Exercise 7.9 · Q3: $\int_0^1\sin^{-1}\frac{2x}{1+x^2}\,dx$

Basics you need

Solution

1. Check the integrand is defined and continuous

   $\sin^{-1}$ needs an input in $[-1, 1]$⁠. For every real $x$⁠, $(1 - x)^2 \ge 0$ and $(1 + x)^2 \ge 0$ give $1 + x^2 \ge 2x$ and $1 + x^2 \ge -2x$⁠, so

   $$-1 \le \frac{2x}{1 + x^2} \le 1.$$

   The inner function is continuous (denominator $\ge 1$⁠), and $\sin^{-1}$ is continuous on $[-1, 1]$⁠, so the composite is continuous on $[0, 1]$⁠.

   Before integrating we must know the integrand exists at every point. The inequality shows the input to $\sin^{-1}$ never leaves $[-1, 1]$⁠; on $[0, 1]$ it actually runs from 0 up to 1 (reached at $x = 1$⁠).
   A composite of continuous functions is continuous (Chapter 5, Section 5.2.1). Lesson: the fundamental theorem

2. Substitute $x = \tan\theta$ and change the limits

   Take $\theta = \tan^{-1}x$⁠, so $x = \tan\theta$ and $dx = \sec^2\theta\,d\theta$⁠.

   New limits. When $x = 0$⁠, $\theta = \tan^{-1}0 = 0$⁠. When $x = 1$⁠, $\theta = \tan^{-1}1 = \frac{\pi}{4}$⁠.

   Why it is valid on the whole interval. On $\left[0, \frac{\pi}{4}\right]$⁠, $x = \tan\theta$ has the continuous derivative $\sec^2\theta$ and is increasing, and it takes every value in $[0, 1]$ exactly once. So as $\theta$ runs from 0 to $\frac{\pi}{4}$⁠, $x$ runs once from 0 to 1.

   The expression $\frac{2x}{1+x^2}$ is the tell-tale shape of $\frac{2\tan\theta}{1+\tan^2\theta}$⁠. Using the principal value $\theta = \tan^{-1}x$ fixes which $\theta$ we mean.
   $\frac{d}{d\theta}\tan\theta = \sec^2\theta$⁠; $\tan^{-1}$ has principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠. Basics: trigonometry

3. Simplify the integrand, checking the range

   $$\begin{aligned}\frac{2x}{1 + x^2} &= \frac{2\tan\theta}{1 + \tan^2\theta} \\ &= \frac{2\sin\theta/\cos\theta}{1/\cos^2\theta} \\ &= 2\sin\theta\cos\theta = \sin 2\theta.\end{aligned}$$

   Since $0 \le \theta \le \frac{\pi}{4}$⁠, we have $0 \le 2\theta \le \frac{\pi}{2}$⁠, which lies inside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. Therefore

   $$\sin^{-1}\left(\frac{2x}{1 + x^2}\right) = \sin^{-1}(\sin 2\theta) = 2\theta.$$

   The integral becomes

   $$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2}\right)dx = \int_0^{\frac{\pi}{4}} 2\theta\,\sec^2\theta\,d\theta.$$
   $\sin^{-1}(\sin y) = y$ is true only when $y$ is in the principal range. Here it is, because $|x| \le 1$⁠. For $x > 1$ we would have $2\theta > \frac{\pi}{2}$ and $\sin^{-1}(\sin 2\theta) = \pi - 2\theta$ instead (the figure shows the split).
   $1 + \tan^2\theta = \sec^2\theta = \frac{1}{\cos^2\theta}$⁠; $\sin 2\theta = 2\sin\theta\cos\theta$⁠. Basics: trigonometric identities

4. Integrate $\theta\sec^2\theta$ by parts

   Take $u = \theta$ (algebraic) and $dv = \sec^2\theta\,d\theta$ (trigonometric), so $du = d\theta$ and $v = \tan\theta$⁠. ILATE puts Algebraic before Trigonometric, so $\theta$ is the one we differentiate.

   $$\begin{aligned}\int\theta\sec^2\theta\,d\theta &= \theta\tan\theta - \int\tan\theta\,d\theta \\ &= \theta\tan\theta - \log|\sec\theta| + C.\end{aligned}$$

   On $\left[0, \frac{\pi}{4}\right]$⁠, $\sec\theta \ge 1 > 0$⁠, so $\log|\sec\theta| = \log\sec\theta$⁠.

