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Chapter 7 · Integrals

Exercise 7.6 · Q7: ∫xsin⁡−1x dx\int x\sin^{-1}x\,dx

Exercise 7.6Integration by parts · Section 7.6
The question

Integrate the functions in Exercises 1 to 22.

7. xsin⁡−1xx\sin^{-1}x

Exam answer

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

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

I=sin⁡−1x⋅x22−∫11−x2⋅x22 dx=x22sin⁡−1x−12J\begin{aligned} I &= \sin^{-1}x\cdot\frac{x^2}{2} \\ &\quad - \int\frac{1}{\sqrt{1 - x^2}}\cdot\frac{x^2}{2}\,dx \\ &= \frac{x^2}{2}\sin^{-1}x - \frac12 J \end{aligned}

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

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 (as cos⁡θ>0\cos\theta > 0⁠).

J=∫sin⁡2θcos⁡θ⋅cos⁡θ dθ=∫1−cos⁡2θ2 dθ=θ2−sin⁡2θ4=θ2−sin⁡θcos⁡θ2=12sin⁡−1x−12x1−x2\begin{aligned} J &= \int\frac{\sin^2\theta}{\cos\theta}\cdot\cos\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\sin^{-1}x - \frac12x\sqrt{1 - x^2} \end{aligned} ∴ I=x22sin⁡−1x−14sin⁡−1x+14x1−x2+C\begin{aligned} \therefore\ I &= \frac{x^2}{2}\sin^{-1}x - \frac14\sin^{-1}x \\ &\quad + \frac14x\sqrt{1 - x^2} + C \end{aligned}

∴ I=2x2−14sin⁡−1x\therefore\ I = \dfrac{2x^2 - 1}{4}\sin^{-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

We cannot integrate sin⁡−1x\sin^{-1}x directly at this stage, but we can differentiate it: its derivative 11−x2\frac{1}{\sqrt{1-x^2}} is algebraic. So sin⁡−1x\sin^{-1}x is uu and xx is integrated. The integral that is left, ∫x21−x2 dx\int\frac{x^2}{\sqrt{1-x^2}}\,dx⁠, contains no inverse function any more, and the substitution x=sin⁡θx = \sin\theta finishes it.

Given, to find, and the plan
Given
The function xsin⁡−1xx\sin^{-1}x⁠, defined for −1≤x≤1-1 \le x \le 1⁠. The working below divides by 1−x2\sqrt{1-x^2}⁠, so we work on the open interval −1<x<1-1 < x < 1⁠.
To find
∫xsin⁡−1x dx\int x\sin^{-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 ∫x21−x2 dx\int\frac{x^2}{\sqrt{1-x^2}}\,dx by the substitution x=sin⁡θx = \sin\theta⁠. (5) Combine and simplify. (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: when the other function is inverse trigonometric, it is the first function. Memory aid ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential; not in the book): I comes before A.

3Derivative of sin⁡−1x\sin^{-1}x

ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}} for −1<x<1-1 < x < 1 (standard formula (viii) read as a derivative). The principal range of sin⁡−1\sin^{-1} is [−π2,π2]\left[-\frac\pi2, \frac\pi2\right] (Chapter 2). Basics: trigonometric identities

4Lowering a power of sine

sin⁡2θ=1−cos⁡2θ2\sin^2\theta = \frac{1-\cos 2\theta}{2} and sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta (as used in Section 7.3.2). Basics: trigonometric identities

Solution

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

    u=sin⁡−1x(first),dvdx=x(second).\begin{aligned} u &= \sin^{-1}x \quad (\text{first}), \\ \frac{dv}{dx} &= x \quad (\text{second}). \end{aligned}
    ILATE: Inverse trigonometric before Algebraic. We have no formula for ∫sin⁡−1x dx\int\sin^{-1}x\,dx here, so sin⁡−1x\sin^{-1}x cannot be the function we integrate. Differentiating it, though, gives an algebraic expression.
    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 of sin⁡−1x\sin^{-1}x exists for −1<x<1-1 < x < 1 (this is why we work on the open interval). vv is the power rule with n=1n = 1⁠; no constant (Remark (ii) after Example 17).
    Standard formulae (viii) and (i), Section 7.2. Basics: powers of x

