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Chapter 7 · Integrals

Exercise 7.6 · Q10: ∫(sin⁡−1x)2 dx\int(\sin^{-1}x)^2\,dx

Exercise 7.6Integration by parts · Section 7.6
The question

Integrate the functions in Exercises 1 to 22.

10. (sin⁡−1x)2(\sin^{-1}x)^2

Exam answer

Let I=∫(sin⁡−1x)2⋅1 dxI = \displaystyle\int(\sin^{-1}x)^2\cdot 1\,dx⁠.

Taking (sin⁡−1x)2(\sin^{-1}x)^2 as the first function, by parts:

I=(sin⁡−1x)2⋅x−∫2sin⁡−1x1−x2⋅x dx=x(sin⁡−1x)2−2K\begin{aligned} I &= (\sin^{-1}x)^2\cdot x \\ &\quad - \int\frac{2\sin^{-1}x}{\sqrt{1 - x^2}}\cdot x\,dx \\ &= x(\sin^{-1}x)^2 - 2K \end{aligned}

where K=∫xsin⁡−1x1−x2 dxK = \displaystyle\int\frac{x\sin^{-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 sin⁡−1x\sin^{-1}x as the first function, by parts:

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

∴ I=x(sin⁡−1x)2\therefore\ I = x(\sin^{-1}x)^2 +21−x2 sin⁡−1x+ 2\sqrt{1 - x^2}\,\sin^{-1}x −2x+C- 2x + C⁠.

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

The one idea this problem is built on

There is only one function here, but we can still integrate by parts by writing it as (sin⁡−1x)2⋅1(\sin^{-1}x)^2\cdot 1⁠, exactly as the book does for log⁡x\log x in Example 18. Differentiating (sin⁡−1x)2(\sin^{-1}x)^2 lowers the power of sin⁡−1x\sin^{-1}x from 2 to 1; a second integration by parts (the book's Example 20) removes it completely.

Given, to find, and the plan
Given
The function (sin⁡−1x)2(\sin^{-1}x)^2⁠, 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
∫(sin⁡−1x)2 dx\int(\sin^{-1}x)^2\,dx⁠.
Plan
(1) Name uu and dvdx\frac{dv}{dx}⁠. (2) Find dudx\frac{du}{dx} (chain rule) and vv⁠. (3) First integration by parts. (4) Second integration by parts for the leftover integral. (5) Combine. (6) Confirm with θ=sin⁡−1x\theta = \sin^{-1}x⁠. (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). When there is no second factor, use the constant 1 (Example 18).

3Derivative of sin⁡−1x\sin^{-1}x and the chain rule

ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}} for −1<x<1-1 < x < 1⁠, so ddx(sin⁡−1x)2=2sin⁡−1x1−x2\frac{d}{dx}(\sin^{-1}x)^2 = \frac{2\sin^{-1}x}{\sqrt{1-x^2}}⁠. Basics: substitution (the chain rule backwards)

4A substitution for x1−x2\frac{x}{\sqrt{1-x^2}}

With t=1−x2t = 1 - x^2⁠, dt=−2x dxdt = -2x\,dx⁠: ∫x dx1−x2=−1−x2+C\int\frac{x\,dx}{\sqrt{1-x^2}} = -\sqrt{1-x^2} + C (Section 7.3.1, as in Example 20). Basics: substitution

Solution

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

    u=(sin⁡−1x)2(first),dvdx=1(second).\begin{aligned} u &= (\sin^{-1}x)^2 \quad (\text{first}), \\ \frac{dv}{dx} &= 1 \quad (\text{second}). \end{aligned}
    We cannot integrate (sin⁡−1x)2(\sin^{-1}x)^2 directly, but we can differentiate it, and its derivative contains only the first power of sin⁡−1x\sin^{-1}x⁠. The constant 1 is the simplest possible second function: its integral is xx⁠. In ILATE terms, the inverse trigonometric factor comes first and 1 (Algebraic) second.
    Remark (iii) after Example 17 and Example 18, Section 7.6. Example 18

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

    dudx=2sin⁡−1x⋅11−x2,v=∫1 dx=x.\frac{du}{dx} = 2\sin^{-1}x\cdot\frac{1}{\sqrt{1-x^2}}, \qquad v = \int 1\,dx = x.
    Chain rule: the outer function is "square", with derivative 2(…)2(\ldots)⁠; the inner function is sin⁡−1x\sin^{-1}x⁠, with derivative 11−x2\frac{1}{\sqrt{1-x^2}} (valid for −1<x<1-1 < x < 1⁠). No constant in vv (Remark (ii) after Example 17).
    Chain rule (Chapter 5, Section 5.3.1); standard formula (viii). Basics: substitution (the chain rule backwards)

