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Chapter 7 · Integrals

Exercise 7.2 · Q21: ∫tan⁡2(2x−3) dx\int\tan^2(2x-3)\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function

tan⁡2(2x−3).\tan^2(2x - 3).
Exam answer

Let I=∫tan⁡2(2x−3) dxI = \displaystyle\int\tan^2(2x - 3)\,dx⁠.

Using tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1⁠,

I=∫sec⁡2(2x−3) dx−∫dxI = \int\sec^2(2x - 3)\,dx - \int dx

Put 2x−3=t2x - 3 = t⁠. Then 2 dx=dt2\,dx = dt⁠.

I=12∫sec⁡2t dt−x=12tan⁡t−x+C\begin{aligned} I &= \frac12\int\sec^2 t\,dt - x \\ &= \frac12\tan t - x + C \end{aligned}

∴ I=12tan⁡(2x−3)−x+C\therefore\ I = \dfrac12\tan(2x - 3) - x + C⁠.

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

The one idea this problem is built on

There is no standard formula for ∫tan⁡2\int\tan^2⁠, but there is one for ∫sec⁡2\int\sec^2⁠. The identity tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1 converts one into the other. Then ∫sec⁡2(2x−3) dx\int\sec^2(2x-3)\,dx needs only the linear substitution t=2x−3t = 2x - 3⁠, which costs a factor 12\frac12⁠.

Given, to find, and the plan
Given
f(x)=tan⁡2(2x−3)f(x) = \tan^2(2x - 3)⁠, defined where cos⁡(2x−3)≠0\cos(2x - 3) \ne 0⁠.
To find
∫tan⁡2(2x−3) dx\displaystyle\int \tan^2(2x - 3)\,dx⁠.
Plan
(1) Rewrite with tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1⁠. (2) Split into two integrals. (3) Substitute t=2x−3t = 2x - 3 in the sec⁡2\sec^2 part. (4) Combine. (5) Differentiate back and state the domain.

Basics you need

1The identity 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta

Divide sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 by cos⁡2θ\cos^2\theta (allowed when cos⁡θ≠0\cos\theta \ne 0⁠). So tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1⁠. Basics: trigonometric identities for integrals

2The integral of sec⁡2\sec^2

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

3Linear substitution

With t=ax+bt = ax + b⁠, dx=1a dtdx = \frac1a\,dt⁠. Basics: substitution

4Sum rule

∫(f−g)=∫f−∫g\int(f - g) = \int f - \int g (Properties (III) and (IV), Section 7.2.1).

Solution

  1. Replace tan⁡2\tan^2 by sec⁡2−1\sec^2 - 1

    With θ=2x−3\theta = 2x - 3⁠, wherever cos⁡(2x−3)≠0\cos(2x - 3) \ne 0⁠,

    tan⁡2(2x−3)=sec⁡2(2x−3)−1.\tan^2(2x - 3) = \sec^2(2x - 3) - 1.
    We know how to integrate sec⁡2\sec^2 (it is the derivative of tan⁡\tan⁠) and how to integrate 11⁠, but not tan⁡2\tan^2 directly. The identity holds at every point where the integrand is defined, so swapping one for the other changes nothing.
    1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta⁠, from sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1⁠. Basics: trigonometric identities

  2. Split the integral

    ∫tan⁡2(2x−3) dx=∫sec⁡2(2x−3) dx−∫dx.\begin{aligned} &\int\tan^2(2x - 3)\,dx \\ &\quad= \int\sec^2(2x - 3)\,dx - \int dx. \end{aligned}
    Properties (III) and (IV), Section 7.2.1 (difference of integrals).

  3. Substitute t=2x−3t = 2x - 3 in the first integral

    Let t=2x−3t = 2x - 3⁠. Then dt=2 dxdt = 2\,dx⁠, so dx=12 dtdx = \frac12\,dt⁠:

    ∫sec⁡2(2x−3) dx=12∫sec⁡2t dt=12tan⁡t+C1=12tan⁡(2x−3)+C1.\begin{aligned} \int\sec^2(2x - 3)\,dx &= \frac12\int\sec^2 t\,dt \\ &= \frac12\tan t + C_1 \\ &= \frac12\tan(2x - 3) + C_1. \end{aligned}
    The 12\frac12 appears because tt changes twice as fast as xx⁠. The second integral is simply ∫dx=x+C2\int dx = x + C_2⁠.
    Change of variable, Section 7.3.1; standard formula (iv), Section 7.2. Lesson: integration by substitution

  4. Combine

    ∫tan⁡2(2x−3) dx=12tan⁡(2x−3)−x+C,\int\tan^2(2x - 3)\,dx = \frac12\tan(2x - 3) - x + C,

    where C=C1−C2C = C_1 - C_2 is a single arbitrary constant.

