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Chapter 7 · Integrals

Exercise 7.2 · Q22: ∫sec⁡2(7−4x) dx\int\sec^2(7-4x)\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function

sec⁡2(7−4x).\sec^2(7 - 4x).
Exam answer

Let I=∫sec⁡2(7−4x) dxI = \displaystyle\int\sec^2(7 - 4x)\,dx⁠.

Put 7−4x=t7 - 4x = t⁠. Then −4 dx=dt-4\,dx = dt⁠.

I=−14∫sec⁡2t dt=−14tan⁡t+C\begin{aligned} I &= -\frac14\int\sec^2 t\,dt \\ &= -\frac14\tan t + C \end{aligned}

∴ I=−14tan⁡(7−4x)+C\therefore\ I = -\dfrac14\tan(7 - 4x) + C⁠.

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

The one idea this problem is built on

sec⁡2\sec^2 is the derivative of tan⁡\tan⁠, and the inside 7−4x7 - 4x is linear with slope −4-4⁠. So put t=7−4xt = 7 - 4x⁠: then dx=−14 dtdx = -\frac14\,dt⁠, and the answer is −14tan⁡(7−4x)+C-\frac14\tan(7 - 4x) + C⁠. The minus sign comes from the negative slope.

Given, to find, and the plan
Given
f(x)=sec⁡2(7−4x)f(x) = \sec^2(7 - 4x)⁠, defined where cos⁡(7−4x)≠0\cos(7 - 4x) \ne 0⁠.
To find
∫sec⁡2(7−4x) dx\displaystyle\int\sec^2(7 - 4x)\,dx⁠.
Plan
(1) Substitute t=7−4xt = 7 - 4x and express dxdx through dtdt⁠. (2) Use ∫sec⁡2t dt=tan⁡t+C\int\sec^2 t\,dt = \tan t + C⁠. (3) Return to xx⁠. (4) Differentiate back and state the domain.

Basics you need

1The integral of sec⁡2\sec^2

ddttan⁡t=sec⁡2t\frac{d}{dt}\tan t = \sec^2 t⁠, so ∫sec⁡2t dt=tan⁡t+C\int\sec^2 t\,dt = \tan t + C (standard formula (iv), Section 7.2). Basics: integration undoes differentiation

2Linear substitution

With t=ax+bt = ax + b and a≠0a \ne 0⁠, dt=a dxdt = a\,dx⁠, so dx=1a dtdx = \frac1a\,dt⁠. Here a=−4a = -4 is negative. Basics: substitution

Solution

  1. Substitute t=7−4xt = 7 - 4x

    Let t=7−4xt = 7 - 4x⁠. Then dtdx=−4\frac{dt}{dx} = -4⁠, so

    dt=−4 dx,dx=−14 dt.dt = -4\,dx, \qquad dx = -\frac14\,dt.

    Therefore

    ∫sec⁡2(7−4x) dx=−14∫sec⁡2t dt.\int\sec^2(7 - 4x)\,dx = -\frac14\int\sec^2 t\,dt.
    The standard formula needs sec⁡2\sec^2 of the plain variable. The substitution supplies that, at the cost of the factor −14-\frac14⁠: tt moves 4 times as fast as xx⁠, and in the opposite direction (as xx increases, tt decreases).
    Change of variable, Section 7.3.1; constants come outside, Property (IV), Section 7.2.1. Lesson: integration by substitution

  2. Integrate and return to xx

    −14∫sec⁡2t dt=−14tan⁡t+C=−14tan⁡(7−4x)+C.\begin{aligned} -\frac14\int\sec^2 t\,dt &= -\frac14\tan t + C \\ &= -\frac14\tan(7 - 4x) + C. \end{aligned}
    Standard formula (iv), Section 7.2.