  3. Apply the formula

    ∫xsin⁡−1x dx=uv−∫v dudx dx=x22sin⁡−1x−12∫x21−x2 dx.\begin{aligned} \int x\sin^{-1}x\,dx &= uv - \int v\,\frac{du}{dx}\,dx \\ &= \frac{x^2}{2}\sin^{-1}x - \frac12\int\frac{x^2}{\sqrt{1-x^2}}\,dx. \end{aligned}
    The inverse sine has gone from the remaining integral; what is left is algebraic. Call it J=∫x21−x2 dxJ = \int\frac{x^2}{\sqrt{1-x^2}}\,dx⁠.
    Section 7.6, equation (1). Lesson: integration by parts

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

    Put x=sin⁡θx = \sin\theta with −π2<θ<π2-\frac\pi2 < \theta < \frac\pi2⁠, that is θ=sin⁡−1x\theta = \sin^{-1}x⁠. Then dx=cos⁡θ dθdx = \cos\theta\,d\theta⁠, and since cos⁡θ>0\cos\theta > 0 here, 1−x2=cos⁡2θ=cos⁡θ\sqrt{1-x^2} = \sqrt{\cos^2\theta} = \cos\theta⁠. So

    J=∫sin⁡2θcos⁡θcos⁡θ dθ=∫sin⁡2θ dθ=∫1−cos⁡2θ2 dθ=θ2−sin⁡2θ4=θ2−sin⁡θcos⁡θ2.\begin{aligned} J &= \int\frac{\sin^2\theta}{\cos\theta}\cos\theta\,d\theta = \int\sin^2\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}. \end{aligned}

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

    J=12sin⁡−1x−12 x1−x2.J = \frac12\sin^{-1}x - \frac12\,x\sqrt{1-x^2}.
    The substitution is legitimate: as xx runs over (−1,1)(-1, 1)⁠, θ=sin⁡−1x\theta = \sin^{-1}x runs over (−π2,π2)\left(-\frac\pi2, \frac\pi2\right)⁠, inside the principal range, so sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta) = \theta⁠; and there cos⁡θ>0\cos\theta > 0⁠, so cos⁡2θ=cos⁡θ\sqrt{\cos^2\theta} = \cos\theta with no sign change. The square of the sine is lowered with the double-angle identity, as in Section 7.3.2. (A second route: x21−x2=11−x2−1−x2\frac{x^2}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}} - \sqrt{1-x^2} and the formula for ∫a2−x2 dx\int\sqrt{a^2-x^2}\,dx of Section 7.6.2 give the same JJ⁠.)
    Substitution (Section 7.3.1); sin⁡2θ=1−cos⁡2θ2\sin^2\theta = \frac{1-\cos2\theta}{2}⁠; principal range of sin⁡−1\sin^{-1} (Chapter 2). Basics: trigonometric identities

  5. Combine and simplify

    ∫xsin⁡−1x dx=x22sin⁡−1x−12J=x22sin⁡−1x−14sin⁡−1x+14x1−x2+C=2x2−14sin⁡−1x+x1−x24+C.\begin{aligned} \int x\sin^{-1}x\,dx &= \frac{x^2}{2}\sin^{-1}x - \frac12 J \\ &= \frac{x^2}{2}\sin^{-1}x - \frac14\sin^{-1}x + \frac14x\sqrt{1-x^2} + C \\ &= \frac{2x^2 - 1}{4}\sin^{-1}x + \frac{x\sqrt{1-x^2}}{4} + C. \end{aligned}
    Multiply JJ by −12-\frac12 (both signs flip), then collect the two sin⁡−1x\sin^{-1}x terms: x22−14=2x2−14\frac{x^2}{2} - \frac14 = \frac{2x^2 - 1}{4}⁠. One constant CC at the end.
    Collecting like terms. Lesson: integration by parts

