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Chapter 7 · Integrals

Exercise 7.2 · Q25: ∫dxcos⁡2x (1−tan⁡x)2\int\frac{dx}{\cos^2x\,(1-\tan x)^2}

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function

1cos⁡2x (1−tan⁡x)2.\frac{1}{\cos^2x\,(1 - \tan x)^2}.
Exam answer

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

I=∫sec⁡2x dx(1−tan⁡x)2I = \int\frac{\sec^2x\,dx}{(1 - \tan x)^2}

Put 1−tan⁡x=t1 - \tan x = t⁠. Then −sec⁡2x dx=dt-\sec^2x\,dx = dt⁠.

I=−∫dtt2=−t−1−1+C=1t+C\begin{aligned} I &= -\int\frac{dt}{t^2} = -\frac{t^{-1}}{-1} + C \\ &= \frac1t + C \end{aligned}

∴ I=11−tan⁡x+C\therefore\ I = \dfrac{1}{1 - \tan x} + C⁠.

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

The one idea this problem is built on

1cos⁡2x=sec⁡2x\frac{1}{\cos^2x} = \sec^2x is the derivative of tan⁡x\tan x⁠. So the integrand is sec⁡2x(1−tan⁡x)2\frac{\sec^2x}{(1 - \tan x)^2}⁠: a power of 1−tan⁡x1 - \tan x times (minus) its derivative. Put t=1−tan⁡xt = 1 - \tan x⁠; then sec⁡2x dx=−dt\sec^2x\,dx = -dt and the integral becomes −∫t−2 dt-\int t^{-2}\,dt⁠.

Given, to find, and the plan
Given
f(x)=1cos⁡2x (1−tan⁡x)2f(x) = \dfrac{1}{\cos^2x\,(1 - \tan x)^2}⁠, defined where cos⁡x≠0\cos x \ne 0 and tan⁡x≠1\tan x \ne 1⁠.
To find
∫dxcos⁡2x (1−tan⁡x)2\displaystyle\int \frac{dx}{\cos^2x\,(1 - \tan x)^2}⁠.
Plan
(1) Rewrite 1cos⁡2x\frac{1}{\cos^2x} as sec⁡2x\sec^2x⁠. (2) Substitute t=1−tan⁡xt = 1 - \tan x⁠. (3) Integrate t−2t^{-2} by the power formula. (4) Return to xx⁠. (5) Differentiate back and state the domain.

Basics you need

1sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x} and the derivative of tan⁡x\tan x

ddxtan⁡x=sec⁡2x=1cos⁡2x\frac{d}{dx}\tan x = \sec^2x = \frac{1}{\cos^2x} wherever cos⁡x≠0\cos x \ne 0 (Table 5.3; formula (iv), Section 7.2). Basics: trigonometric identities

2Substitution

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

3The power formula with a negative power

∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n+1}}{n+1} + C for n≠−1n \ne -1⁠; with n=−2n = -2 this gives ∫t−2 dt=−t−1+C\int t^{-2}\,dt = -t^{-1} + C⁠. Basics: the power rule

Solution

  1. Rewrite 1cos⁡2x\frac{1}{\cos^2 x} as sec⁡2x\sec^2x

    1cos⁡2x (1−tan⁡x)2=sec⁡2x(1−tan⁡x)2.\frac{1}{\cos^2x\,(1 - \tan x)^2} = \frac{\sec^2x}{(1 - \tan x)^2}.
    In this form the integrand shows a function of tan⁡x\tan x multiplied by sec⁡2x\sec^2x⁠, the derivative of tan⁡x\tan x⁠. That is the substitution pattern.
    sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}⁠, so sec⁡2x=1cos⁡2x\sec^2x = \frac{1}{\cos^2x}⁠. Basics: trigonometric identities

  2. Substitute t=1−tan⁡xt = 1 - \tan x

    Let t=1−tan⁡xt = 1 - \tan x⁠. Then dtdx=−sec⁡2x\frac{dt}{dx} = -\sec^2x⁠, so

    sec⁡2x dx=−dt.\sec^2x\,dx = -dt.

    The integral becomes

    ∫sec⁡2x dx(1−tan⁡x)2=−∫dtt2=−∫t−2 dt.\int\frac{\sec^2x\,dx}{(1 - \tan x)^2} = -\int\frac{dt}{t^2} = -\int t^{-2}\,dt.
    Taking the whole bracket 1−tan⁡x1 - \tan x as tt (rather than just tan⁡x\tan x⁠) makes the denominator a plain power t2t^2⁠. The minus sign comes from differentiating −tan⁡x-\tan x⁠.
    Change of variable, Section 7.3.1. Lesson: integration by substitution

  3. Integrate t−2t^{-2}

    Power formula with n=−2n = -2 (allowed, since n≠−1n \ne -1⁠):

    −∫t−2 dt=−t−1−1+C=1t+C.\begin{aligned} -\int t^{-2}\,dt &= -\frac{t^{-1}}{-1} + C \\ &= \frac{1}{t} + C. \end{aligned}
    The two minus signs cancel: one from sec⁡2x dx=−dt\sec^2x\,dx = -dt⁠, one from dividing by n+1=−1n + 1 = -1⁠.
    Standard formula (i), Section 7.2. Basics: the power rule

  4. Return to xx

    ∫dxcos⁡2x (1−tan⁡x)2=11−tan⁡x+C.\int \frac{dx}{\cos^2x\,(1 - \tan x)^2} = \frac{1}{1 - \tan x} + C.

