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Chapter 7 · Integrals

Miscellaneous · Q22: ∫tan⁡−11−x1+x dx\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx

Miscellaneous ExerciseSimplify with x = cos θ to (1/2)cos⁻¹x, then by parts · Miscellaneous Exercise
The question

Integrate the function:

tan⁡−11−x1+x\tan^{-1}\sqrt{\frac{1-x}{1+x}}
Exam answer

Let I=∫tan⁡−11−x1+x dxI = \displaystyle\int\tan^{-1}\sqrt{\frac{1 - x}{1 + x}}\,dx⁠.

Put x=cos⁡θx = \cos\theta⁠, where θ=cos⁡−1x\theta = \cos^{-1}x⁠, 0<θ<π0 < \theta < \pi⁠.

1−x1+x=1−cos⁡θ1+cos⁡θ=2sin⁡2θ22cos⁡2θ2=tan⁡θ2\begin{aligned} \sqrt{\frac{1 - x}{1 + x}} &= \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} \\ &= \sqrt{\frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}} = \tan\frac{\theta}{2} \end{aligned}

(as 0<θ2<π20 < \frac{\theta}{2} < \frac{\pi}{2}⁠, tan⁡θ2>0\tan\frac{\theta}{2} > 0⁠).

∴ tan⁡−11−x1+x=θ2=12cos⁡−1x\begin{aligned} \therefore\ \tan^{-1}\sqrt{\frac{1 - x}{1 + x}} &= \frac{\theta}{2} \\ &= \frac12\cos^{-1}x \end{aligned}

So I=12∫cos⁡−1x dxI = \frac12\displaystyle\int\cos^{-1}x\,dx⁠. Taking cos⁡−1x\cos^{-1}x as the first function and 1 as the second, by parts:

I=12[xcos⁡−1x−∫x(−11−x2)dx]=12[xcos⁡−1x+∫x dx1−x2]\begin{aligned} I &= \frac12\bigg[x\cos^{-1}x \\ &\qquad - \int x\left(-\frac{1}{\sqrt{1 - x^2}}\right)dx\bigg] \\ &= \frac12\bigg[x\cos^{-1}x \\ &\qquad + \int\frac{x\,dx}{\sqrt{1 - x^2}}\bigg] \end{aligned}

Put 1−x2=t1 - x^2 = t⁠. Then −2x dx=dt-2x\,dx = dt⁠.

∫x dx1−x2=−12∫dtt=−t=−1−x2\begin{aligned} \int\frac{x\,dx}{\sqrt{1 - x^2}} &= -\frac12\int\frac{dt}{\sqrt t} \\ &= -\sqrt t = -\sqrt{1 - x^2} \end{aligned}

∴ I=12[xcos⁡−1x−1−x2]\therefore\ I = \dfrac12\left[x\cos^{-1}x - \sqrt{1 - x^2}\right] +C+ C⁠.

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

The one idea this problem is built on

Before integrating anything, simplify the function. With x=cos⁡θx = \cos\theta⁠, the quotient 1−cos⁡θ1+cos⁡θ\frac{1-\cos\theta}{1+\cos\theta} is tan⁡2θ2\tan^2\frac{\theta}{2}⁠, so the integrand is tan⁡−1(tan⁡θ2)=θ2=12cos⁡−1x\tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2} = \frac12\cos^{-1}x⁠. An inverse trigonometric function on its own is integrated by parts, with 1 as the second function.

Given, to find, and the plan
Given
f(x)=tan⁡−11−x1+xf(x) = \tan^{-1}\sqrt{\frac{1-x}{1+x}}⁠. We need x≠−1x \ne -1 and 1−x1+x≥0\frac{1-x}{1+x} \ge 0⁠; this holds exactly for −1<x≤1-1 < x \le 1⁠.
To find
∫tan⁡−11−x1+x dx\displaystyle\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx⁠.
Plan
(1) Write x=cos⁡θx = \cos\theta with θ\theta in [0,π)[0, \pi)⁠. (2) Simplify the root. (3) Simplify tan⁡−1\tan^{-1}⁠, so f(x)=12cos⁡−1xf(x) = \frac12\cos^{-1}x⁠. (4) Integrate cos⁡−1x\cos^{-1}x by parts. (5) Differentiate back. (6) Compare with the key.

Basics you need

1Principal ranges (Chapter 2)

cos⁡−1\cos^{-1} takes values in [0,π][0, \pi]⁠; tan⁡−1\tan^{-1} takes values in (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})⁠. So tan⁡−1(tan⁡u)=u\tan^{-1}(\tan u) = u only when −π2<u<π2-\frac{\pi}{2} < u < \frac{\pi}{2}⁠.

