Miscellaneous · Q37: $\int_0^1\sin^{-1}x\,dx=\frac{\pi}{2}-1$

Basics you need

Solution

1. The integrand is continuous on $[0, 1]$

   $\sin^{-1}x$ is defined and continuous on $[-1, 1]$⁠, so it is continuous on $[0, 1]$⁠, including at the endpoint $x = 1$⁠, where it equals $\frac{\pi}{2}$⁠.

   So the integral exists in the ordinary sense of Theorem 2, Section 7.8.3. Keep in mind that the derivative of $\sin^{-1}x$ does not exist at $x = 1$⁠; this matters in Step 5.
   Lesson: definite integrals and the fundamental theorem

2. Integration by parts with $u = \sin^{-1}x$

   $$\begin{aligned} u &= \sin^{-1}x, & du &= \frac{dx}{\sqrt{1-x^2}}, \\ dv &= 1\,dx, & v &= x . \end{aligned}$$

   So, for $-1 < x < 1$⁠,

   $$\begin{aligned} &\int\sin^{-1}x\,dx \\ &= x\sin^{-1}x - \int\frac{x}{\sqrt{1-x^2}}\,dx . \end{aligned}$$
   Remark (iii) of Section 7.6: an inverse trigonometric function is taken as the first function. We cannot integrate $\sin^{-1}x$ yet, but we can differentiate it into an algebraic expression, and the second function 1 integrates to $x$⁠. The new integral contains no inverse function at all.
   Section 7.6, formula (1); Example 18 uses the same "times 1" trick for $\log x$⁠. Lesson: integration by parts

3. The leftover integral: put $t = 1 - x^2$

   With $t = 1 - x^2$⁠, $dt = -2x\,dx$⁠, so $x\,dx = -\frac12\,dt$⁠:

   $$\begin{aligned} \int\frac{x\,dx}{\sqrt{1-x^2}} &= -\frac12\int t^{-1/2}\,dt \\ &= -\frac12\cdot 2t^{1/2} = -\sqrt{1-x^2} . \end{aligned}$$

   Therefore

   $$\int\sin^{-1}x\,dx = x\sin^{-1}x + \sqrt{1-x^2} + C .$$
   The numerator $x\,dx$ is (a constant times) the differential of the expression $1 - x^2$ under the root, which is exactly when substitution works. The power rule with $n = -\frac12$ gives $\frac{t^{1/2}}{1/2} = 2t^{1/2}$⁠. Subtracting $-\sqrt{1-x^2}$ gives $+\sqrt{1-x^2}$⁠.
   Section 7.3.1; standard formula (i), Section 7.2. Lesson: integration by substitution

4. Check: differentiate the antiderivative

   Let $F(x) = x\sin^{-1}x + \sqrt{1-x^2}$⁠. For $-1 < x < 1$⁠:

   $$\begin{aligned} F'(x) &= \sin^{-1}x + \frac{x}{\sqrt{1-x^2}} + \frac{-2x}{2\sqrt{1-x^2}} \\ &= \sin^{-1}x \ ✓ \end{aligned}$$
   Product rule on $x\sin^{-1}x$⁠, chain rule on $\sqrt{1-x^2}$ (outer square root, inner $1 - x^2$⁠). The two fractions cancel, leaving the integrand.
   Basics: integration undoes differentiation

5. The endpoint $x = 1$

   Step 4 holds for $-1 < x < 1$ only, because $\frac{1}{\sqrt{1-x^2}}$ is undefined at $x = 1$⁠. But $F$ itself is continuous on all of $[0, 1]$⁠: at $x = 1$ it equals $1\cdot\frac{\pi}{2} + 0 = \frac{\pi}{2}$⁠, and $x\sin^{-1}x \to \frac{\pi}{2}$⁠, $\sqrt{1-x^2} \to 0$ as $x \to 1$⁠.

   That is enough for $I = F(1) - F(0)$⁠. The area function $A(x) = \int_0^x\sin^{-1}t\,dt$ has $A'(x) = \sin^{-1}x$ (Theorem 1, Section 7.8.2), so $A - F$ has derivative 0 on $[0, 1)$ and is constant there (functions with the same derivative on an interval differ by a constant: Remark in Section 7.2). Since $A(0) = 0$⁠, $A(x) = F(x) - F(0)$ for $0 \le x < 1$⁠. Now let $x \to 1$⁠. $F$ is continuous at 1 (shown above). $A$ is continuous at 1 too: Theorem 1 gives $A'(x) = \sin^{-1}x$ at every $x$ of the closed interval $[0, 1]$⁠, including $x = 1$ where $\sin^{-1}1 = \frac{\pi}{2}$⁠, and a function differentiable at a point is continuous there (Theorem 3, Chapter 5, Section 5.3). Taking the limit of both sides gives $I = A(1) = F(1) - F(0)$⁠.

   Theorem 2 asks for an antiderivative on the closed interval. Here our formula for $F'$ fails exactly at the right end, so we check that nothing is lost there. (The book does not raise this point.) The continuity of $F$ at $x = 1$ is what saves us.
   Theorem 1, Section 7.8.2 (it needs only that $\sin^{-1}x$ is continuous on $[0, 1]$⁠, Step 1); Remark in Section 7.2; Theorem 3, Section 5.3. Lesson: definite integrals

6. Evaluate

   $F(1) = 1\cdot\sin^{-1}1 + \sqrt{0} = \frac{\pi}{2}$⁠, and $F(0) = 0 + \sqrt1 = 1$⁠.

   $$I = F(1) - F(0) = \frac{\pi}{2} - 1 \approx 0.570796 .$$

   This is the required result.

   Principal values: $\sin^{-1}1 = \frac{\pi}{2}$⁠, because $\sin\frac{\pi}{2} = 1$ and $\frac{\pi}{2}$ is in $[-\frac{\pi}{2}, \frac{\pi}{2}]$⁠. Note $F(0) = 1$⁠, not 0: the term $\sqrt{1-x^2}$ does not vanish at $x = 0$⁠.
   Theorem 2, Section 7.8.3.

7. The same answer from a picture

   The graph of an inverse function is the mirror image of the original graph, so "area under $\sin^{-1}x$" and "area beside $\sin y$" fill the rectangle together. The area beside the curve is an easy integral, $\big[-\cos y\big]_0^{\pi/2} = 0 - (-1) = 1$⁠. This is an independent check of the whole calculation.
   Basics: area as a sum of thin strips

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