Class 12 Maths, Worked in Full Past papersGet updates
Chapter 7 · Integrals

Exercise 7.6 · Q13: ∫tan⁡−1x dx\int\tan^{-1}x\,dx

Exercise 7.6Integration by parts · Section 7.6
The question

Integrate the functions in Exercises 1 to 22.

13.

tan⁡−1x\tan^{-1}x
Exam answer

Let I=∫tan⁡−1x⋅1 dxI = \displaystyle\int\tan^{-1}x\cdot 1\,dx⁠.

Taking tan⁡−1x\tan^{-1}x as the first function, by parts:

I=tan⁡−1x⋅x−∫11+x2⋅x dx=xtan⁡−1x−∫x dx1+x2\begin{aligned} I &= \tan^{-1}x\cdot x - \int\frac{1}{1 + x^2}\cdot x\,dx \\ &= x\tan^{-1}x - \int\frac{x\,dx}{1 + x^2} \end{aligned}

Put 1+x2=t1 + x^2 = t⁠. Then 2x dx=dt2x\,dx = dt⁠.

I=xtan⁡−1x−12∫dtt=xtan⁡−1x−12log⁡∣t∣+C\begin{aligned} I &= x\tan^{-1}x - \frac12\int\frac{dt}{t} \\ &= x\tan^{-1}x - \frac12\log|t| + C \end{aligned}

∴ I=xtan⁡−1x−12log⁡(1+x2)\therefore\ I = x\tan^{-1}x - \dfrac12\log(1 + x^2) +C+ 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⁡−1x dx\int\tan^{-1}x\,dx⁠, but its derivative 11+x2\frac{1}{1+x^2} is a simple algebraic function. So write tan⁡−1x=tan⁡−1x⋅1\tan^{-1}x = \tan^{-1}x\cdot 1 and integrate by parts with tan⁡−1x\tan^{-1}x as the first function and the constant 11 as the second: the parts formula swaps the hard function for its easy derivative.

Given, to find, and the plan
Given
tan⁡−1x\tan^{-1}x⁠, defined and continuous for every real xx⁠.
To find
∫tan⁡−1x dx\int\tan^{-1}x\,dx⁠.
Plan
(1) Choose tan⁡−1x\tan^{-1}x first and 11 second. (2) Apply the parts formula. (3) Find the leftover ∫x1+x2 dx\int\frac{x}{1+x^2}\,dx by the substitution t=1+x2t = 1 + x^2⁠. (4) Put it together. (5) Differentiate back.

Basics you need

1Integration by parts

∫fg dx=f∫g dx−∫[f′∫g dx]dx\int f g\,dx = f\int g\,dx - \int\big[f'\int g\,dx\big]dx⁠, with ff the first function and gg the second (Section 7.6). Lesson: integration by parts

2Derivative of tan⁡−1x\tan^{-1}x

ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2} (standard formula (x), Section 7.2). Basics: integration undoes differentiation

3Substitution and ∫dtt\int\frac{dt}{t}

If t=1+x2t = 1 + x^2 then dt=2x dxdt = 2x\,dx⁠; and ∫1t dt=log⁡∣t∣+C\int\frac1t\,dt = \log|t| + C (standard formula (xii)). Basics: substitution

Solution

  1. Choose the first and second functions

    Write the integrand as a product tan⁡−1x⋅1\tan^{-1}x\cdot 1⁠. Take

    f(x)=tan⁡−1x (first),g(x)=1 (second).f(x) = \tan^{-1}x \ (\text{first}),\qquad g(x) = 1\ (\text{second}).

    Then f′(x)=11+x2f'(x) = \frac{1}{1+x^2} and ∫g dx=x\int g\,dx = x⁠.

    The parts formula needs the integral of the second function, so the second function must be one we can integrate. We cannot integrate tan⁡−1x\tan^{-1}x directly (that is the whole question), but we can differentiate it, and its derivative is simpler. Remark (iii) in Section 7.6 says exactly this: inverse trigonometric and logarithmic functions are taken as the first function. (Many students remember the order with the word ILATE: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential; it is a memory aid, not a rule printed in the book.) Example 18 used the same trick for log⁡x\log x⁠.
    Choice of first function: Remark (iii), Section 7.6. Lesson: integration by parts

  2. Apply the integration by parts formula

    ∫tan⁡−1x⋅1 dx=tan⁡−1x⋅x−∫11+x2⋅x dx=xtan⁡−1x−∫x1+x2 dx.\begin{aligned} \int\tan^{-1}x\cdot 1\,dx &= \tan^{-1}x\cdot x - \int\frac{1}{1+x^2}\cdot x\,dx \\ &= x\tan^{-1}x - \int\frac{x}{1+x^2}\,dx. \end{aligned}
    First function times the integral of the second (tan⁡−1x⋅x\tan^{-1}x\cdot x⁠), minus the integral of (derivative of the first) times (integral of the second). No constant is needed when integrating the second function (Remark (ii), Section 7.6); one constant is added at the end.
    ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}⁠; ∫1 dx=x\int 1\,dx = x⁠. Lesson: integration by parts