    Two arbitrary constants combine into one, exactly as the book does in Example 6(ii) and (iii).

  5. Check: differentiate the answer back

    By the chain rule, ddxtan⁡(2x−3)=2sec⁡2(2x−3)\frac{d}{dx}\tan(2x - 3) = 2\sec^2(2x - 3)⁠. Hence

    ddx[12tan⁡(2x−3)−x]=sec⁡2(2x−3)−1=tan⁡2(2x−3).\begin{aligned} \frac{d}{dx}\left[\frac12\tan(2x - 3) - x\right] &= \sec^2(2x - 3) - 1 \\ &= \tan^2(2x - 3). \end{aligned}

    This is the integrand, so the answer is correct.

    ddxtan⁡x=sec⁡2x\frac{d}{dx}\tan x = \sec^2 x⁠; chain rule, Theorem 4, Section 5.3.1; identity of Step 1.

  6. Where the answer is valid

    Both tan⁡2(2x−3)\tan^2(2x - 3) and tan⁡(2x−3)\tan(2x - 3) need cos⁡(2x−3)≠0\cos(2x - 3) \ne 0⁠, that is,

    2x−3≠(2k+1)π2,k∈Z,2x - 3 \ne (2k + 1)\frac{\pi}{2}, \quad k \in \Z,

    or equivalently x≠32+(2k+1)π4x \ne \frac32 + (2k+1)\frac{\pi}{4}⁠. The answer holds on each open interval between two consecutive excluded points.

    At the excluded points tan⁡\tan has vertical asymptotes, so no antiderivative can cross them. On each separate interval the constant CC may, strictly, be different.
    cos⁡θ=0\cos\theta = 0 exactly when θ\theta is an odd multiple of π2\frac{\pi}{2}⁠. Basics: trigonometric identities
Result ∫tan⁡2(2x−3) dx=12tan⁡(2x−3)−x+C,\int\tan^2(2x - 3)\,dx = \frac12\tan(2x - 3) - x + C,

valid wherever cos⁡(2x−3)≠0\cos(2x - 3) \ne 0⁠. (Textbook answer: 12tan⁡(2x−3)−x+C\frac12\tan(2x - 3) - x + C⁠.)

Check it with numbers

Let F(x)=12tan⁡(2x−3)−xF(x) = \frac12\tan(2x - 3) - x and look near x=1x = 1 (where 2x−3=−12x - 3 = -1⁠) with h=0.0001h = 0.0001⁠:

  • F(1.0001)≈−1.7784614171F(1.0001) \approx -1.7784614171 and F(0.9999)≈−1.7789465209F(0.9999) \approx -1.7789465209⁠.
  • Slope estimate: F(1.0001)−F(0.9999)0.0002\dfrac{F(1.0001) - F(0.9999)}{0.0002} ≈2.425519\approx 2.425519⁠.
  • Integrand at x=1x = 1⁠: tan⁡2(−1)=tan⁡21≈2.425519\tan^2(-1) = \tan^2 1 \approx 2.425519⁠.

They agree to seven significant figures.

Common mistakes
  1. Forgetting the −x-x⁠. It comes from the −1-1 in sec⁡2−1\sec^2 - 1⁠.
  2. Forgetting the 12\frac12⁠. ddxtan⁡(2x−3)=2sec⁡2(2x−3)\frac{d}{dx}\tan(2x - 3) = 2\sec^2(2x - 3)⁠, so the antiderivative needs 12\frac12⁠.
  3. Writing tan⁡3(2x−3)3\frac{\tan^3(2x-3)}{3}⁠. The power rule applies to a power of the variable, not to a power of tan⁡\tan without the derivative sec⁡2\sec^2 beside it.
  4. Using tan⁡2θ=1−sec⁡2θ\tan^2\theta = 1 - \sec^2\theta⁠. The sign is the other way round: sec⁡2θ−1\sec^2\theta - 1⁠.
What to take away
  1. ∫tan⁡2x dx=tan⁡x−x+C\int\tan^2 x\,dx = \tan x - x + C⁠, and with a linear inside, divide the tan⁡\tan term by the slope.
  2. When a trigonometric power has no direct formula, look for an identity that turns it into one that does (Section 7.3.2 develops this idea).

Every fact used, and where it comes from

FactUsed inSource
tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1Steps 1, 5From sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 (Class XI)
Sum and constant-multiple rulesStep 2Properties (III), (IV), Section 7.2.1
Change of variable (substitution)Step 3Section 7.3.1 (Lesson: integration by substitution)
∫sec⁡2t dt=tan⁡t+C\int\sec^2 t\,dt = \tan t + C⁠; ∫dx=x+C\int dx = x + CStep 3Standard formulae (iv) and (i), Section 7.2
Combining constantsStep 4Example 6(ii), (iii), Section 7.3.1
Chain ruleStep 5Theorem 4, Section 5.3.1
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