  3. First integration by parts

    ∫(sin⁡−1x)2 dx=uv−∫v dudx dx=x(sin⁡−1x)2−2∫xsin⁡−1x1−x2 dx.\begin{aligned} \int(\sin^{-1}x)^2\,dx &= uv - \int v\,\frac{du}{dx}\,dx \\ &= x(\sin^{-1}x)^2 - 2\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx. \end{aligned}
    The power of sin⁡−1x\sin^{-1}x has dropped from 2 to 1. The leftover integral K=∫xsin⁡−1x1−x2 dxK = \int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx is exactly the book's Example 20.
    Section 7.6, equation (1). Lesson: integration by parts

  4. Second integration by parts: KK

    Take u=sin⁡−1xu = \sin^{-1}x and dvdx=x1−x2\frac{dv}{dx} = \frac{x}{\sqrt{1-x^2}}⁠. Then dudx=11−x2\frac{du}{dx} = \frac{1}{\sqrt{1-x^2}} and, by t=1−x2t = 1 - x^2⁠, v=−1−x2v = -\sqrt{1-x^2}⁠. So

    K=−1−x2 sin⁡−1x−∫−1−x21−x2 dx=−1−x2 sin⁡−1x+∫1 dx=x−1−x2 sin⁡−1x.\begin{aligned} K &= -\sqrt{1-x^2}\,\sin^{-1}x - \int\frac{-\sqrt{1-x^2}}{\sqrt{1-x^2}}\,dx \\ &= -\sqrt{1-x^2}\,\sin^{-1}x + \int 1\,dx \\ &= x - \sqrt{1-x^2}\,\sin^{-1}x. \end{aligned}
    Again the inverse sine is uu⁠. The second function x1−x2\frac{x}{\sqrt{1-x^2}} integrates by a substitution because x dxx\,dx is −12-\frac12 times the differential of 1−x21 - x^2⁠. The square roots then cancel (they are non-zero on (−1,1)(-1, 1)⁠), leaving ∫1 dx\int 1\,dx⁠.
    Example 20, Section 7.6; substitution, Section 7.3.1. Example 20

  5. Combine

    ∫(sin⁡−1x)2 dx=x(sin⁡−1x)2−2K=x(sin⁡−1x)2−2x+21−x2 sin⁡−1x+C.\begin{aligned} \int(\sin^{-1}x)^2\,dx &= x(\sin^{-1}x)^2 - 2K \\ &= x(\sin^{-1}x)^2 - 2x \\ &\quad + 2\sqrt{1-x^2}\,\sin^{-1}x + C. \end{aligned}
    Multiply KK by −2-2⁠: xx becomes −2x-2x and −1−x2sin⁡−1x-\sqrt{1-x^2}\sin^{-1}x becomes +21−x2sin⁡−1x+2\sqrt{1-x^2}\sin^{-1}x⁠. One constant CC at the end.
    Distributive law. Lesson: integration by parts

  6. Confirm with θ=sin⁡−1x\theta = \sin^{-1}x

    For −1<x<1-1 < x < 1⁠, θ=sin⁡−1x\theta = \sin^{-1}x lies in (−π2,π2)\left(-\frac\pi2, \frac\pi2\right)⁠, x=sin⁡θx = \sin\theta⁠, dx=cos⁡θ dθdx = \cos\theta\,d\theta⁠, and cos⁡θ=1−x2\cos\theta = \sqrt{1-x^2} (positive). The integral becomes ∫θ2cos⁡θ dθ\int\theta^2\cos\theta\,d\theta⁠. By parts twice (u=θ2u = \theta^2⁠, then u=θu = \theta⁠):

    ∫θ2cos⁡θ dθ=θ2sin⁡θ−2∫θsin⁡θ dθ=θ2sin⁡θ+2θcos⁡θ−2sin⁡θ+C.\begin{aligned} \int\theta^2\cos\theta\,d\theta &= \theta^2\sin\theta - 2\int\theta\sin\theta\,d\theta \\ &= \theta^2\sin\theta + 2\theta\cos\theta - 2\sin\theta + C. \end{aligned}

    Back in xx (sin⁡θ=x\sin\theta = x⁠, cos⁡θ=1−x2\cos\theta = \sqrt{1-x^2}⁠): x(sin⁡−1x)2x(\sin^{-1}x)^2 +21−x2 sin⁡−1x+ 2\sqrt{1-x^2}\,\sin^{-1}x −2x+C- 2x + C⁠, the same.