  5. Check: differentiate the answer back

    Write the answer as (1−tan⁡x)−1(1 - \tan x)^{-1} and use the chain rule (outer u−1u^{-1}⁠, inner u=1−tan⁡xu = 1 - \tan x with u′=−sec⁡2xu' = -\sec^2x⁠):

    ddx(1−tan⁡x)−1=−(1−tan⁡x)−2⋅(−sec⁡2x)=sec⁡2x(1−tan⁡x)2=1cos⁡2x (1−tan⁡x)2.\begin{aligned} &\frac{d}{dx}(1 - \tan x)^{-1} \\ &\quad= -(1 - \tan x)^{-2}\cdot(-\sec^2x) \\ &\quad= \frac{\sec^2x}{(1 - \tan x)^2} \\ &\quad= \frac{1}{\cos^2x\,(1 - \tan x)^2}. \end{aligned}

    This is the integrand, so the answer is correct.

    Chain rule, Theorem 4, Section 5.3.1; dduu−1=−u−2\frac{d}{du}u^{-1} = -u^{-2}⁠.

  6. Where the answer is valid

    We need cos⁡x≠0\cos x \ne 0 (for tan⁡x\tan x and 1cos⁡2x\frac{1}{\cos^2x}⁠) and tan⁡x≠1\tan x \ne 1 (denominator). So

    x≠π2+kπandx≠π4+kπ,k∈Z.x \ne \frac{\pi}{2} + k\pi \quad\text{and}\quad x \ne \frac{\pi}{4} + k\pi, \qquad k \in \Z.

    The answer holds on each open interval between consecutive excluded points.

    Between two consecutive excluded points the integrand is continuous, and the antiderivative 11−tan⁡x\frac{1}{1 - \tan x} is valid there; strictly, the constant may differ from one interval to the next.
Result ∫dxcos⁡2x (1−tan⁡x)2=11−tan⁡x+C,\int \frac{dx}{\cos^2x\,(1 - \tan x)^2} = \frac{1}{1 - \tan x} + C,

valid wherever cos⁡x≠0\cos x \ne 0 and tan⁡x≠1\tan x \ne 1⁠. (Textbook answer: 11−tan⁡x+C\dfrac{1}{1 - \tan x} + C⁠.)

Check it with numbers

Let F(x)=11−tan⁡xF(x) = \frac{1}{1 - \tan x} and look near x=0.3x = 0.3 with h=0.0001h = 0.0001⁠:

  • F(0.3001)≈1.4481123013F(0.3001) \approx 1.4481123013 and F(0.2999)≈1.4476529087F(0.2999) \approx 1.4476529087⁠.
  • Slope estimate: F(0.3001)−F(0.2999)0.0002\dfrac{F(0.3001) - F(0.2999)}{0.0002} ≈2.296963\approx 2.296963⁠.
  • Integrand at x=0.3x = 0.3⁠: 1cos⁡20.3 (1−tan⁡0.3)2≈2.296963\dfrac{1}{\cos^2 0.3\,(1 - \tan 0.3)^2} \approx 2.296963⁠.

They agree.

Common mistakes
  1. Getting the sign wrong. dt=−sec⁡2x dxdt = -\sec^2x\,dx⁠; together with ∫t−2 dt=−t−1\int t^{-2}\,dt = -t^{-1}⁠, the answer is +11−tan⁡x+\frac{1}{1 - \tan x}⁠.
  2. Writing log⁡∣1−tan⁡x∣\log|1 - \tan x|⁠. The logarithm is for ∫dtt\int\frac{dt}{t}⁠; here the power is t−2t^{-2}⁠, not t−1t^{-1}⁠.
  3. Expanding (1−tan⁡x)2(1 - \tan x)^2⁠. That only makes the integral harder; keep the bracket as one block and substitute it.
What to take away
  1. 1cos⁡2x\frac{1}{\cos^2x} (that is, sec⁡2x\sec^2x⁠) next to a function of tan⁡x\tan x signals a substitution in tan⁡x\tan x⁠.
  2. Choose tt as the whole bracket when the bracket is raised to a power.

Every fact used, and where it comes from

FactUsed inSource
sec⁡2x=1cos⁡2x\sec^2x = \frac{1}{\cos^2x}Steps 1, 5Definition of sec⁡\sec (Class XI)
ddxtan⁡x=sec⁡2x\frac{d}{dx}\tan x = \sec^2xSteps 2, 5Table 5.3, Section 5.3; formula (iv), Section 7.2
Change of variable (substitution)Step 2Section 7.3.1 (Lesson: integration by substitution)
∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n+1}}{n+1} + C⁠, n≠−1n \ne -1Step 3Standard formula (i), Section 7.2
Chain ruleStep 5Theorem 4, Section 5.3.1
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