  3. Check: differentiate the answer back

    By the chain rule (outer tan⁡u\tan u⁠, inner u=7−4xu = 7 - 4x with u′=−4u' = -4⁠):

    ddx[−14tan⁡(7−4x)]=−14sec⁡2(7−4x)⋅(−4)=sec⁡2(7−4x).\begin{aligned} &\frac{d}{dx}\left[-\frac14\tan(7 - 4x)\right] \\ &\quad= -\frac14\sec^2(7 - 4x)\cdot(-4) \\ &\quad= \sec^2(7 - 4x). \end{aligned}

    This is the integrand, so the answer is correct.

    The two negative factors, −14-\frac14 and −4-4⁠, multiply to +1+1⁠. That is why the answer must carry a minus sign.
    ddxtan⁡x=sec⁡2x\frac{d}{dx}\tan x = \sec^2 x⁠; chain rule, Theorem 4, Section 5.3.1.

  4. Where the answer is valid

    Both sec⁡2(7−4x)\sec^2(7 - 4x) and tan⁡(7−4x)\tan(7 - 4x) need cos⁡(7−4x)≠0\cos(7 - 4x) \ne 0⁠, that is,

    7−4x≠(2k+1)π2,k∈Z.7 - 4x \ne (2k + 1)\frac{\pi}{2}, \quad k \in \Z.

    The answer holds on each open interval between consecutive excluded points; strictly, CC may differ from one interval to the next.

    cos⁡θ=0\cos\theta = 0 exactly when θ\theta is an odd multiple of π2\frac{\pi}{2}⁠. Basics: trigonometric identities
Result ∫sec⁡2(7−4x) dx=−14tan⁡(7−4x)+C,\int\sec^2(7 - 4x)\,dx = -\frac14\tan(7 - 4x) + C,

valid wherever cos⁡(7−4x)≠0\cos(7 - 4x) \ne 0⁠. (Textbook answer: −14tan⁡(7−4x)+C-\frac14\tan(7 - 4x) + C⁠.)

Check it with numbers

Let F(x)=−14tan⁡(7−4x)F(x) = -\frac14\tan(7 - 4x) and look near x=1.5x = 1.5 (where 7−4x=17 - 4x = 1⁠) with h=0.0001h = 0.0001⁠:

  • F(1.5001)≈−0.3890095925F(1.5001) \approx -0.3890095925 and F(1.4999)≈−0.3896946966F(1.4999) \approx -0.3896946966⁠.
  • Slope estimate: F(1.5001)−F(1.4999)0.0002\dfrac{F(1.5001) - F(1.4999)}{0.0002} ≈3.425520\approx 3.425520⁠.
  • Integrand at x=1.5x = 1.5⁠: sec⁡21=1cos⁡21≈3.425519\sec^2 1 = \dfrac{1}{\cos^2 1} \approx 3.425519⁠.

They agree to six significant figures.

Common mistakes
  1. Dropping the minus sign. The slope of 7−4x7 - 4x is −4-4⁠, not 44⁠; +14tan⁡(7−4x)+\frac14\tan(7 - 4x) differentiates to −sec⁡2(7−4x)-\sec^2(7 - 4x)⁠.
  2. Multiplying by −4-4 instead of dividing. Integration divides by the slope; differentiation multiplies by it.
  3. Dividing by 7. The constant term of the linear inside plays no part in the factor.
What to take away
  1. ∫sec⁡2(ax+b) dx=1atan⁡(ax+b)+C\displaystyle\int\sec^2(ax + b)\,dx = \frac1a\tan(ax + b) + C⁠, and the sign of aa matters.
  2. Always check the sign by differentiating back.

Every fact used, and where it comes from

FactUsed inSource
Change of variable (substitution)Step 1Section 7.3.1 (Lesson: integration by substitution)
∫kf=k∫f\int k f = k\int fStep 1Property (IV), Section 7.2.1
∫sec⁡2t dt=tan⁡t+C\int\sec^2 t\,dt = \tan t + CStep 2Standard formula (iv), Section 7.2
ddxtan⁡x=sec⁡2x\frac{d}{dx}\tan x = \sec^2 x⁠; chain ruleStep 3Formula (iv), Section 7.2; Theorem 4, Section 5.3.1
Zeros of cos⁡\cosStep 4Trigonometric functions (Class XI)
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