Exercise 7.6 · Q22: $\int\sin^{-1}\frac{2x}{1+x^2}\,dx$

Basics you need

Solution

1. The integrand is defined for every real $x$

   $\sin^{-1}$ needs its input in $[-1, 1]$⁠. For every real $x$⁠,

   $$1 + x^2 - 2|x| = (|x| - 1)^2 \ge 0,$$

   so $2|x| \le 1 + x^2$⁠, that is $\left|\frac{2x}{1+x^2}\right| \le 1$⁠. Equality holds only at $x = \pm 1$⁠.

   So the integrand is defined, and continuous (a composition of continuous functions), on the whole real line. That matters at the end: a continuous function on an interval has an antiderivative on that whole interval (the area function of the First fundamental theorem, Theorem 1 in Section 7.8.2, stated there without proof), so we should be able to give one formula that works everywhere.
   Completing the square: $1 + x^2 - 2|x| = (|x| - 1)^2$⁠, and a square is never negative. Basics: completing the square

2. Simplify with $x = \tan\theta$

   Put $x = \tan\theta$ with $\theta = \tan^{-1}x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠; every real $x$ arises this way. Then

   $$\frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta,$$

   so the integrand is $\sin^{-1}(\sin 2\theta)$⁠. This equals $2\theta$ exactly when $-\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2}$⁠, that is $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$⁠, that is $-1 \le x \le 1$⁠. Hence

   $$\sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x \quad\text{for } -1 \le x \le 1.$$
   $\sin^{-1}$ always returns an angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. If $2\theta$ is already in that interval, $\sin^{-1}$ gives $2\theta$ back; if not, it gives a different angle with the same sine (Step 5). Since $\tan$ is increasing on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, $|\theta| \le \frac{\pi}{4}$ is the same as $|x| \le 1$⁠.
   Principal value branch of $\sin^{-1}$ (Chapter 2, Section 2.2). Basics: trigonometric identities

3. Integrate $2\tan^{-1}x$ by parts (for $-1 \le x \le 1$⁠)

   Take $\tan^{-1}x$ as the first function and $1$ as the second, as in Question 13:

   $$\begin{aligned} \int\tan^{-1}x\,dx &= x\tan^{-1}x - \int\frac{x}{1+x^2}\,dx \\ &= x\tan^{-1}x - \frac12\log(1+x^2). \end{aligned}$$

   Multiplying by 2,

   $$\int\sin^{-1}\frac{2x}{1+x^2}\,dx = 2x\tan^{-1}x - \log(1+x^2) + C$$

   for $-1 \le x \le 1$⁠.

   The inverse tangent is the first function because we can differentiate it but not yet integrate it (Remark (iii), Section 7.6; in the memory aid ILATE, not in the book, I comes before A). The derivative of the first function is $\frac{1}{1+x^2}$ and the integral of the second is $x$⁠. The leftover $\int\frac{x}{1+x^2}\,dx$ is $\frac12\log(1+x^2)$ by the substitution $t = 1 + x^2$⁠, $dt = 2x\,dx$⁠; no absolute value is needed because $1 + x^2 > 0$⁠.
   Integration by parts; inverse trigonometric function as first function (Remark (iii), Section 7.6). Lesson: integration by parts

4. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\big[2x\tan^{-1}x - \log(1+x^2)\big] \\ &= 2\tan^{-1}x + \frac{2x}{1+x^2} - \frac{2x}{1+x^2} \\ &= 2\tan^{-1}x. \end{aligned}$$

   By Step 2 this equals $\sin^{-1}\frac{2x}{1+x^2}$ for $-1 \le x \le 1$⁠, so the answer is correct on that interval.

   Product rule on $2x\tan^{-1}x$ and chain rule on $\log(1+x^2)$⁠; the two fractions cancel. Note that the derivative is $2\tan^{-1}x$ for every $x$⁠, but it equals the integrand only on $[-1, 1]$⁠. That is exactly why the formula fails outside.
   Product rule; chain rule (Chapter 5). Basics: integration undoes differentiation

5. What happens for $|x| > 1$

   For $x > 1$⁠: $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$⁠, so $2\theta \in \left(\frac{\pi}{2}, \pi\right)$⁠. The angle $\pi - 2\theta$ lies in $\left(0, \frac{\pi}{2}\right)$ and has the same sine, so

   $$\sin^{-1}\frac{2x}{1+x^2} = \pi - 2\tan^{-1}x.$$

   For $x < -1$⁠: $2\theta \in \left(-\pi, -\frac{\pi}{2}\right)$⁠. The angle $-\pi - 2\theta$ lies in $\left(-\frac{\pi}{2}, 0\right)$ and has the same sine, so

   $$\sin^{-1}\frac{2x}{1+x^2} = -\pi - 2\tan^{-1}x.$$

   Both cases read $\pi\,\text{sgn}(x) - 2\tan^{-1}x$⁠. Integrating (with Step 3 for the $\tan^{-1}$ part, and $\int\pi\,\text{sgn}(x)\,dx = \pi|x|$⁠):

   $$\begin{aligned} &\int\sin^{-1}\frac{2x}{1+x^2}\,dx \\ &= \pi|x| - 2x\tan^{-1}x + \log(1+x^2) + C \end{aligned}$$

   for $|x| > 1$ (on each of the two intervals).

   $\sin(\pi - \phi) = \sin\phi$ and $\sin(-\pi - \phi) = \sin\phi$⁠, and $\sin^{-1}$ picks the one angle with that sine that lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. Here $\text{sgn}(x)$ is $1$ for $x > 0$ and $-1$ for $x < 0$⁠. The textbook formula is not valid here: its derivative $2\tan^{-1}x$ is not the integrand (see the figure).
   Principal value branch of $\sin^{-1}$ (Chapter 2). Basics: trigonometric identities

6. One antiderivative for all real $x$

   Since the integrand is continuous everywhere (Step 1), the pieces can be joined. Choose the constants so the values match at $x = \pm 1$⁠: the formula of Step 3 gives $\frac{\pi}{2} - \log 2$ at both $x = 1$ and $x = -1$⁠, and the formula of Step 5 gives $\frac{\pi}{2} + \log 2$ there, so subtract $2\log 2$ from the latter:

   for $|x| \le 1$⁠,

   $$G(x) = 2x\tan^{-1}x - \log(1+x^2),$$

   and for $|x| > 1$⁠,

   $$G(x) = \pi|x| - 2x\tan^{-1}x + \log\frac{1+x^2}{4}.$$

   Then $\int\sin^{-1}\frac{2x}{1+x^2}\,dx = G(x) + C$ for every real $x$⁠.

   $G$ is continuous at $\pm 1$ by the choice of constant, and its one-sided derivatives at $x = 1$ are $2\tan^{-1}1 = \frac{\pi}{2}$ (left) and $\pi - 2\tan^{-1}1 = \frac{\pi}{2}$ (right), equal to the integrand's value $\sin^{-1}1 = \frac{\pi}{2}$⁠; similarly at $x = -1$⁠. So $G'$ equals the integrand everywhere. ($\log\frac{1+x^2}{4} = \log(1+x^2) - 2\log 2$⁠.)
   Antiderivatives on an interval differ by a constant (Section 7.2). Basics: integration undoes differentiation

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