  6. Check: differentiate the answer

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

    ddx[2x2−14sin⁡−1x]=xsin⁡−1x+2x2−141−x2,ddx[x1−x24]=14(1−x2−x21−x2)=1−2x241−x2.\begin{aligned} \frac{d}{dx}\Big[\frac{2x^2-1}{4}\sin^{-1}x\Big] &= x\sin^{-1}x + \frac{2x^2 - 1}{4\sqrt{1-x^2}}, \\ \frac{d}{dx}\Big[\frac{x\sqrt{1-x^2}}{4}\Big] &= \frac14\left(\sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\right) \\ &= \frac{1 - 2x^2}{4\sqrt{1-x^2}}. \end{aligned}

    Adding, the two fractions cancel and the sum is xsin⁡−1xx\sin^{-1}x⁠.

    The derivative equals the integrand at every point of (−1,1)(-1, 1)⁠. In the second line, 1−x2=1−x21−x2\sqrt{1-x^2} = \frac{1-x^2}{\sqrt{1-x^2}} puts everything over one denominator.
    Product rule and chain rule (Chapter 5). Basics: integration undoes differentiation
Result ∫xsin⁡−1x dx=2x2−14sin⁡−1x+x1−x24+C,\begin{aligned} &\int x\sin^{-1}x\,dx \\ &= \frac{2x^2 - 1}{4}\sin^{-1}x + \frac{x\sqrt{1 - x^2}}{4} + C, \end{aligned}

valid for −1<x<1-1 < x < 1⁠. (Textbook answer: 14(2x2−1)sin⁡−1x+x1−x24+C\frac14(2x^2 - 1)\sin^{-1}x + \frac{x\sqrt{1-x^2}}{4} + 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)=−18⋅π6+316−0=316−π48=0.0428033285…\begin{aligned} F\left(\tfrac12\right) - F(0) &= -\frac18\cdot\frac\pi6 + \frac{\sqrt3}{16} - 0 \\ &= \frac{\sqrt3}{16} - \frac{\pi}{48} = 0.0428033285\ldots \end{aligned}

Numerical integration of xsin⁡−1xx\sin^{-1}x over [0,12]\left[0, \frac12\right] gives 0.0428033285…0.0428033285\ldots⁠, the same.

Common mistakes
  1. Taking u=xu = x⁠. Then v=∫sin⁡−1x dxv = \int\sin^{-1}x\,dx is needed, which we do not yet have.
  2. cos⁡2θ=cos⁡θ\sqrt{\cos^2\theta} = \cos\theta without a reason. It holds because θ\theta lies in (−π2,π2)\left(-\frac\pi2, \frac\pi2\right)⁠, where cos⁡θ>0\cos\theta > 0⁠.
  3. Signs when multiplying JJ by −12-\frac12⁠. The x1−x2x\sqrt{1-x^2} term ends up with a plus sign.
What to take away
  1. Inverse trigonometric function times a power of xx⁠: the inverse function is uu⁠.
  2. The leftover integral often needs a second method; x=sin⁡θx = \sin\theta suits 1−x2\sqrt{1-x^2}⁠.
  3. For every trigonometric substitution, state the range of θ\theta and why cos⁡2θ=cos⁡θ\sqrt{\cos^2\theta} = \cos\theta there.

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
ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}⁠, ∣x∣<1|x| < 1Steps 2, 6Standard formula (viii), Section 7.2; Chapter 5, Section 5.3.3
Substitution x=sin⁡θx = \sin\thetaStep 4Section 7.3.1 (Lesson: substitution)
Principal range [−π2,π2]\left[-\frac\pi2, \frac\pi2\right] of sin⁡−1\sin^{-1}Step 4Chapter 2, Section 2.2
sin⁡2θ=1−cos⁡2θ2\sin^2\theta = \frac{1-\cos 2\theta}{2}⁠, sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta = 2\sin\theta\cos\thetaStep 4Class XI trigonometry; used in Section 7.3.2
Product and chain rulesStep 6Chapter 5, Sections 5.3 and 5.3.1
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