   Differentiating $\theta$ gives 1, which removes the awkward factor, while $\sec^2\theta$ has the simple integral $\tan\theta$⁠. The leftover $\int\tan\theta\,d\theta$ is a standard result.
   Integration by parts (Section 7.6); $\int\tan\theta\,d\theta = \log|\sec\theta| + C$ (Section 7.3.1). Lesson: integration by parts

5. Evaluate between the limits

   $$\begin{aligned}\int_0^{\frac{\pi}{4}} 2\theta\sec^2\theta\,d\theta &= 2\Big[\theta\tan\theta - \log\sec\theta\Big]_0^{\frac{\pi}{4}} \\ &= 2\left[\frac{\pi}{4}\cdot 1 - \log\sqrt{2}\right] - 2[0 - \log 1] \\ &= \frac{\pi}{2} - 2\cdot\frac{1}{2}\log 2 \\ &= \frac{\pi}{2} - \log 2.\end{aligned}$$

   In decimals, $\frac{\pi}{2} \approx 1.5707963$ and $\log 2 \approx 0.6931472$⁠, so the value is about 0.8776491.

   At $\frac{\pi}{4}$⁠: $\tan\frac{\pi}{4} = 1$ and $\sec\frac{\pi}{4} = \sqrt{2}$⁠, with $\log\sqrt{2} = \frac{1}{2}\log 2$⁠. At 0: $\tan 0 = 0$ and $\sec 0 = 1$⁠, with $\log 1 = 0$⁠.
   $\log a^k = k\log a$⁠, so $\log 2^{1/2} = \frac{1}{2}\log 2$⁠. Basics: logarithms

6. Same result by staying in $x$

   Put $\theta = \tan^{-1}x$ back: $\theta\tan\theta = x\tan^{-1}x$ and $\sec\theta = \sqrt{1 + \tan^2\theta} = \sqrt{1 + x^2}$⁠. So an antiderivative of the integrand on $[0, 1]$ is

   $$F(x) = 2x\tan^{-1}x - \log(1 + x^2).$$

   Check: differentiate.

   $$\begin{aligned}F'(x) &= 2\tan^{-1}x + \frac{2x}{1 + x^2} - \frac{2x}{1 + x^2} \\ &= 2\tan^{-1}x,\end{aligned}$$

   and Step 3 showed $2\tan^{-1}x = 2\theta = \sin^{-1}\frac{2x}{1+x^2}$ for $0 \le x \le 1$⁠. Then $F(1) - F(0) = 2\cdot\frac{\pi}{4} - \log 2 - 0 = \frac{\pi}{2} - \log 2$⁠.

   The product rule on $2x\tan^{-1}x$ gives $2\tan^{-1}x + \frac{2x}{1+x^2}$⁠, and the second term cancels the derivative of $\log(1+x^2)$⁠. The check holds on $[0, 1]$⁠, exactly where we need it.
   Product rule and $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ (Chapter 5). Basics: integration undoes differentiation

7. The region, and where the identity stops

   The figure shows why the range check in Step 3 matters: the simplification to $2\theta$ is right on our interval and wrong beyond $x = 1$⁠. The shaded region fits under the line $y = \frac{\pi}{2}$ (rectangle area 1.571) and above the triangle with corners (0, 0), (1, 0), (1, $\frac{\pi}{2}$⁠) (area 0.785), consistent with 0.878.
   For a function that is never negative, the definite integral is the area under its graph. Basics: area as a sum of thin strips

Every fact used, and where it comes from