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Chapter 7 · Integrals

Exercise 7.6 · Q9: ∫xcos⁡−1x dx\int x\cos^{-1}x\,dx

Exercise 7.6Integration by parts · Section 7.6
The question

Integrate the functions in Exercises 1 to 22.

9. xcos⁡−1xx\cos^{-1}x

Exam answer

Let I=∫xcos⁡−1x dxI = \displaystyle\int x\cos^{-1}x\,dx⁠.

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

I=cos⁡−1x⋅x22−∫(−11−x2)x22 dx=x22cos⁡−1x+12J\begin{aligned} I &= \cos^{-1}x\cdot\frac{x^2}{2} \\ &\quad - \int\left(-\frac{1}{\sqrt{1 - x^2}}\right)\frac{x^2}{2}\,dx \\ &= \frac{x^2}{2}\cos^{-1}x + \frac12 J \end{aligned}

where J=∫x2 dx1−x2J = \displaystyle\int\frac{x^2\,dx}{\sqrt{1 - x^2}}⁠.

Put x=cos⁡θx = \cos\theta⁠, 0<θ<π0 < \theta < \pi⁠. Then dx=−sin⁡θ dθdx = -\sin\theta\,d\theta⁠, and 1−x2=sin⁡θ\sqrt{1 - x^2} = \sin\theta (as sin⁡θ>0\sin\theta > 0⁠).

J=∫cos⁡2θsin⁡θ⋅(−sin⁡θ) dθ=−∫1+cos⁡2θ2 dθ=−θ2−sin⁡2θ4=−θ2−sin⁡θcos⁡θ2=−12cos⁡−1x−12x1−x2\begin{aligned} J &= \int\frac{\cos^2\theta}{\sin\theta}\cdot(-\sin\theta)\,d\theta \\ &= -\int\frac{1 + \cos 2\theta}{2}\,d\theta \\ &= -\frac\theta2 - \frac{\sin 2\theta}{4} \\ &= -\frac\theta2 - \frac{\sin\theta\cos\theta}{2} \\ &= -\frac12\cos^{-1}x - \frac12x\sqrt{1 - x^2} \end{aligned} ∴ I=x22cos⁡−1x−14cos⁡−1x−14x1−x2+C\begin{aligned} \therefore\ I &= \frac{x^2}{2}\cos^{-1}x - \frac14\cos^{-1}x \\ &\quad - \frac14x\sqrt{1 - x^2} + C \end{aligned}

∴ I=2x2−14cos⁡−1x\therefore\ I = \dfrac{2x^2 - 1}{4}\cos^{-1}x −x1−x24+C- \dfrac{x\sqrt{1 - x^2}}{4} + C⁠.

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

The one idea this problem is built on

This is Question 7 with cos⁡−1x\cos^{-1}x instead of sin⁡−1x\sin^{-1}x⁠. The derivative of cos⁡−1x\cos^{-1}x is the derivative of sin⁡−1x\sin^{-1}x with a minus sign, so the same leftover integral ∫x21−x2 dx\int\frac{x^2}{\sqrt{1-x^2}}\,dx appears, with the opposite sign. At the end the identity sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac\pi2 turns our answer into the book's, at the cost of a constant π8\frac\pi8⁠, which the arbitrary constant CC absorbs.

Given, to find, and the plan
Given
The function xcos⁡−1xx\cos^{-1}x⁠, defined for −1≤x≤1-1 \le x \le 1⁠. The working divides by 1−x2\sqrt{1-x^2}⁠, so we work on −1<x<1-1 < x < 1⁠.
To find
∫xcos⁡−1x dx\int x\cos^{-1}x\,dx⁠.
Plan
(1) Name uu and dvdx\frac{dv}{dx}⁠. (2) Find dudx\frac{du}{dx} and vv⁠. (3) Apply the formula. (4) Find J=∫x21−x2 dxJ = \int\frac{x^2}{\sqrt{1-x^2}}\,dx (as in Question 7). (5) Combine. (6) Rewrite in terms of cos⁡−1x\cos^{-1}x only. (7) Differentiate back.

