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

Example 27: ∫01tan⁡−1x1+x2 dx\int_0^1\frac{\tan^{-1}x}{1+x^2}\,dx

ExamplesDefinite integrals by substitution · Section 7.9
The question

Evaluate ∫01tan⁡−1x1+x2  dx\displaystyle\int_0^1 \frac{\tan^{-1}x}{1 + x^2}\; dx⁠.

Exam answer

Let I=∫01tan⁡−1x1+x2 dxI = \displaystyle\int_0^1\frac{\tan^{-1}x}{1 + x^2}\,dx⁠.

Put tan⁡−1x=t\tan^{-1}x = t⁠. Then dx1+x2=dt\frac{dx}{1 + x^2} = dt⁠.

When x=0x = 0⁠, t=0t = 0⁠; when x=1x = 1⁠, t=π4t = \frac{\pi}{4}⁠.

I=∫0π/4t dt=[t22]0π/4=12⋅π216\begin{aligned} I &= \int_0^{\pi/4} t\,dt = \left[\frac{t^2}{2}\right]_0^{\pi/4} \\ &= \frac12\cdot\frac{\pi^2}{16} \end{aligned}

∴ I=π232\therefore\ I = \dfrac{\pi^2}{32}⁠.

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

The one idea this problem is built on

The factor 11+x2\frac{1}{1 + x^2} is the derivative of tan⁡−1x\tan^{-1}x⁠. So with t=tan⁡−1xt = \tan^{-1}x we get dt=dx1+x2dt = \frac{dx}{1 + x^2}⁠, and the integral becomes simply ∫t dt\int t\,dt⁠. Changing the limits at the same time (x=0→t=0x = 0 \to t = 0⁠, x=1→t=π4x = 1 \to t = \frac{\pi}{4}⁠) finishes it in one line.

Given, to find, and the plan
Given
The integrand tan⁡−1x1+x2\frac{\tan^{-1}x}{1+x^2} on [0,1][0, 1]⁠.
To find
The exact value of the integral.
Plan
Check continuity; substitute t=tan⁡−1xt = \tan^{-1}x and change the limits; integrate tt⁠; confirm with an antiderivative in xx and numerically.

Basics you need

1Derivative of tan⁡−1x\tan^{-1}x

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

2Principal values

tan⁡−1\tan^{-1} takes values in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, so tan⁡−10=0\tan^{-1}0 = 0 and tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4} (Chapter 2). Basics: trigonometric identities

3Substitution with new limits

When gg has a continuous derivative on [a,b][a, b] and ff is continuous at every value that gg takes there (Section 7.9), ∫abf(g(x)) g′(x) dx=∫g(a)g(b)f(t) dt.\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(t)\,dt. Lesson: integration by substitution

4Power rule

∫t dt=t22+C\int t\,dt = \frac{t^2}{2} + C⁠. Basics: the power rule

Solution

  1. Check continuity on [0,1][0, 1]

    tan⁡−1x\tan^{-1}x is continuous for all real xx⁠, and 1+x2≥11 + x^2 \ge 1 is never 0. So the quotient is continuous on [0,1][0, 1]⁠, and Theorem 2 applies.

    The fundamental theorem and the substitution rule both need a continuous integrand on the closed interval. Here nothing can go wrong: the denominator is always at least 1.
    A quotient of continuous functions is continuous where the denominator is not 0. Lesson: the fundamental theorem

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

    Then

    dt=11+x2 dx,sotan⁡−1x1+x2 dx=t dt.dt = \frac{1}{1 + x^2}\,dx, \quad\text{so}\quad \frac{\tan^{-1}x}{1 + x^2}\,dx = t\,dt.

    New limits. When x=0x = 0⁠, t=tan⁡−10=0t = \tan^{-1}0 = 0⁠. When x=1x = 1⁠, t=tan⁡−11=π4t = \tan^{-1}1 = \frac{\pi}{4}⁠. As xx runs from 0 to 1, tt runs from 0 to π4\frac{\pi}{4}⁠.

    The integrand is "a function times the derivative of what is inside it", the signal for substitution. The substitution is legitimate on the whole interval: tan⁡−1x\tan^{-1}x is differentiable for every real xx and increasing, so tt moves once from 0 to π4\frac{\pi}{4}⁠.
    tan⁡π4=1\tan\frac{\pi}{4} = 1 and π4\frac{\pi}{4} is in the principal range (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, so tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4} (not 5π4\frac{5\pi}{4}⁠). Basics: trigonometry

  3. Integrate in tt

    ∫01tan⁡−1x1+x2 dx=∫0π4t dt=[t22]0π4=12[π216−0]=π232.\begin{aligned}\int_0^1\frac{\tan^{-1}x}{1 + x^2}\,dx &= \int_0^{\frac{\pi}{4}} t\,dt = \left[\frac{t^2}{2}\right]_0^{\frac{\pi}{4}} \\ &= \frac12\left[\frac{\pi^2}{16} - 0\right] = \frac{\pi^2}{32}.\end{aligned}

    In decimals, π232≈0.3084251\frac{\pi^2}{32} \approx 0.3084251⁠.