2Half angles

1−cos⁡θ=2sin⁡2θ21 - \cos\theta = 2\sin^2\frac{\theta}{2}⁠, 1+cos⁡θ=2cos⁡2θ21 + \cos\theta = 2\cos^2\frac{\theta}{2}⁠. Basics: trigonometric identities

3Integration by parts (Section 7.6)

∫u v dx=u∫v dx−∫(u′∫v dx)dx\int u\,v\,dx = u\int v\,dx - \int\left(u'\int v\,dx\right)dx⁠; for an inverse trigonometric function alone, take it as the first function and 1 as the second. Lesson: integration by parts

Solution

  1. Write x=cos⁡θx = \cos\theta⁠, 0≤θ<π0 \le \theta < \pi

    For each xx in (−1,1](-1, 1] there is exactly one θ\theta in [0,π)[0, \pi) with cos⁡θ=x\cos\theta = x⁠, namely θ=cos⁡−1x\theta = \cos^{-1}x⁠.

    Why cos⁡θ\cos\theta⁠: the function contains 1−x1 - x and 1+x1 + x⁠, and with x=cos⁡θx = \cos\theta these have the half-angle forms that make a perfect square. Why [0,π)[0, \pi)⁠: it is the principal range of cos⁡−1\cos^{-1} with π\pi removed (because x=−1x = -1 is not in the domain), so θ=cos⁡−1x\theta = \cos^{-1}x exactly. This is a rewriting of the integrand, not a change of the variable of integration: we keep integrating in xx⁠.
    Principal range of cos⁡−1\cos^{-1} (Chapter 2).

  2. Simplify the root

    1−x1+x=1−cos⁡θ1+cos⁡θ=2sin⁡2θ22cos⁡2θ2=tan⁡2θ2.\begin{aligned}\frac{1-x}{1+x} &= \frac{1-\cos\theta}{1+\cos\theta}\\ &= \frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} = \tan^2\frac{\theta}{2} .\end{aligned}

    Because 0≤θ2<π20 \le \frac{\theta}{2} < \frac{\pi}{2}⁠, tan⁡θ2≥0\tan\frac{\theta}{2} \ge 0⁠, so 1−x1+x=tan⁡θ2\sqrt{\frac{1-x}{1+x}} = \tan\frac{\theta}{2}⁠.

    u2=∣u∣\sqrt{u^2} = |u|⁠; the modulus can be dropped because θ2\frac{\theta}{2} is in the first quadrant. Also cos⁡θ2≠0\cos\frac{\theta}{2} \ne 0 there, so the fraction is defined.
    Basics: trigonometric identities

  3. Simplify tan⁡−1\tan^{-1}⁠: the integrand is 12cos⁡−1x\frac12\cos^{-1}x

    tan⁡−11−x1+x=tan⁡−1(tan⁡θ2)=θ2=12cos⁡−1x.\tan^{-1}\sqrt{\frac{1-x}{1+x}} = \tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2} = \frac12\cos^{-1}x .
    tan⁡−1(tan⁡u)=u\tan^{-1}(\tan u) = u needs uu in (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})⁠, the principal range of tan⁡−1\tan^{-1}⁠. Here u=θ2u = \frac{\theta}{2} lies in [0,π2)[0, \frac{\pi}{2})⁠, so the cancellation is legitimate for every xx in (−1,1](-1, 1]⁠.
    Principal range of tan⁡−1\tan^{-1} (Chapter 2).

  4. Integrate cos⁡−1x\cos^{-1}x by parts

    First function cos⁡−1x\cos^{-1}x (derivative −11−x2-\frac{1}{\sqrt{1-x^2}}⁠), second function 1 (integral xx⁠):

    ∫cos⁡−1x dx=xcos⁡−1x−∫x⋅(−11−x2)dx=xcos⁡−1x+∫x dx1−x2.\begin{aligned}\int\cos^{-1}x\,dx &= x\cos^{-1}x - \int x\cdot\left(-\frac{1}{\sqrt{1-x^2}}\right)dx\\ &= x\cos^{-1}x + \int\frac{x\,dx}{\sqrt{1-x^2}} .\end{aligned}

    For the last integral put t=1−x2t = 1 - x^2⁠, dt=−2x dxdt = -2x\,dx⁠:

    ∫x dx1−x2=−12∫t−1/2 dt=−t=−1−x2.\int\frac{x\,dx}{\sqrt{1-x^2}} = -\frac12\int t^{-1/2}\,dt = -\sqrt t = -\sqrt{1-x^2} .

    Hence, multiplying by 12\frac12⁠,

    ∫tan⁡−11−x1+x dx=12[xcos⁡−1x−1−x2]+C.\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx = \frac12\left[x\cos^{-1}x - \sqrt{1-x^2}\right] + C .
    We know how to differentiate cos⁡−1x\cos^{-1}x but not how to integrate it directly, so it must be the first function. The book's rule is Remark (iii) of Section 7.6: when the other factor is a power of xx (here 1=x01 = x^0⁠) and one factor is an inverse trigonometric function, the inverse trigonometric function is taken as the first function. (The memory aid ILATE, not used in the book, orders the choice Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential.) Example 18 uses the same device of 1 as the second function for ∫log⁡x dx\int\log x\,dx⁠. Differentiating cos⁡−1x\cos^{-1}x produces an algebraic function, which a substitution then handles.
    Section 7.6, Remark (iii); Example 18 (log⁡x\log x with 1 as the second function); Section 7.3.1. Lesson: integration by parts