Basics you need

1Integration by parts

Section 7.6, equation (1):

∫u dvdx dx=uv−∫v dudx dx,\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx,

with uu the first function (differentiated) and dvdx\frac{dv}{dx} the second (integrated to give vv⁠). Lesson: integration by parts

2Which factor should be uu⁠?

Remark (iii) after Example 17: an inverse trigonometric function is the first function. Memory aid ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential; not in the book): I before A.

3Derivative of cos⁡−1x\cos^{-1}x

ddxcos⁡−1x=−11−x2\frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1-x^2}} for −1<x<1-1 < x < 1 (standard formula (ix) read as a derivative). Basics: trigonometric identities

4Complementary inverse functions

sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac\pi2 for −1≤x≤1-1 \le x \le 1⁠. Reason: let θ=sin⁡−1x\theta = \sin^{-1}x⁠, so θ\theta is in [−π2,π2]\left[-\frac\pi2, \frac\pi2\right] and sin⁡θ=x\sin\theta = x⁠. Then π2−θ\frac\pi2 - \theta is in [0,π][0, \pi]⁠, the principal range of cos⁡−1\cos^{-1}⁠, and cos⁡(π2−θ)=sin⁡θ=x\cos\left(\frac\pi2 - \theta\right) = \sin\theta = x⁠. Hence cos⁡−1x=π2−θ\cos^{-1}x = \frac\pi2 - \theta⁠. Basics: trigonometric identities

Solution

  1. Name uu and dvdx\frac{dv}{dx}

    u=cos⁡−1x(first),dvdx=x(second).\begin{aligned} u &= \cos^{-1}x \quad (\text{first}), \\ \frac{dv}{dx} &= x \quad (\text{second}). \end{aligned}
    ILATE: Inverse trigonometric before Algebraic. We cannot integrate cos⁡−1x\cos^{-1}x yet, but its derivative is algebraic.
    Remark (iii) after Example 17, Section 7.6. Lesson: integration by parts

  2. Find dudx\frac{du}{dx} and vv

    dudx=−11−x2,v=∫x dx=x22.\frac{du}{dx} = -\frac{1}{\sqrt{1-x^2}}, \qquad v = \int x\,dx = \frac{x^2}{2}.
    The derivative exists for −1<x<1-1 < x < 1⁠; note its minus sign. vv by the power rule, with no constant (Remark (ii) after Example 17).
    Standard formulae (ix) and (i), Section 7.2. Basics: powers of x

  3. Apply the formula

    ∫xcos⁡−1x dx=uv−∫v dudx dx=x22cos⁡−1x+12∫x21−x2 dx.\begin{aligned} \int x\cos^{-1}x\,dx &= uv - \int v\,\frac{du}{dx}\,dx \\ &= \frac{x^2}{2}\cos^{-1}x + \frac12\int\frac{x^2}{\sqrt{1-x^2}}\,dx. \end{aligned}
    The minus in the formula and the minus in dudx\frac{du}{dx} make a plus. The leftover integral is J=∫x21−x2 dxJ = \int\frac{x^2}{\sqrt{1-x^2}}\,dx⁠, the same one as in Question 7.
    Section 7.6, equation (1). Lesson: integration by parts

  4. Find JJ with x=sin⁡θx = \sin\theta

    Put x=sin⁡θx = \sin\theta⁠, −π2<θ<π2-\frac\pi2 < \theta < \frac\pi2⁠. Then dx=cos⁡θ dθdx = \cos\theta\,d\theta and 1−x2=cos⁡θ\sqrt{1-x^2} = \cos\theta (positive here), so

    J=∫sin⁡2θ dθ=∫1−cos⁡2θ2 dθ=θ2−sin⁡θcos⁡θ2=12sin⁡−1x−12 x1−x2.\begin{aligned} J &= \int\sin^2\theta\,d\theta = \int\frac{1-\cos 2\theta}{2}\,d\theta \\ &= \frac\theta2 - \frac{\sin\theta\cos\theta}{2} \\ &= \frac12\sin^{-1}x - \frac12\,x\sqrt{1-x^2}. \end{aligned}
    For xx in (−1,1)(-1, 1)⁠, θ=sin⁡−1x\theta = \sin^{-1}x lies in (−π2,π2)\left(-\frac\pi2, \frac\pi2\right)⁠, inside the principal range, where cos⁡θ>0\cos\theta > 0⁠; so cos⁡2θ=cos⁡θ\sqrt{\cos^2\theta} = \cos\theta⁠. Question 7, Step 4 has every detail.
    Substitution (Section 7.3.1); sin⁡2θ=1−cos⁡2θ2\sin^2\theta = \frac{1-\cos 2\theta}{2}⁠. Exercise 7.6 Q7

  5. Combine

    ∫xcos⁡−1x dx=x22cos⁡−1x+12J=x22cos⁡−1x+14sin⁡−1x−14 x1−x2+C.\begin{aligned} \int x\cos^{-1}x\,dx &= \frac{x^2}{2}\cos^{-1}x + \frac12 J \\ &= \frac{x^2}{2}\cos^{-1}x + \frac14\sin^{-1}x \\ &\quad - \frac14\,x\sqrt{1-x^2} + C. \end{aligned}
    This is already a correct answer: it differentiates back to xcos⁡−1xx\cos^{-1}x⁠. It mixes sin⁡−1x\sin^{-1}x and cos⁡−1x\cos^{-1}x⁠, though, and the book writes its answer with cos⁡−1x\cos^{-1}x only.
    Collecting terms. Lesson: integration by parts

  6. Rewrite with cos⁡−1x\cos^{-1}x only

    Use sin⁡−1x=π2−cos⁡−1x\sin^{-1}x = \frac\pi2 - \cos^{-1}x⁠:

    x22cos⁡−1x+14sin⁡−1x=x22cos⁡−1x+π8−14cos⁡−1x=2x2−14cos⁡−1x+π8.\begin{aligned} \frac{x^2}{2}\cos^{-1}x + \frac14\sin^{-1}x &= \frac{x^2}{2}\cos^{-1}x + \frac\pi8 - \frac14\cos^{-1}x \\ &= \frac{2x^2 - 1}{4}\cos^{-1}x + \frac\pi8. \end{aligned}

    So

    ∫xcos⁡−1x dx=2x2−14cos⁡−1x−x1−x24+π8+C.\begin{aligned} \int x\cos^{-1}x\,dx &= \frac{2x^2-1}{4}\cos^{-1}x - \frac{x\sqrt{1-x^2}}{4} \\ &\quad + \frac\pi8 + C. \end{aligned}

    Since CC is an arbitrary constant, so is π8+C\frac\pi8 + C⁠; renaming it CC gives the book's form.

    Two antiderivatives of the same function may differ by a constant. Here the difference is exactly π8\frac\pi8⁠, because 14sin⁡−1x\frac14\sin^{-1}x and −14cos⁡−1x-\frac14\cos^{-1}x differ by 14⋅π2\frac14\cdot\frac\pi2⁠.
    sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac\pi2 (Basics item 4); Property (II), Section 7.2.1. Basics: trigonometric identities

  7. Check: differentiate the answer

    Product rule, with ddxcos⁡−1x=−11−x2\frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1-x^2}} and ddx1−x2=−x1−x2\frac{d}{dx}\sqrt{1-x^2} = \frac{-x}{\sqrt{1-x^2}}⁠:

    ddx[2x2−14cos⁡−1x]=xcos⁡−1x−2x2−141−x2,ddx[−x1−x24]=2x2−141−x2.\begin{aligned} \frac{d}{dx}\Big[\frac{2x^2-1}{4}\cos^{-1}x\Big] &= x\cos^{-1}x - \frac{2x^2-1}{4\sqrt{1-x^2}}, \\ \frac{d}{dx}\Big[-\frac{x\sqrt{1-x^2}}{4}\Big] &= \frac{2x^2-1}{4\sqrt{1-x^2}}. \end{aligned}

    Adding, the fractions cancel and the sum is xcos⁡−1xx\cos^{-1}x⁠.

    The second line: −14(1−x2−x21−x2)-\frac14\left(\sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\right) =−1−2x241−x2= -\frac{1-2x^2}{4\sqrt{1-x^2}}⁠. The derivative equals the integrand at every point of (−1,1)(-1, 1)⁠.
    Product rule and chain rule (Chapter 5). Basics: integration undoes differentiation
Result ∫xcos⁡−1x dx=2x2−14cos⁡−1x−x1−x24+C,\begin{aligned} &\int x\cos^{-1}x\,dx \\ &= \frac{2x^2 - 1}{4}\cos^{-1}x - \frac{x\sqrt{1 - x^2}}{4} + C, \end{aligned}

valid for −1<x<1-1 < x < 1⁠. (Textbook answer: (2x2−1)cos⁡−1x4−x41−x2+C(2x^2 - 1)\frac{\cos^{-1}x}{4} - \frac x4\sqrt{1-x^2} + C⁠, the same.) The form in Step 5 is also correct; it differs from this one by the constant π8\frac\pi8⁠.

Check it with numbers

Call the answer F(x)F(x)⁠. On [0,12]\left[0, \frac12\right]⁠, with cos⁡−10=π2\cos^{-1}0 = \frac\pi2⁠, cos⁡−112=π3\cos^{-1}\frac12 = \frac\pi3⁠:

F(12)=−18⋅π3−316,F(0)=−π8,F(12)−F(0)=π12−316=0.1535462123…\begin{aligned} F\left(\tfrac12\right) &= -\frac18\cdot\frac\pi3 - \frac{\sqrt3}{16}, \quad F(0) = -\frac\pi8, \\ F\left(\tfrac12\right) - F(0) &= \frac{\pi}{12} - \frac{\sqrt3}{16} = 0.1535462123\ldots \end{aligned}

Numerical integration of xcos⁡−1xx\cos^{-1}x over [0,12]\left[0, \frac12\right] gives 0.1535462123…0.1535462123\ldots⁠, the same.

Common mistakes
  1. Missing the minus in ddxcos⁡−1x\frac{d}{dx}\cos^{-1}x⁠. It flips the sign of the whole leftover integral.
  2. Thinking the two forms disagree. Step 5 and the book differ only by π8\frac\pi8⁠, a constant, so both are right.
  3. Using cos⁡2θ=cos⁡θ\sqrt{\cos^2\theta} = \cos\theta without checking that cos⁡θ≥0\cos\theta \ge 0 on the range of θ\theta⁠.
What to take away
  1. cos⁡−1x\cos^{-1}x behaves like sin⁡−1x\sin^{-1}x with an extra minus sign in the derivative.
  2. sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac\pi2 converts between forms; the π2\frac\pi2 only changes the constant.
  3. Answers to indefinite integrals that look different may differ by a constant; check by differentiating.

Every fact used, and where it comes from

FactUsed inSource
Integration by partsStep 3Section 7.6, equation (1) (Lesson: integration by parts)
Inverse trigonometric function as the first function (ILATE)Step 1Remark (iii) after Example 17, Section 7.6
No constant needed in vvStep 2Remark (ii) after Example 17, Section 7.6
ddxcos⁡−1x=−11−x2\frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1-x^2}}⁠, ∣x∣<1|x| < 1Steps 2, 7Standard formula (ix), Section 7.2; Chapter 5, Section 5.3.3
Substitution x=sin⁡θx = \sin\theta and sin⁡2θ=1−cos⁡2θ2\sin^2\theta = \frac{1-\cos2\theta}{2}Step 4Section 7.3.1; Section 7.3.2 (Lesson: substitution)
sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac\pi2Step 6Proved in Basics item 4 from the principal ranges, Chapter 2, Section 2.2
Antiderivatives differ by a constantStep 6Property (II), Section 7.2.1
Product and chain rulesStep 7Chapter 5, Sections 5.3 and 5.3.1
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