    ∫θsin⁡θ dθ=−θcos⁡θ+sin⁡θ\int\theta\sin\theta\,d\theta = -\theta\cos\theta + \sin\theta (as in Question 1). The substitution is legitimate because θ\theta stays in the principal range, where sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta) = \theta and cos⁡θ>0\cos\theta > 0⁠. Two routes agreeing is a strong check.
    Substitution (Section 7.3.1); principal range of sin⁡−1\sin^{-1} (Chapter 2). Exercise 7.6 Q1

  7. Check: differentiate the answer

    ddx[x(sin⁡−1x)2]=(sin⁡−1x)2+2xsin⁡−1x1−x2,ddx[21−x2 sin⁡−1x]=−2xsin⁡−1x1−x2+2,ddx[−2x]=−2.\begin{aligned} \tfrac{d}{dx}\big[x(\sin^{-1}x)^2\big] &= (\sin^{-1}x)^2 + \frac{2x\sin^{-1}x}{\sqrt{1-x^2}}, \\ \tfrac{d}{dx}\big[2\sqrt{1-x^2}\,\sin^{-1}x\big] &= -\frac{2x\sin^{-1}x}{\sqrt{1-x^2}} + 2, \\ \tfrac{d}{dx}\big[-2x\big] &= -2. \end{aligned}

    Adding the three lines gives (sin⁡−1x)2(\sin^{-1}x)^2⁠.

    Product rule in the first two lines (in the second, 1−x2⋅11−x2=1\sqrt{1-x^2}\cdot\frac{1}{\sqrt{1-x^2}} = 1⁠, times 2). The fractions cancel and so do +2+2 and −2-2⁠, for every xx in (−1,1)(-1, 1)⁠.
    Product rule and chain rule (Chapter 5). Basics: integration undoes differentiation
Result ∫(sin⁡−1x)2 dx=x(sin⁡−1x)2+21−x2 sin⁡−1x−2x+C,\begin{aligned} &\int(\sin^{-1}x)^2\,dx \\ &= x(\sin^{-1}x)^2 + 2\sqrt{1 - x^2}\,\sin^{-1}x \\ &\quad - 2x + C, \end{aligned}

valid for −1<x<1-1 < x < 1⁠. (Textbook answer: (sin⁡−1x)2x(\sin^{-1}x)^2x +21−x2 sin⁡−1x+ 2\sqrt{1-x^2}\,\sin^{-1}x −2x+C- 2x + C⁠, the same.)

Check it with numbers

Call the answer F(x)F(x)⁠. On [0,12]\left[0, \frac12\right]⁠, with sin⁡−112=π6\sin^{-1}\frac12 = \frac\pi6 and 1−14=32\sqrt{1-\frac14} = \frac{\sqrt3}{2}⁠:

F(12)−F(0)=12⋅π236+3⋅π6−1=π272+3 π6−1=0.0439775210…\begin{aligned} F\left(\tfrac12\right) - F(0) &= \frac12\cdot\frac{\pi^2}{36} + \sqrt3\cdot\frac\pi6 - 1 \\ &= \frac{\pi^2}{72} + \frac{\sqrt3\,\pi}{6} - 1 \\ &= 0.0439775210\ldots \end{aligned}

Numerical integration of (sin⁡−1x)2(\sin^{-1}x)^2 over [0,12]\left[0, \frac12\right] gives 0.0439775210…0.0439775210\ldots⁠, the same.

Common mistakes
  1. Forgetting the chain rule. ddx(sin⁡−1x)2=2sin⁡−1x1−x2\frac{d}{dx}(\sin^{-1}x)^2 = \frac{2\sin^{-1}x}{\sqrt{1-x^2}}⁠, not 2sin⁡−1x2\sin^{-1}x⁠.
  2. Sign of vv in Step 4. ∫x1−x2 dx=−1−x2\int\frac{x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}⁠, with a minus.
  3. Signs when multiplying KK by −2-2⁠. The 1−x2sin⁡−1x\sqrt{1-x^2}\sin^{-1}x term ends up positive and the xx term negative.
What to take away
  1. A single hard-to-integrate function f(x)f(x) can be written as f(x)⋅1f(x)\cdot 1 and integrated by parts.
  2. Each integration by parts lowers the power of sin⁡−1x\sin^{-1}x by one.
  3. The substitution θ=sin⁡−1x\theta = \sin^{-1}x turns this into a familiar ∫θ2cos⁡θ dθ\int\theta^2\cos\theta\,d\theta⁠.

Every fact used, and where it comes from

FactUsed inSource
Integration by partsSteps 3, 4, 6Section 7.6, equation (1) (Lesson: integration by parts)
Inverse trigonometric function first; 1 as the second functionStep 1Remark (iii) after Example 17; Example 18, Section 7.6
No constant needed in vvSteps 2, 4Remark (ii) after Example 17, Section 7.6
ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}⁠; chain ruleSteps 2, 4, 7Standard formula (viii), Section 7.2; Chapter 5, Sections 5.3.1 and 5.3.3
∫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
Substitutions t=1−x2t = 1 - x^2 and θ=sin⁡−1x\theta = \sin^{-1}xSteps 4, 6Section 7.3.1 (Lesson: substitution)
Principal range of sin⁡−1\sin^{-1}Step 6Chapter 2, Section 2.2
Product ruleStep 7Chapter 5, Section 5.3
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