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Chapter 7 · Integrals

Exercise 7.2 · Q31: ∫sin⁡x(1+cos⁡x)2 dx\int\frac{\sin x}{(1+\cos x)^2}\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function (Exercise 7.2, "Integrate the functions in Exercises 1 to 37"):

sin⁡x(1+cos⁡x)2\frac{\sin x}{(1 + \cos x)^2}
Exam answer

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

Put 1+cos⁡x=t1 + \cos x = t⁠. Then −sin⁡x dx=dt-\sin x\,dx = dt⁠.

I=−∫t−2 dt=−t−1−1+C=1t+C\begin{aligned} I &= -\int t^{-2}\,dt = -\frac{t^{-1}}{-1} + C \\ &= \frac1t + C \end{aligned}

∴ I=11+cos⁡x+C\therefore\ I = \dfrac{1}{1 + \cos x} + C⁠.

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

The one idea this problem is built on

As in Q30, the derivative of 1+cos⁡x1 + \cos x is −sin⁡x-\sin x⁠, the numerator up to sign. With t=1+cos⁡xt = 1 + \cos x⁠, the integral becomes −∫t−2 dt-\int t^{-2}\,dt⁠. This time the bottom is squared, so the power is −2-2⁠, not −1-1⁠: the power rule applies and there is no logarithm.

Given, to find, and the plan
Given
f(x)=sin⁡x(1+cos⁡x)2f(x) = \dfrac{\sin x}{(1 + \cos x)^2}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠, and where the answer is valid.
Plan
(1) Find where ff is defined. (2) Substitute t=1+cos⁡xt = 1 + \cos x⁠, sin⁡x dx=−dt\sin x\,dx = -dt⁠. (3) Integrate t−2t^{-2} by the power rule. (4) Return to xx⁠. (5) Differentiate back. (6) Confirm with a second route.

Basics you need

1Substitution

If t=g(x)t = g(x)⁠, then dt=g′(x) dxdt = g'(x)\,dx⁠. Basics: substitution (the chain rule backwards)

2Power rule with a negative power

∫t−2 dt=t−1−1+C=−1t+C\int t^{-2}\,dt = \frac{t^{-1}}{-1} + C = -\frac1t + C⁠, valid for t≠0t \ne 0 (standard formula (i), n=−2≠−1n = -2 \ne -1⁠). Basics: powers and the power rule

3Range of cos⁡x\cos x

cos⁡x=−1\cos x = -1 exactly at odd multiples of π\pi⁠; elsewhere 1+cos⁡x>01 + \cos x > 0⁠. Basics: trigonometric facts

4Integration undoes differentiation

Check every answer by differentiating it. Basics: integration undoes differentiation

Solution

  1. Where is the integrand defined?

    The denominator (1+cos⁡x)2(1 + \cos x)^2 is zero exactly when cos⁡x=−1\cos x = -1⁠, that is, at

    x=(2n+1)π,n an integer.x = (2n+1)\pi, \quad n \text{ an integer}.

    At every other real xx⁠, the integrand is defined, and 1+cos⁡x>01 + \cos x > 0⁠.

    We must never divide by zero; the answer is claimed only on the intervals between these points (Note, Section 7.2).

  2. Substitute t=1+cos⁡xt = 1 + \cos x

    Put t=1+cos⁡xt = 1 + \cos x⁠. Then dtdx=−sin⁡x\frac{dt}{dx} = -\sin x⁠, so

    sin⁡x dx=−dt.\sin x\,dx = -dt.

    The integral becomes

    ∫sin⁡x dx(1+cos⁡x)2=∫−dtt2=−∫t−2 dt.\int \frac{\sin x\,dx}{(1 + \cos x)^2} = \int \frac{-dt}{t^2} = -\int t^{-2}\,dt.
    The bracket that is squared, 1+cos⁡x1 + \cos x⁠, has derivative −sin⁡x-\sin x⁠, and sin⁡x\sin x is the numerator. That is the pattern "function of g(x)g(x) times g′(x)g'(x)" that substitution handles (Section 7.3.1). Writing sin⁡x dx=−dt\sin x\,dx = -dt first keeps the minus sign safe.
    Lesson: integration by substitution

  3. Integrate t−2t^{-2}

    Power rule with n=−2n = -2⁠, n+1=−1n + 1 = -1⁠:

    −∫t−2 dt=−t−1−1+C=t−1+C=1t+C.\begin{aligned} -\int t^{-2}\,dt &= -\frac{t^{-1}}{-1} + C \\ &= t^{-1} + C = \frac1t + C. \end{aligned}
    The power rule is allowed because n=−2≠−1n = -2 \ne -1⁠. The two minus signs (from sin⁡x dx=−dt\sin x\,dx = -dt and from dividing by −1-1⁠) cancel, so the answer is positive.
    Standard formula (i), Section 7.2. Basics: powers

  4. Return to xx

    ∫sin⁡x(1+cos⁡x)2 dx=11+cos⁡x+C,\int \frac{\sin x}{(1 + \cos x)^2}\,dx = \frac{1}{1 + \cos x} + C,

    valid wherever cos⁡x≠−1\cos x \ne -1⁠.

