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Chapter 7 · Integrals

Exercise 7.2 · Q23: ∫sin⁡−1x1−x2 dx\int\frac{\sin^{-1}x}{\sqrt{1-x^2}}\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function

sin⁡−1x1−x2.\frac{\sin^{-1}x}{\sqrt{1 - x^2}}.
Exam answer

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

Put sin⁡−1x=t\sin^{-1}x = t⁠. Then dx1−x2=dt\frac{dx}{\sqrt{1 - x^2}} = dt⁠.

I=∫t dt=t22+CI = \int t\,dt = \frac{t^2}{2} + C

∴ I=12(sin⁡−1x)2+C\therefore\ I = \dfrac12\left(\sin^{-1}x\right)^2 + C⁠.

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

The one idea this problem is built on

The factor 11−x2\frac{1}{\sqrt{1-x^2}} is exactly the derivative of sin⁡−1x\sin^{-1}x⁠. So with t=sin⁡−1xt = \sin^{-1}x we get dt=dx1−x2dt = \frac{dx}{\sqrt{1-x^2}}⁠, and the integral is simply ∫t dt=t22+C\int t\,dt = \frac{t^2}{2} + C⁠.

Given, to find, and the plan
Given
f(x)=sin⁡−1x1−x2f(x) = \dfrac{\sin^{-1}x}{\sqrt{1-x^2}}⁠. It needs −1≤x≤1-1 \le x \le 1 for sin⁡−1x\sin^{-1}x⁠, and 1−x2>01 - x^2 > 0 for the square root in the denominator, so its domain is −1<x<1-1 < x < 1⁠.
To find
∫sin⁡−1x1−x2 dx\displaystyle\int \frac{\sin^{-1}x}{\sqrt{1-x^2}}\,dx⁠.
Plan
(1) Recognise the derivative of sin⁡−1x\sin^{-1}x⁠. (2) Substitute t=sin⁡−1xt = \sin^{-1}x⁠, noting the range of tt⁠. (3) Integrate tt⁠. (4) Return to xx⁠. (5) Differentiate back.

Basics you need

1The derivative 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 (Example 24, Section 5.3.3; formula (viii), Section 7.2). The principal value sin⁡−1x\sin^{-1}x lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. Basics: integration undoes differentiation

2Substitution

If t=g(x)t = g(x)⁠, then dt=g′(x) dxdt = g'(x)\,dx and ∫f(g(x)) g′(x) dx=∫f(t) dt\int f(g(x))\,g'(x)\,dx = \int f(t)\,dt⁠. Basics: substitution

3The power formula

∫t dt=t22+C\int t\,dt = \frac{t^2}{2} + C (standard formula (i), Section 7.2, with n=1n = 1⁠). Basics: the power rule

Solution

  1. Spot the inner function and its derivative

    Write the integrand as

    sin⁡−1x⋅11−x2.\sin^{-1}x\cdot\frac{1}{\sqrt{1-x^2}}.

    The second factor is the derivative of the first.

    This is the pattern f(g(x)) g′(x)f(g(x))\,g'(x) with g(x)=sin⁡−1xg(x) = \sin^{-1}x and outer function f(t)=tf(t) = t⁠. Recognising 11−x2\frac{1}{\sqrt{1-x^2}} as a derivative is the key; it would be hard to integrate the quotient any other way.
    Section 5.3.3, derivatives of inverse trigonometric functions.

  2. Substitute t=sin⁡−1xt = \sin^{-1}x

    Let t=sin⁡−1xt = \sin^{-1}x⁠. For −1<x<1-1 < x < 1⁠, tt lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, and

    dt=dx1−x2.dt = \frac{dx}{\sqrt{1-x^2}}.

