Example 20: $\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx$

Basics you need

Solution

1. Choose the first and second functions

   Write the integrand as a product and take

   $$f(x) = \sin^{-1}x, \qquad g(x) = \frac{x}{\sqrt{1-x^2}}.$$

   Then $f'(x) = \frac{1}{\sqrt{1-x^2}}$⁠.

   We do not have a formula for $\int\sin^{-1}x\,dx$ at this stage, so $\sin^{-1}x$ cannot be the second function; but its derivative is a simple algebraic expression. The book's Remark (iii) says the same: an inverse trigonometric function is taken as the first function.
   $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}$⁠, $|x| < 1$⁠. Lesson: integration by parts

2. Integrate the second function

   Put $t = 1 - x^2$⁠. Then $dt = -2x\,dx$⁠, so $x\,dx = -\frac12\,dt$⁠. On $(-1, 1)$⁠, $t$ lies in $(0, 1]$⁠, so $t > 0$ and $\sqrt t$ is defined.

   $$\begin{aligned} \int\frac{x\,dx}{\sqrt{1-x^2}} &= -\frac12\int\frac{dt}{\sqrt t} = -\frac12\cdot 2\sqrt t \\ &= -\sqrt t = -\sqrt{1-x^2}. \end{aligned}$$
   The top $x\,dx$ is, up to the factor $-\frac12$⁠, the differential of $1 - x^2$⁠, the expression under the root. That is the signal for the substitution $t = 1 - x^2$⁠. As in Example 17, no constant is needed here.
   Substitution (Section 7.3.1) and the power rule with $n = -\frac12$⁠. Basics: substitution

3. Apply integration by parts

   $$\begin{aligned} &\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx \\ &= (\sin^{-1}x)\left(-\sqrt{1-x^2}\right) \\ &\quad - \int\frac{1}{\sqrt{1-x^2}}\left(-\sqrt{1-x^2}\right)dx \\ &= -\sqrt{1-x^2}\,\sin^{-1}x + \int 1\,dx \\ &= x - \sqrt{1-x^2}\,\sin^{-1}x + C. \end{aligned}$$
   First function times integral of second, minus the integral of (derivative of first) times (integral of second). The product $\frac{1}{\sqrt{1-x^2}}\cdot\left(-\sqrt{1-x^2}\right)$ is exactly $-1$ (the roots cancel because $\sqrt{1-x^2} \ne 0$ on $(-1, 1)$⁠), and $-\int(-1)\,dx = +x$⁠.
   $\int 1\,dx = x + C$ (formula (i)). Lesson: integration by parts

4. Confirm with the substitution $\theta = \sin^{-1}x$

   The book notes that we may instead put $\theta = \sin^{-1}x$⁠. For $-1 < x < 1$ this gives $-\frac\pi2 < \theta < \frac\pi2$⁠, with $x = \sin\theta$ and $dx = \cos\theta\,d\theta$⁠. On this interval $\cos\theta > 0$⁠, so $\sqrt{1-x^2} = \sqrt{\cos^2\theta} = \cos\theta$⁠. Then

   $$\begin{aligned} \int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx &= \int\frac{\sin\theta\cdot\theta}{\cos\theta}\cos\theta\,d\theta \\ &= \int\theta\sin\theta\,d\theta. \end{aligned}$$

   By parts with $\theta$ first and $\sin\theta$ second:

   $$\begin{aligned} \int\theta\sin\theta\,d\theta &= -\theta\cos\theta + \int\cos\theta\,d\theta \\ &= -\theta\cos\theta + \sin\theta + C. \end{aligned}$$

   Back in $x$⁠: $-\sqrt{1-x^2}\,\sin^{-1}x + x + C$⁠, the same answer.

   The substitution is legitimate because $\sin^{-1}(\sin\theta) = \theta$ exactly when $\theta$ lies in the principal range $\left[-\frac\pi2, \frac\pi2\right]$⁠, and there $\cos\theta \ge 0$⁠, which is what lets us replace $\sqrt{\cos^2\theta}$ by $\cos\theta$ without a sign change. Two different routes giving the same result is a strong check.
   Principal value branch of $\sin^{-1}$ (Chapter 2); $\sqrt{a^2} = |a|$⁠. Basics: trigonometric identities

5. Check: differentiate the answer

   By the product rule, with $\frac{d}{dx}\sqrt{1-x^2} = \frac{-x}{\sqrt{1-x^2}}$⁠:

   $$\begin{aligned} &\frac{d}{dx}\Big[x - \sqrt{1-x^2}\,\sin^{-1}x\Big] \\ &= 1 - \left[\frac{-x}{\sqrt{1-x^2}}\sin^{-1}x + \sqrt{1-x^2}\cdot\frac{1}{\sqrt{1-x^2}}\right] \\ &= 1 + \frac{x\sin^{-1}x}{\sqrt{1-x^2}} - 1 = \frac{x\sin^{-1}x}{\sqrt{1-x^2}}. \end{aligned}$$
   The derivative is exactly the integrand at every point of $(-1, 1)$⁠, so the antiderivative is correct there.
   Product rule and chain rule (Chapter 5). Basics: integration undoes differentiation

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