  3. The leftover integral by substitution

    Put t=1+x2t = 1 + x^2⁠. Then dt=2x dxdt = 2x\,dx⁠, so x dx=12 dtx\,dx = \frac12\,dt⁠, and

    ∫x1+x2 dx=12∫dtt=12log⁡∣t∣=12log⁡(1+x2).\begin{aligned} \int\frac{x}{1+x^2}\,dx &= \frac12\int\frac{dt}{t} = \frac12\log|t| \\ &= \frac12\log(1+x^2). \end{aligned}
    The top xx is half the derivative of the bottom 1+x21 + x^2⁠, which is the signal for this substitution. The absolute value can be dropped because 1+x2≥1>01 + x^2 \ge 1 > 0 for every real xx⁠.
    Substitution (Section 7.3.1) and ∫dtt=log⁡∣t∣\int\frac{dt}{t} = \log|t| (formula (xii)). Basics: substitution

  4. Put the pieces together

    ∫tan⁡−1x dx=xtan⁡−1x−12log⁡(1+x2)+C.\int\tan^{-1}x\,dx = x\tan^{-1}x - \frac12\log(1+x^2) + C.
    Every function involved (tan⁡−1x\tan^{-1}x⁠, xx⁠, log⁡(1+x2)\log(1+x^2)⁠) is defined and differentiable for all real xx⁠, so the answer holds on the whole real line.
    Antiderivatives differ by a constant, so one CC covers all of them (Section 7.2). Basics: integration undoes differentiation

  5. Check: differentiate the answer

    ddx[xtan⁡−1x−12log⁡(1+x2)]=tan⁡−1x+x1+x2−12⋅2x1+x2=tan⁡−1x.\begin{aligned} &\frac{d}{dx}\left[x\tan^{-1}x - \frac12\log(1+x^2)\right] \\ &= \tan^{-1}x + \frac{x}{1+x^2} - \frac12\cdot\frac{2x}{1+x^2} \\ &= \tan^{-1}x. \end{aligned}
    The product rule on xtan⁡−1xx\tan^{-1}x gives 1⋅tan⁡−1x+x⋅11+x21\cdot\tan^{-1}x + x\cdot\frac{1}{1+x^2}⁠. The chain rule on log⁡(1+x2)\log(1+x^2) gives 2x1+x2\frac{2x}{1+x^2}⁠, and half of that cancels the extra x1+x2\frac{x}{1+x^2} exactly. What is left is the integrand.
    Product rule; chain rule (Chapter 5, Section 5.3.1). Basics: integration undoes differentiation
Result ∫tan⁡−1x dx=xtan⁡−1x−12log⁡(1+x2)+C,\int\tan^{-1}x\,dx = x\tan^{-1}x - \frac12\log(1+x^2) + C,

valid for every real xx⁠. (Textbook answer: xtan⁡−1x−12log⁡(1+x2)+Cx\tan^{-1}x - \frac12\log(1+x^2) + C⁠, the same.)

Check it with numbers

On [0,1][0, 1]⁠, with F(x)=xtan⁡−1x−12log⁡(1+x2)F(x) = x\tan^{-1}x - \frac12\log(1+x^2)⁠:

F(1)−F(0)=π4−12log⁡2−0=0.4388245731…\begin{aligned} F(1) - F(0) &= \frac{\pi}{4} - \frac12\log 2 - 0 \\ &= 0.4388245731\ldots \end{aligned}

Numerical integration of tan⁡−1x\tan^{-1}x over [0,1][0, 1] gives 0.4388245731…0.4388245731\ldots⁠, the same.

Common mistakes
  1. Not seeing a product. There is only one function, so students do not think of parts. Write it as tan⁡−1x⋅1\tan^{-1}x\cdot 1⁠.
  2. Taking 11 as the first function. Then you need ∫tan⁡−1x dx\int\tan^{-1}x\,dx as the integral of the second function, which is the question itself.
  3. Losing the 12\frac12 in ∫x1+x2 dx=12log⁡(1+x2)\int\frac{x}{1+x^2}\,dx = \frac12\log(1+x^2)⁠: the derivative of the bottom is 2x2x⁠, not xx⁠.
  4. Sign of the second term. The parts formula has a minus in front of the remaining integral, so the log term is subtracted.
What to take away
  1. To integrate a lone inverse trigonometric or logarithmic function, multiply by 11 and integrate by parts with the 11 as the second function.
  2. ∫tan⁡−1x dx=xtan⁡−1x−12log⁡(1+x2)+C\int\tan^{-1}x\,dx = x\tan^{-1}x - \frac12\log(1+x^2) + C⁠; it is used again in Question 22.

Every fact used, and where it comes from

FactUsed inSource
Integration by partsSteps 1, 2Section 7.6 (Lesson: integration by parts)
Inverse trigonometric functions are taken as the first function; no constant for the secondSteps 1, 2Remarks (iii) and (ii), Section 7.6; Example 18
ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}Steps 1, 2, 5Standard formula (x), Section 7.2
Substitution; ∫dtt=log⁡∣t∣+C\int\frac{dt}{t} = \log|t| + CStep 3Section 7.3.1; standard formula (xii), Section 7.2
Product rule, chain ruleStep 5Chapter 5, Sections 5.3 and 5.3.1
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →