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Chapter 7 · Integrals

Exercise 7.6 · Q11: ∫xcos⁡−1x1−x2 dx\int\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx

Exercise 7.6Integration by parts · Section 7.6
The question

Integrate the functions in Exercises 1 to 22.

11. xcos⁡−1x1−x2\dfrac{x\cos^{-1}x}{\sqrt{1-x^2}}

Exam answer

Let I=∫xcos⁡−1x1−x2 dxI = \displaystyle\int\frac{x\cos^{-1}x}{\sqrt{1 - x^2}}\,dx⁠.

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}

Taking cos⁡−1x\cos^{-1}x as the first function and x1−x2\frac{x}{\sqrt{1 - x^2}} as the second, by parts:

I=cos⁡−1x(−1−x2)−∫(−11−x2)×(−1−x2)dx=−1−x2 cos⁡−1x−∫1 dx\begin{aligned} I &= \cos^{-1}x\left(-\sqrt{1 - x^2}\right) \\ &\quad - \int\left(-\frac{1}{\sqrt{1 - x^2}}\right) \\ &\qquad \times\left(-\sqrt{1 - x^2}\right)dx \\ &= -\sqrt{1 - x^2}\,\cos^{-1}x - \int 1\,dx \end{aligned}

∴ I=−1−x2 cos⁡−1x−x+C\therefore\ I = -\sqrt{1 - x^2}\,\cos^{-1}x - x + C⁠.

The answer key prints −1−x2 cos⁡−1x+x+C-\sqrt{1-x^2}\,\cos^{-1}x + x + C⁠; read literally this is not an antiderivative, but read with a pair of brackets, −(1−x2 cos⁡−1x+x)+C-\big(\sqrt{1-x^2}\,\cos^{-1}x + x\big) + C⁠, it is the same as the answer above.

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

The one idea this problem is built on

Split the integrand as cos⁡−1x\cos^{-1}x times x1−x2\frac{x}{\sqrt{1-x^2}}⁠. The inverse cosine is easy to differentiate, to −11−x2-\frac{1}{\sqrt{1-x^2}}⁠; the other factor integrates by a substitution to −1−x2-\sqrt{1-x^2}⁠. In the formula these multiply to exactly +1+1⁠, so the leftover integral is just ∫1 dx\int 1\,dx⁠. This is the book's Example 20 with cos⁡−1\cos^{-1} in place of sin⁡−1\sin^{-1}⁠, and the extra minus sign in the derivative changes the sign of the xx term.

Given, to find, and the plan
Given
xcos⁡−1x1−x2\frac{x\cos^{-1}x}{\sqrt{1-x^2}}⁠. The square root in the denominator must be defined and non-zero, so 1−x2>01 - x^2 > 0⁠: the domain is −1<x<1-1 < x < 1⁠.
To find
Its indefinite integral on (−1,1)(-1, 1)⁠.
Plan
(1) Name uu and dvdx\frac{dv}{dx}⁠. (2) Find dudx\frac{du}{dx}⁠, and vv by the substitution t=1−x2t = 1 - x^2⁠. (3) Apply the formula. (4) Confirm by a second route through Example 20. (5) Compare with the answer key. (6) 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).

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

4Substitution t=1−x2t = 1 - x^2

dt=−2x dxdt = -2x\,dx⁠, so x dx=−12 dtx\,dx = -\frac12\,dt⁠, and ∫t−1/2 dt=2t+C\int t^{-1/2}\,dt = 2\sqrt t + C (Section 7.3.1 and formula (i)). Basics: substitution (the chain rule backwards)

Solution

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

    u=cos⁡−1x(first),dvdx=x1−x2(second).\begin{aligned} u &= \cos^{-1}x \quad (\text{first}), \\ \frac{dv}{dx} &= \frac{x}{\sqrt{1-x^2}} \quad (\text{second}). \end{aligned}
    ILATE: the Inverse trigonometric factor is uu⁠. We cannot integrate cos⁡−1x\cos^{-1}x at this stage, but its derivative is algebraic. The rest of the integrand, x1−x2\frac{x}{\sqrt{1-x^2}}⁠, can be integrated by a substitution, so it is a good second function.
    Remark (iii) after Example 17, Section 7.6; same split as Example 20. Example 20

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

    dudx=−11−x2\frac{du}{dx} = -\frac{1}{\sqrt{1-x^2}}⁠. For vv⁠, put t=1−x2t = 1 - x^2 (so tt is in (0,1](0, 1] and t\sqrt t is defined), x dx=−12 dtx\,dx = -\frac12\,dt⁠:

    v=∫x dx1−x2=−12∫t−1/2 dt=−12⋅2t=−1−x2.\begin{aligned} v &= \int\frac{x\,dx}{\sqrt{1-x^2}} = -\frac12\int t^{-1/2}\,dt \\ &= -\frac12\cdot 2\sqrt t = -\sqrt{1-x^2}. \end{aligned}
    The numerator x dxx\,dx is −12-\frac12 times the differential of the expression under the root, which is the signal for this substitution. Quick test: ddx(−1−x2)=x1−x2\frac{d}{dx}\left(-\sqrt{1-x^2}\right) = \frac{x}{\sqrt{1-x^2}}⁠. No constant in vv (Remark (ii) after Example 17).
    Standard formula (ix); substitution, Section 7.3.1. Basics: substitution (the chain rule backwards)

  3. Apply the formula

    ∫xcos⁡−1x1−x2 dx=cos⁡−1x⋅(−1−x2)−∫(−1−x2)(−11−x2)dx=−1−x2 cos⁡−1x−∫1 dx=−1−x2 cos⁡−1x−x+C.\begin{aligned} &\int\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx \\ &= \cos^{-1}x\cdot\left(-\sqrt{1-x^2}\right) \\ &\quad - \int\left(-\sqrt{1-x^2}\right)\left(-\frac{1}{\sqrt{1-x^2}}\right)dx \\ &= -\sqrt{1-x^2}\,\cos^{-1}x - \int 1\,dx \\ &= -\sqrt{1-x^2}\,\cos^{-1}x - x + C. \end{aligned}
    v⋅dudxv\cdot\frac{du}{dx} is a product of two negative quantities, so it equals +1+1 (the square roots cancel since they are non-zero on (−1,1)(-1, 1)⁠). The formula subtracts its integral, so we get −∫1 dx=−x-\int 1\,dx = -x⁠. Compare Example 20, where dudx\frac{du}{dx} is positive and the result has +x+x⁠.
    Section 7.6, equation (1); ∫1 dx=x+C\int 1\,dx = x + C⁠. Lesson: integration by parts

  4. Second route through Example 20

    Since cos⁡−1x=π2−sin⁡−1x\cos^{-1}x = \frac\pi2 - \sin^{-1}x on [−1,1][-1, 1] (proof: if θ=sin⁡−1x\theta = \sin^{-1}x then π2−θ\frac\pi2 - \theta lies in [0,π][0, \pi] and its cosine is sin⁡θ=x\sin\theta = x⁠), the integrand is

    π2⋅x1−x2−xsin⁡−1x1−x2.\frac\pi2\cdot\frac{x}{\sqrt{1-x^2}} - \frac{x\sin^{-1}x}{\sqrt{1-x^2}}.

    Integrate each piece (Step 2 for the first; Example 20 for the second):

    −π21−x2−(x−1−x2 sin⁡−1x)=−1−x2(π2−sin⁡−1x)−x=−1−x2 cos⁡−1x−x.\begin{aligned} &-\frac\pi2\sqrt{1-x^2} - \left(x - \sqrt{1-x^2}\,\sin^{-1}x\right) \\ &= -\sqrt{1-x^2}\left(\frac\pi2 - \sin^{-1}x\right) - x \\ &= -\sqrt{1-x^2}\,\cos^{-1}x - x. \end{aligned}
    An independent route, using the book's own worked result, lands on the same answer, including the sign of xx⁠.
    Example 20, Section 7.6; principal ranges, Chapter 2, Section 2.2. Example 20

  5. Compare with the answer key

    The answer key (book page 444), as printed in this reprint, reads −1−x2 cos⁡−1x+x+C-\sqrt{1-x^2}\,\cos^{-1}x + x + C⁠. Read literally, put G(x)=−1−x2 cos⁡−1x+xG(x) = -\sqrt{1-x^2}\,\cos^{-1}x + x and differentiate:

    G′(x)=xcos⁡−1x1−x2+1−x2⋅11−x2+1=xcos⁡−1x1−x2+2.\begin{aligned} G'(x) &= \frac{x\cos^{-1}x}{\sqrt{1-x^2}} + \sqrt{1-x^2}\cdot\frac{1}{\sqrt{1-x^2}} + 1 \\ &= \frac{x\cos^{-1}x}{\sqrt{1-x^2}} + 2. \end{aligned}

    This is the integrand plus 2, so the printed expression, read literally, is not an antiderivative.

    The printed line also has wide blank gaps just after the leading minus sign and just before "+ C+\,C". In the PDF these gaps hold blank characters, stacked in three pieces at each end, exactly where a tall pair of brackets would stand, but nothing visible is drawn there. With the brackets in place the key reads

    −(1−x2 cos⁡−1x+x)+C,-\left(\sqrt{1-x^2}\,\cos^{-1}x + x\right) + C,

    which is exactly our answer. So the key most likely intends our answer; as printed, without the brackets, it shows +x+x where the correct sign is −x-x⁠.

    The literal reading and our answer differ by 2x2x⁠, which is not a constant, so they cannot both be antiderivatives. Steps 3 and 4 (two routes) and Step 6 (differentiating back) all give −x-x⁠. Opening the brackets of the bracketed reading gives −1−x2 cos⁡−1x-\sqrt{1-x^2}\,\cos^{-1}x and −x-x⁠: our answer.
    Property (II), Section 7.2.1: antiderivatives of one function differ only by a constant. Basics: integration undoes differentiation

  6. Check: differentiate the answer

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

    ddx[−1−x2 cos⁡−1x−x]=xcos⁡−1x1−x2−1−x2(−11−x2)−1=xcos⁡−1x1−x2+1−1=xcos⁡−1x1−x2.\begin{aligned} &\frac{d}{dx}\Big[-\sqrt{1-x^2}\,\cos^{-1}x - x\Big] \\ &= \frac{x\cos^{-1}x}{\sqrt{1-x^2}} - \sqrt{1-x^2}\left(-\frac{1}{\sqrt{1-x^2}}\right) - 1 \\ &= \frac{x\cos^{-1}x}{\sqrt{1-x^2}} + 1 - 1 \\ &= \frac{x\cos^{-1}x}{\sqrt{1-x^2}}. \end{aligned}
    The +1+1 from the product rule is cancelled by the −1-1 from −x-x⁠, and exactly the integrand remains at every point of (−1,1)(-1, 1)⁠.
    Product rule and chain rule (Chapter 5). Basics: integration undoes differentiation
Result ∫xcos⁡−1x1−x2 dx=−1−x2 cos⁡−1x−x+C,\int\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx = -\sqrt{1 - x^2}\,\cos^{-1}x - x + C,

valid for −1<x<1-1 < x < 1⁠. (Textbook answer, as printed: −1−x2 cos⁡−1x+x+C-\sqrt{1-x^2}\,\cos^{-1}x + x + C⁠. Read literally this is not an antiderivative: its derivative is the integrand plus 2 (Step 5). The printed line leaves blank the places where a pair of brackets would stand; read as −(1−x2cos⁡−1x+x)+C-\big(\sqrt{1-x^2}\cos^{-1}x + x\big) + C⁠, it is the same as our answer.)

Check it with numbers

With F(x)=−1−x2 cos⁡−1x−xF(x) = -\sqrt{1-x^2}\,\cos^{-1}x - x⁠, on [0,12]\left[0, \frac12\right]⁠, where cos⁡−10=π2\cos^{-1}0 = \frac\pi2 and cos⁡−112=π3\cos^{-1}\frac12 = \frac\pi3⁠:

F(12)−F(0)=(−32⋅π3−12)+π2=π2−3 π6−12=0.1638966447…\begin{aligned} F\left(\tfrac12\right) - F(0) &= \left(-\frac{\sqrt3}{2}\cdot\frac\pi3 - \frac12\right) + \frac\pi2 \\ &= \frac\pi2 - \frac{\sqrt3\,\pi}{6} - \frac12 \\ &= 0.1638966447\ldots \end{aligned}

Numerical integration of the integrand over [0,12]\left[0, \frac12\right] gives 0.1638966447…0.1638966447\ldots⁠, the same. The key's expression read literally (without brackets) would give 1.1638966447…1.1638966447\ldots⁠, too large by exactly 2⋅12=12\cdot\frac12 = 1⁠.

Common mistakes
  1. Copying Example 20's +x+x⁠. With cos⁡−1x\cos^{-1}x⁠, dudx\frac{du}{dx} is negative, which turns +x+x into −x-x⁠.
  2. Sign of vv⁠. ∫x1−x2 dx=−1−x2\int\frac{x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}⁠.
  3. Ignoring the domain. The integrand and the answer exist only for −1<x<1-1 < x < 1⁠.
What to take away
  1. When dudx⋅v\frac{du}{dx}\cdot v is a constant, the leftover integral is immediate; watch its sign.
  2. sin⁡−1x\sin^{-1}x and cos⁡−1x\cos^{-1}x answers differ in signs because their derivatives differ in sign.
  3. Always differentiate back, even when comparing with an answer key: a printed answer can lose a bracket or a sign.

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, 5, 6Standard formula (ix), Section 7.2; Chapter 5, Section 5.3.3
Substitution t=1−x2t = 1 - x^2Step 2Section 7.3.1 (Lesson: substitution)
∫xsin⁡−1x1−x2 dx\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx =x−1−x2sin⁡−1x+C= x - \sqrt{1-x^2}\sin^{-1}x + CStep 4Example 20, Section 7.6
cos⁡−1x=π2−sin⁡−1x\cos^{-1}x = \frac\pi2 - \sin^{-1}xStep 4Proved in Step 4 from the principal ranges, Chapter 2, Section 2.2
Antiderivatives differ only by a constantStep 5Property (II), Section 7.2.1
Answer key entryStep 5Answers, Exercise 7.6 Q11 (book page 444)
Product and chain rulesSteps 5, 6Chapter 5, Sections 5.3 and 5.3.1
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