  5. Check: differentiate the answer

    ddx 12[xcos⁡−1x−1−x2]=12[cos⁡−1x−x1−x2+x1−x2]=12cos⁡−1x=tan⁡−11−x1+x ✓\begin{aligned}&\frac{d}{dx}\,\frac12\left[x\cos^{-1}x - \sqrt{1-x^2}\right]\\ &= \frac12\left[\cos^{-1}x - \frac{x}{\sqrt{1-x^2}} + \frac{x}{\sqrt{1-x^2}}\right]\\ &= \frac12\cos^{-1}x = \tan^{-1}\sqrt{\frac{1-x}{1+x}}\ \checkmark\end{aligned}
    Product rule on xcos⁡−1xx\cos^{-1}x⁠, and ddx1−x2=−2x21−x2\frac{d}{dx}\sqrt{1-x^2} = \frac{-2x}{2\sqrt{1-x^2}}⁠. The last equality is Step 3. Since 1−x2\sqrt{1-x^2} has no derivative at x=±1x = \pm1⁠, the answer is valid on (−1,1)(-1, 1) (and continuous up to x=1x = 1⁠).
    Derivatives of cos⁡−1x\cos^{-1}x and u\sqrt{u} (Chapter 5). Basics: integration undoes differentiation

  6. Compare with the textbook's answer

    The key gives 12[xcos⁡−1x−1−x2]+C\frac12\left[x\cos^{-1}x - \sqrt{1-x^2}\right] + C⁠, exactly our answer.

    The simplification of Step 3 is the whole trick; it holds on the entire domain (−1,1](-1, 1]⁠, so no case split is needed.
Result

For −1<x<1-1 < x < 1⁠:

∫tan⁡−11−x1+x dx=12[xcos⁡−1x−1−x2]+C\begin{aligned}&\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx\\ &= \frac12\left[x\cos^{-1}x - \sqrt{1-x^2}\right] + C\end{aligned}

(Textbook answer: 12[xcos⁡−1x−1−x2]+C\frac12\left[x\cos^{-1}x - \sqrt{1-x^2}\right] + C⁠.)

Check it with numbers

The simplification of Step 3 at three points:

xxtan⁡−11−x1+x\tan^{-1}\sqrt{\frac{1-x}{1+x}}12cos⁡−1x\frac12\cos^{-1}x
00tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}12⋅π2=π4\frac12\cdot\frac{\pi}{2} = \frac{\pi}{4}
12\frac12tan⁡−113=π6\tan^{-1}\frac{1}{\sqrt3} = \frac{\pi}{6}12⋅π3=π6\frac12\cdot\frac{\pi}{3} = \frac{\pi}{6}
−0.6-0.6tan⁡−12≈1.1071487178\tan^{-1}2 \approx 1.107148717812×2.2142974356≈1.1071487178\frac12\times 2.2142974356 \approx 1.1071487178

Area from 0 to 1: our answer FF has F(1)=0F(1) = 0 and F(0)=−12F(0) = -\frac12⁠, so F(1)−F(0)=12F(1) - F(0) = \frac12⁠. Numerical integration of the integrand from 0 to 1 also gives 0.50.5⁠.

Common mistakes
  1. Integrating straight away. Parts on tan⁡−11−x1+x\tan^{-1}\sqrt{\frac{1-x}{1+x}} as it stands is messy; simplify the function first.
  2. Getting cos⁡−1x\cos^{-1}x instead of 12cos⁡−1x\frac12\cos^{-1}x⁠. The half angle θ2\frac{\theta}{2} is what tan⁡−1\tan^{-1} returns.
  3. Ignoring the ranges. tan⁡−1(tan⁡u)=u\tan^{-1}(\tan u) = u and tan⁡2u=tan⁡u\sqrt{\tan^2u} = \tan u both need 0≤u<π20 \le u < \frac{\pi}{2} here, which is why θ\theta is taken in [0,π)[0, \pi)⁠.
What to take away
  1. For 1−x1+x\frac{1-x}{1+x} think x=cos⁡θx = \cos\theta⁠; for 1−x1+x\frac{1-\sqrt x}{1+\sqrt x} (Q19) think x=cos⁡θ\sqrt x = \cos\theta⁠.
  2. ∫cos⁡−1x dx=xcos⁡−1x−1−x2+C\int\cos^{-1}x\,dx = x\cos^{-1}x - \sqrt{1-x^2} + C⁠, by parts with 1 as the second function.

Every fact used, and where it comes from

FactUsed inSource
Principal ranges of cos⁡−1\cos^{-1} and tan⁡−1\tan^{-1}Steps 1, 3Chapter 2
Half-angle identitiesStep 2Class XI trigonometry (Basics)
Integration by parts; inverse trigonometric function as the first functionStep 4Section 7.6, Remark (iii); Example 18 does the same for log⁡x\log x (Lesson: integration by parts)
Substitution t=1−x2t = 1 - x^2⁠; power formulaStep 4Section 7.3.1; standard formula (i), Section 7.2
ddxcos⁡−1x=−11−x2\frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1-x^2}}⁠; product ruleSteps 4, 5Chapter 5
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