    The integral becomes

    ∫sin⁡−1x⋅dx1−x2=∫t dt.\int\sin^{-1}x\cdot\frac{dx}{\sqrt{1-x^2}} = \int t\,dt.
    The substitution is valid on the whole domain −1<x<1-1 < x < 1⁠, because sin⁡−1\sin^{-1} is differentiable at every point of that open interval (it is not differentiable at ±1\pm1⁠, but those points are excluded anyway).
    Change of variable, Section 7.3.1. Lesson: integration by substitution

  3. Integrate and return to xx

    ∫t dt=t22+C=12(sin⁡−1x)2+C.\int t\,dt = \frac{t^2}{2} + C = \frac12\left(\sin^{-1}x\right)^2 + C.
    Standard formula (i), Section 7.2, with n=1n = 1⁠. Basics: the power rule

  4. Check: differentiate the answer back

    By the chain rule (outer 12u2\frac12u^2⁠, inner u=sin⁡−1xu = \sin^{-1}x⁠):

    ddx[12(sin⁡−1x)2]=12⋅2sin⁡−1x⋅11−x2=sin⁡−1x1−x2.\begin{aligned} \frac{d}{dx}\left[\frac12\left(\sin^{-1}x\right)^2\right] &= \frac12\cdot 2\sin^{-1}x\cdot\frac{1}{\sqrt{1-x^2}} \\ &= \frac{\sin^{-1}x}{\sqrt{1-x^2}}. \end{aligned}

    This is the integrand, so the answer is correct for −1<x<1-1 < x < 1⁠.

    Chain rule, Theorem 4, Section 5.3.1; Property (I), Section 7.2.1.
Result ∫sin⁡−1x1−x2 dx=12(sin⁡−1x)2+C,\int \frac{\sin^{-1}x}{\sqrt{1 - x^2}}\,dx = \frac12\left(\sin^{-1}x\right)^2 + C,

valid for −1<x<1-1 < x < 1⁠. (Textbook answer: 12(sin⁡−1x)2+C\frac12\left(\sin^{-1}x\right)^2 + C⁠.)

Check it with numbers

Let F(x)=12(sin⁡−1x)2F(x) = \frac12\left(\sin^{-1}x\right)^2 and look near x=0.5x = 0.5 with h=0.001h = 0.001⁠:

  • F(0.501)≈0.137683308F(0.501) \approx 0.137683308 and F(0.499)≈0.136474107F(0.499) \approx 0.136474107⁠.
  • Slope estimate: F(0.501)−F(0.499)0.002\dfrac{F(0.501) - F(0.499)}{0.002} ≈0.604601\approx 0.604601⁠.
  • Integrand at x=0.5x = 0.5⁠: sin⁡−10.5=π6\sin^{-1}0.5 = \frac{\pi}{6}⁠, so it is π/60.75≈0.604600\dfrac{\pi/6}{\sqrt{0.75}} \approx 0.604600⁠.

They agree to five significant figures (the last digit differs only through the error of the difference quotient).

Common mistakes
  1. Reading sin⁡−1x\sin^{-1}x as 1sin⁡x\frac{1}{\sin x}⁠. Here it is the inverse sine; 1sin⁡x\frac{1}{\sin x} is cosec⁡x\operatorname{cosec}x⁠.
  2. Writing (sin⁡−1x)2(\sin^{-1}x)^2 without the 12\frac12⁠. Differentiating (sin⁡−1x)2(\sin^{-1}x)^2 gives twice the integrand.
  3. Confusing (sin⁡−1x)2(\sin^{-1}x)^2 with sin⁡−1(x2)\sin^{-1}(x^2)⁠. The square is on the angle, not on xx⁠.
  4. Ignoring the domain. The answer only makes sense for −1<x<1-1 < x < 1⁠.
What to take away
  1. ∫g(x) g′(x) dx=12g(x)2+C\int g(x)\,g'(x)\,dx = \frac12 g(x)^2 + C⁠: a function times its own derivative integrates to half its square.
  2. 11−x2\frac{1}{\sqrt{1-x^2}} signals t=sin⁡−1xt = \sin^{-1}x⁠; 11+x2\frac{1}{1+x^2} signals t=tan⁡−1xt = \tan^{-1}x (Question 18).

Every fact used, and where it comes from

FactUsed inSource
ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}⁠, −1<x<1-1 < x < 1Steps 1, 2, 4Example 24, Section 5.3.3; formula (viii), Section 7.2
Principal range of sin⁡−1\sin^{-1} is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]Step 2Chapter 2, Section 2.2
Change of variable (substitution)Step 2Section 7.3.1 (Lesson: integration by substitution)
∫t dt=t22+C\int t\,dt = \frac{t^2}{2} + CStep 3Standard formula (i), Section 7.2
Chain ruleStep 4Theorem 4, Section 5.3.1
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