    Once in tt⁠, the integral is the simplest power: ∫t dt=t22\int t\,dt = \frac{t^2}{2}⁠. Squaring π4\frac{\pi}{4} gives π216\frac{\pi^2}{16}⁠, and half of that is π232\frac{\pi^2}{32}⁠.
    Power rule with n=1n = 1⁠, then F(upper) minus F(lower). Basics: the power rule

  4. Same result by staying in xx

    Returning to xx instead: ∫t dt=t22\int t\,dt = \frac{t^2}{2} gives the antiderivative

    F(x)=(tan⁡−1x)22.F(x) = \frac{\left(\tan^{-1}x\right)^2}{2}.

    Check: by the chain rule F′(x)=2tan⁡−1x2⋅11+x2=tan⁡−1x1+x2F'(x) = \frac{2\tan^{-1}x}{2}\cdot\frac{1}{1 + x^2} = \frac{\tan^{-1}x}{1+x^2}⁠. Then F(1)−F(0)=12(π4)2−0=π232F(1) - F(0) = \frac{1}{2}\left(\frac{\pi}{4}\right)^2 - 0 = \frac{\pi^2}{32}⁠, as before.

    The two routes of Section 7.9 must agree; doing both is a cheap check that the limits were changed correctly.
    Chain rule: outer function u22\frac{u^2}{2}⁠, inner function u=tan⁡−1xu = \tan^{-1}x⁠. Basics: substitution

  5. The region

    1 2 3 0.2 π/8 x y (1, π/8) area ≈ 0.308 y = tan⁻¹x / (1 + x²)
    y=tan⁡−1x1+x2y = \frac{\tan^{-1}x}{1+x^2} starts at 0, and at x=1x = 1 its height is π/42=π8≈0.393\frac{\pi/4}{2} = \frac{\pi}{8} \approx 0.393⁠. The shaded area from 0 to 1 is π232≈0.308\frac{\pi^2}{32} \approx 0.308⁠.
    On [0,1][0, 1] the integrand is ≥0\ge 0 (tan⁡−1x≥0\tan^{-1}x \ge 0 for x≥0x \ge 0⁠), so the integral is an area. The region fits inside a 1 by 0.42 box, consistent with an area of about 0.31.
    Definite integral of a non-negative function = area under its graph. Basics: area as a sum of thin strips
Result

∫01tan⁡−1x1+x2 dx=π232≈0.3084251,\int_0^1\frac{\tan^{-1}x}{1 + x^2}\,dx = \frac{\pi^2}{32} \approx 0.3084251, the same as the book's answer.

Check it with numbers

A numerical integration of tan⁡−1x1+x2\frac{\tan^{-1}x}{1+x^2} over [0,1][0, 1] gives 0.3084251, and π232=9.869604432≈0.3084251\frac{\pi^2}{32} = \frac{9.8696044}{32} \approx 0.3084251⁠. At the midpoint, F(0.5)=(tan⁡−10.5)22≈0.1074846F(0.5) = \frac{(\tan^{-1}0.5)^2}{2} \approx 0.1074846⁠: less than half the total, as the figure suggests (the curve is lower on the left).

Common mistakes
  1. Keeping the xx-limits. After t=tan⁡−1xt = \tan^{-1}x⁠, the upper limit is π4\frac{\pi}{4}⁠, not 1.
  2. Writing tan⁡−11=45\tan^{-1}1 = 45⁠. In calculus angles are in radians: π4\frac{\pi}{4}⁠.
  3. Trying integration by parts first. It works, but the substitution is far shorter once you spot 11+x2=ddxtan⁡−1x\frac{1}{1+x^2} = \frac{d}{dx}\tan^{-1}x⁠.
What to take away
  1. "f(x)×f′(x)f(x)\times f'(x)" integrates to f(x)22\frac{f(x)^2}{2}⁠: here f=tan⁡−1f = \tan^{-1}⁠.
  2. Changing the limits uses the principal values of the inverse function.
  3. Both routes of Section 7.9 (back to xx⁠, or new limits) give the same number.

Every fact used, and where it comes from

FactUsed inSource
ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}Steps 2 and 4Chapter 5; standard formula (x), Section 7.2
tan⁡−10=0\tan^{-1}0 = 0⁠, tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}Step 2Principal values, Chapter 2
Substitution with change of limitsStep 2Section 7.9 (Lesson)
∫t dt=t22+C\int t\,dt = \frac{t^2}{2} + CStep 3Standard formula (i), Section 7.2 (Basics)
∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a)Steps 3 and 4Theorem 2, Section 7.8.3 (Lesson)
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 →