  5. Check: differentiate the answer back

    Let F(x)=(1+cos⁡x)−1F(x) = (1 + \cos x)^{-1}⁠. Chain rule, outer u−1u^{-1}⁠, inner u=1+cos⁡xu = 1 + \cos x⁠:

    F′(x)=−(1+cos⁡x)−2⋅(−sin⁡x)=sin⁡x(1+cos⁡x)2.\begin{aligned} F'(x) &= -(1 + \cos x)^{-2}\cdot(-\sin x) \\ &= \frac{\sin x}{(1 + \cos x)^2}. \end{aligned}

    This is the integrand. ✓

    By Property (I), Section 7.2.1, FF is an antiderivative of ff exactly when F′=fF' = f⁠.
    Chain rule, Theorem 4, Section 5.3.1. Basics: integration undoes differentiation

  6. A second route through tan⁡x2\tan\frac{x}{2}

    With sin⁡x=2sin⁡x2cos⁡x2\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} and 1+cos⁡x=2cos⁡2x21 + \cos x = 2\cos^2\frac{x}{2}⁠:

    sin⁡x(1+cos⁡x)2=2sin⁡x2cos⁡x24cos⁡4x2=12tan⁡x2sec⁡2x2.\begin{aligned} \frac{\sin x}{(1 + \cos x)^2} &= \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{4\cos^4\frac{x}{2}} \\ &= \tfrac12\tan\tfrac{x}{2}\sec^2\tfrac{x}{2}. \end{aligned}

    Put s=tan⁡x2s = \tan\frac{x}{2}⁠, ds=12sec⁡2x2 dxds = \frac12\sec^2\frac{x}{2}\,dx⁠. The integral is ∫s ds=s22\int s\,ds = \frac{s^2}{2}⁠, so it equals 12tan⁡2x2+C\frac12\tan^2\frac{x}{2} + C⁠. And this matches our answer up to a constant:

    12tan⁡2x2=12(sec⁡2x2−1)=12cos⁡2x2−12=11+cos⁡x−12.\begin{aligned} \tfrac12\tan^2\tfrac{x}{2} &= \tfrac12\left(\sec^2\tfrac{x}{2} - 1\right) \\ &= \frac{1}{2\cos^2\frac{x}{2}} - \tfrac12 \\ &= \frac{1}{1 + \cos x} - \tfrac12. \end{aligned}
    Two correct antiderivatives differ by a constant (Property (II), Section 7.2.1); here the constant is −12-\frac12⁠, absorbed into CC⁠. Getting the same family by a different road confirms the result.
    sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta and 1+cos⁡x=2cos⁡2x21 + \cos x = 2\cos^2\frac{x}{2}⁠. Basics: trigonometric identities
Result

∫sin⁡x(1+cos⁡x)2 dx=11+cos⁡x+C,\int \frac{\sin x}{(1 + \cos x)^2}\,dx = \frac{1}{1 + \cos x} + C, valid wherever cos⁡x≠−1\cos x \ne -1⁠. (Textbook answer: 11+cos⁡x+C\frac{1}{1 + \cos x} + C⁠.)

Check it with numbers

Take x=1x = 1 and h=0.001h = 0.001⁠:

  • F(1.001)≈0.6495781853F(1.001) \approx 0.6495781853 and F(0.999)≈0.6488688404F(0.999) \approx 0.6488688404⁠,
  • slope ≈0.6495781853−0.64886884040.002\approx \frac{0.6495781853 - 0.6488688404}{0.002} ≈0.35467\approx 0.35467⁠,
  • integrand: sin⁡1(1+cos⁡1)2=0.35467\frac{\sin 1}{(1 + \cos 1)^2} = 0.35467⁠. ✓
Common mistakes
  1. Answering with a logarithm. The denominator is squared, so the power is −2-2⁠, not −1-1⁠. A logarithm appears only for 1t\frac1t⁠.
  2. Sign slip. There are two minus signs (from dt=−sin⁡x dxdt = -\sin x\,dx and from t−1−1\frac{t^{-1}}{-1}⁠); they cancel. Answering −11+cos⁡x-\frac{1}{1 + \cos x} means one was lost.
  3. Thinking 12tan⁡2x2\frac12\tan^2\frac{x}{2} is a different answer. It equals 11+cos⁡x−12\frac{1}{1 + \cos x} - \frac12 (Step 6), so the two differ only by a constant.
What to take away
  1. ∫g′(x)g(x)2 dx=−1g(x)+C\int\frac{g'(x)}{g(x)^2}\,dx = -\frac{1}{g(x)} + C⁠: a squared denominator whose derivative is on top gives a reciprocal, not a log.
  2. Compare Q30 and Q31: the same substitution, but the power of the denominator decides between log⁡\log and the power rule.
  3. Different methods may give answers that differ by a constant; check by subtracting.

Every fact used, and where it comes from

FactUsed inSource
cos⁡x=−1\cos x = -1 exactly at odd multiples of π\piStep 1Trigonometric functions (Class XI, Chapter 3)
Substitution: t=g(x)t = g(x)⁠, dt=g′(x) dxdt = g'(x)\,dxSteps 2, 6Section 7.3.1 (Lesson: substitution)
∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n+1}}{n+1} + C⁠, n≠−1n \ne -1Steps 3, 6Standard formula (i), Section 7.2
Chain rule; F′=fF' = f checks the answerStep 5Theorem 4, Section 5.3.1; Property (I), Section 7.2.1
Half-angle identities; sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\thetaStep 6Trigonometric functions (Class XI, Chapter 3)
Antiderivatives of the same function differ by a constantStep 6Property (II), Section 7.2.1
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