Exercise 7.8 · Q9: $\int_0^1\frac{dx}{\sqrt{1 - x^2}}$

Basics you need

Solution

1. Where is the integrand defined and continuous?

   $\sqrt{1 - x^2}$ is real when $1 - x^2 \ge 0$⁠, and the fraction needs $\sqrt{1 - x^2} \ne 0$⁠. Together: $1 - x^2 > 0$⁠, that is $-1 < x < 1$⁠. So on our interval:

   • for $0 \le x < 1$⁠, $f(x)$ is defined and continuous (a quotient of continuous functions with non-zero denominator);
   • at $x = 1$⁠, the denominator is $\sqrt{0} = 0$⁠: $f(1)$ is not defined, and $f(x) \to \infty$ as $x \to 1^-$⁠.
   Remark (iv) after Theorem 2 says the integrand must be well defined and continuous on the whole of $[a, b]$⁠. That fails at the single point $x = 1$⁠, so we cannot quote Theorem 2 on $[0, 1]$ directly. We can quote it on $[0, b]$ for any $b$ with $0 < b < 1$⁠, because there $f$ is continuous.
   Quotient of continuous functions: Chapter 5, Theorem 1, Section 5.2.1. Lesson: the fundamental theorem

2. Antiderivative, and check it

   By standard formula (viii),

   $$\int\frac{dx}{\sqrt{1 - x^2}} = \sin^{-1}x + C, \qquad F(x) = \sin^{-1}x.$$

   Check: for $-1 < x < 1$⁠, $F'(x) = \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}} = f(x)$⁠.

   This is the derivative of $\sin^{-1}x$ from Chapter 5 read backwards. It holds on the open interval $(-1, 1)$⁠, which contains $[0, b]$ for every $b < 1$⁠.
   Standard formula (viii), Section 7.2; $\frac{d}{dx}\sin^{-1}x$ is in Chapter 5, Section 5.3.3. Basics: integration undoes differentiation

3. Theorem 2 on $[0, b]$⁠, with $0 < b < 1$

   $$\int_0^b\frac{dx}{\sqrt{1 - x^2}} = \sin^{-1}b - \sin^{-1}0 = \sin^{-1}b.$$
   On $[0, b]$ the integrand is continuous and $F' = f$⁠, so all the conditions of Theorem 2 hold. $\sin^{-1}0 = 0$ because $\sin 0 = 0$ and 0 lies in the principal range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠.
   Theorem 2, Section 7.8.3. Lesson: the fundamental theorem

4. Let $b$ tend to 1

   $\sin^{-1}$ is continuous on $[-1, 1]$⁠, so as $b \to 1^-$⁠,

   $$\begin{aligned}\int_0^1\frac{dx}{\sqrt{1 - x^2}} &= \lim_{b \to 1^-}\sin^{-1}b = \sin^{-1}1 \\ &= \frac{\pi}{2} \approx 1.5707963.\end{aligned}$$
   The value of the integral over $[0, 1]$ is defined as this limit (an "improper integral"; the idea goes beyond this book, but the computation is just a limit). $\sin^{-1}1 = \frac{\pi}{2}$ because $\sin\frac{\pi}{2} = 1$ and $\frac{\pi}{2}$ is in the principal range. The answer key gives only the value $\frac{\pi}{2}$⁠, which is what the usual one-line evaluation $\big[\sin^{-1}x\big]_0^1$ produces. That shortcut gives the same number as the limit because the antiderivative $\sin^{-1}x$ is continuous on the closed interval $[0, 1]$⁠, even though the integrand is not.
   Principal value branch of $\sin^{-1}$⁠: Chapter 2; its continuity: Chapter 5, Section 5.3.3 (stated without proof). Basics: trigonometric values

5. Confirm with $x = \sin\theta$

   For $0 \le \theta < \frac{\pi}{2}$⁠, put $x = \sin\theta$⁠: then $dx = \cos\theta\,d\theta$⁠, and $\sqrt{1 - \sin^2\theta} = \cos\theta$ because $\cos\theta > 0$ for these $\theta$⁠. So

   $$\frac{dx}{\sqrt{1 - x^2}} = \frac{\cos\theta\,d\theta}{\cos\theta} = d\theta.$$

   On $[0, b]$ with $b < 1$ the new limits are $\theta = 0$ and $\theta = \sin^{-1}b$⁠, so the integral is $\int_0^{\sin^{-1}b}d\theta = \sin^{-1}b$⁠, as in Step 3. As $b \to 1^-$⁠, the upper limit $\sin^{-1}b$ rises to $\frac{\pi}{2}$⁠, and the value tends to $\frac{\pi}{2}$⁠.

   The substitution turns the integrand into the constant 1 in $\theta$⁠, so the integral is simply the length of the $\theta$-interval. It is legitimate because $x = \sin\theta$ is increasing on $\left[0, \frac{\pi}{2}\right]$⁠, so $\theta = \sin^{-1}x$ runs exactly once over $\left[0, \frac{\pi}{2}\right)$⁠, and $\cos\theta$ is positive there, so $\sqrt{\cos^2\theta} = \cos\theta$ with no sign change.
   Substitution in definite integrals, Section 7.9; $\sqrt{1 - \sin^2\theta} = |\cos\theta|$⁠. Lesson: integration by substitution

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