain, so 1−sin⁡x1 - \sin x is strictly positive; 2−sin⁡x≥2−1=12 - \sin x \ge 2 - 1 = 1⁠. A positive quotient equals its own absolute value.
    Range of sine. Basics: trigonometric identities

  5. Check: differentiate the answer

    Chain rule on each logarithm (inner functions 2−sin⁡x2 - \sin x and 1−sin⁡x1 - \sin x⁠, each with derivative −cos⁡x-\cos x⁠):

    ddx(log⁡(2−sin⁡x)−log⁡(1−sin⁡x))=−cos⁡x2−sin⁡x+cos⁡x1−sin⁡x=cos⁡x[(2−sin⁡x)−(1−sin⁡x)](1−sin⁡x)(2−sin⁡x)=cos⁡x(1−sin⁡x)(2−sin⁡x). ✓\begin{aligned} &\frac{d}{dx}\Big(\log(2 - \sin x) - \log(1 - \sin x)\Big) \\ &= \frac{-\cos x}{2 - \sin x} + \frac{\cos x}{1 - \sin x} \\ &= \frac{\cos x\big[(2 - \sin x) - (1 - \sin x)\big]}{(1 - \sin x)(2 - \sin x)} \\ &= \frac{\cos x}{(1 - \sin x)(2 - \sin x)}. \ \checkmark \end{aligned}
    Over the common denominator the bracket is (2−sin⁡x)−(1−sin⁡x)=1(2 - \sin x) - (1 - \sin x) = 1⁠, leaving the original integrand.
    Property (I), Section 7.2.1. Basics: integration undoes differentiation
Result ∫cos⁡x dx(1−sin⁡x)(2−sin⁡x)=log⁡∣2−sin⁡x1−sin⁡x∣+C,\int\frac{\cos x\,dx}{(1 - \sin x)(2 - \sin x)} = \log\left|\frac{2 - \sin x}{1 - \sin x}\right| + C,

for sin⁡x≠1\sin x \ne 1 (and the modulus may be dropped, since both brackets are positive). (Textbook answer: log⁡∣2−sin⁡x1−sin⁡x∣+C\log\left|\frac{2 - \sin x}{1 - \sin x}\right| + C⁠, the same.)

Check it with numbers

Definite integral from 00 to π6\frac{\pi}{6} (continuous there, as sin⁡x≤12\sin x \le \frac12⁠): F(π6)−F(0)=log⁡1.50.5−log⁡21F\big(\frac{\pi}{6}\big) - F(0) = \log\frac{1.5}{0.5} - \log\frac21 =log⁡3−log⁡2=log⁡32= \log 3 - \log 2 = \log\frac32 ≈0.4054651\approx 0.4054651⁠. A numerical integration of the integrand over [0,π6]\big[0, \frac{\pi}{6}\big] gives 0.40546510.4054651 too. ✓

Common mistakes
  1. ∫dt1−t=log⁡∣1−t∣\int\frac{dt}{1 - t} = \log|1 - t|⁠. The derivative of 1−t1 - t is −1-1⁠; the correct result is −log⁡∣1−t∣-\log|1 - t|⁠.
  2. Answer upside down. The signs give log⁡∣2−t∣−log⁡∣1−t∣\log|2 - t| - \log|1 - t|⁠, so 2−sin⁡x2 - \sin x is on top.
  3. Leaving the answer in tt⁠. Replace tt by sin⁡x\sin x at the end.
What to take away
  1. When the numerator is the derivative of the quantity the denominator is built from, substitute that quantity first.
  2. Partial fractions in tt work exactly as in xx⁠.
  3. Use the range of the function substituted (here −1≤sin⁡x≤1-1 \le \sin x \le 1⁠) to decide whether moduli can be dropped.

Every fact used, and where it comes from

FactUsed inSource
ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x⁠; substitution t=sin⁡xt = \sin xStep 1Standard formula (ii), Section 7.2; Section 7.3.1
At−a+Bt−b\frac{A}{t - a} + \frac{B}{t - b}Step 2Table 7.2, row 1, Section 7.5
∫dtt=log⁡∣t∣+C\int\frac{dt}{t} = \log|t| + CStep 3Standard formula (xii), Section 7.2
log⁡∣p∣−log⁡∣q∣=log⁡∣pq∣\log|p| - \log|q| = \log\left|\frac pq\right|Step 3Laws of logarithms (Basics: logarithms)
−1≤sin⁡x≤1-1 \le \sin x \le 1Step 4Chapter 3 of Class XI (Basics: trigonometric identities)
Derivative of an antiderivative is the integrandStep 5Property (I), Section 7.2.1
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