Miscellaneous · Q19: $\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\,dx$

Basics you need

Solution

1. Choose the substitution $x = \cos^2\theta$⁠, $0 \le \theta \le \frac{\pi}{2}$

   $$x = \cos^2\theta ,\qquad dx = -2\cos\theta\sin\theta\,d\theta .$$

   For each $x$ in $[0, 1]$ there is exactly one such $\theta$⁠, namely $\theta = \cos^{-1}\sqrt x$⁠, and then

   $$\sqrt x = \cos\theta ,\qquad \sqrt{1-x} = \sin\theta .$$
   Why this substitution: the integrand is built from $1 \pm \sqrt x$⁠, and $1 \pm \cos\theta$ have the half-angle forms that make the quotient a perfect square. Why this range: $\sqrt x \ge 0$ must equal $\cos\theta$⁠, and on $[0, \frac{\pi}{2}]$ the cosine takes every value in $[0, 1]$ exactly once, with $\cos\theta \ge 0$ and $\sin\theta \ge 0$⁠. So $\sqrt{\cos^2\theta} = \cos\theta$ and $\sqrt{1 - \cos^2\theta} = \sin\theta$ without sign worries.
   Chain rule: $\frac{d}{d\theta}\cos^2\theta = 2\cos\theta\cdot(-\sin\theta)$⁠. Basics: substitution

2. Simplify the root

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

   Since $0 \le \frac{\theta}{2} \le \frac{\pi}{4}$⁠, $\tan\frac{\theta}{2} \ge 0$⁠, so

   $$\sqrt{\frac{1-\sqrt x}{1+\sqrt x}} = \tan\frac{\theta}{2} .$$
   $\sqrt{u^2} = |u|$⁠, not $u$⁠. The modulus disappears only because the range chosen in Step 1 keeps $\frac{\theta}{2}$ in the first quadrant.
   Basics: trigonometric identities

3. Rewrite the whole integral in $\theta$

   $$\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\,dx = \int\tan\frac{\theta}{2}\cdot\left(-2\cos\theta\sin\theta\right)d\theta .$$

   Now

   $$\begin{aligned}\tan\frac{\theta}{2}\sin\theta &= \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\cdot 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\ &= 2\sin^2\frac{\theta}{2} = 1 - \cos\theta ,\end{aligned}$$

   so the integral is

   $$\begin{aligned}&-2\int(1 - \cos\theta)\cos\theta\,d\theta\\ &= \int\left(-2\cos\theta + 2\cos^2\theta\right)d\theta\\ &= \int\left(-2\cos\theta + 1 + \cos 2\theta\right)d\theta .\end{aligned}$$
   Writing $\sin\theta$ through half angles cancels the $\cos\frac{\theta}{2}$ in the denominator of $\tan\frac{\theta}{2}$⁠. The square $\cos^2\theta$ is then lowered with $2\cos^2\theta = 1 + \cos 2\theta$ so that every term has a standard integral.
   Section 7.3.2 (integration using trigonometric identities). Basics: trigonometric identities

4. Integrate in $\theta$

   $$\begin{aligned}&\int\left(-2\cos\theta + 1 + \cos 2\theta\right)d\theta\\ &= -2\sin\theta + \theta + \frac12\sin 2\theta + C\\ &= -2\sin\theta + \theta + \sin\theta\cos\theta + C .\end{aligned}$$
   Standard formulae (ii) and (i) of Section 7.2, and $\int\cos 2\theta\,d\theta = \frac12\sin 2\theta$ (substitute $u = 2\theta$⁠). Then $\frac12\sin 2\theta = \sin\theta\cos\theta$⁠, which is easier to translate back to $x$⁠.
   Standard formula (ii), Section 7.2. Lesson: integration by substitution

5. Return to $x$

   From Step 1: $\theta = \cos^{-1}\sqrt x$⁠, $\sin\theta = \sqrt{1-x}$⁠, $\cos\theta = \sqrt x$⁠, so $\sin\theta\cos\theta = \sqrt{x}\sqrt{1-x} = \sqrt{x - x^2}$⁠:

   $$\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\,dx = -2\sqrt{1-x} + \cos^{-1}\sqrt x + \sqrt{x-x^2} + C .$$
   $\theta = \cos^{-1}\sqrt x$ is correct because $\theta$ lies in $[0, \frac{\pi}{2}] \subset [0, \pi]$⁠, the principal range of $\cos^{-1}$⁠, so $\cos^{-1}(\cos\theta) = \theta$⁠. Also $\sqrt x\sqrt{1-x} = \sqrt{x(1-x)}$ because both factors are $\ge 0$⁠.
   Principal range of $\cos^{-1}$ (Chapter 2).

6. Check: differentiate the answer

   For $0 < x < 1$⁠, write $s = \sqrt x$⁠, so $\frac{ds}{dx} = \frac{1}{2s}$ and $\sqrt{x-x^2} = s\sqrt{1-x}$⁠. The three terms give

   $$\begin{aligned}\frac{d}{dx}\left(-2\sqrt{1-x}\right) &= \frac{1}{\sqrt{1-x}} ,\\ \frac{d}{dx}\cos^{-1}\sqrt x &= -\frac{1}{\sqrt{1-x}}\cdot\frac{1}{2s} ,\\ \frac{d}{dx}\sqrt{x-x^2} &= \frac{1-2x}{2s\sqrt{1-x}} .\end{aligned}$$

   Over the common denominator $2s\sqrt{1-x}$ the numerators add to $2s - 1 + 1 - 2s^2 = 2s(1-s)$⁠, so the derivative is

   $$\frac{2s(1-s)}{2s\sqrt{1-x}} = \frac{1-s}{\sqrt{(1-s)(1+s)}} = \sqrt{\frac{1-s}{1+s}}\ \checkmark$$
   We used $1 - x = (1-s)(1+s)$ and, for $0 < s < 1$⁠, $\frac{1-s}{\sqrt{1-s}} = \sqrt{1-s}$⁠. With $s = \sqrt x$ the last expression is the integrand. The answer is valid on $(0, 1)$⁠; it is also continuous on the closed interval $[0, 1]$⁠.
   $\frac{d}{dx}\cos^{-1}u = -\frac{1}{\sqrt{1-u^2}}\cdot\frac{du}{dx}$ (Chapter 5). Basics: integration undoes differentiation

7. Compare with the textbook, and an equivalent form

   The key gives $-2\sqrt{1-x} + \cos^{-1}\sqrt x + \sqrt{x-x^2} + C$⁠, exactly our answer.

   A solution that substitutes $x = \sin^2\phi$ instead ends with $-\sin^{-1}\sqrt x$ in place of $\cos^{-1}\sqrt x$⁠. Since $\cos^{-1}\sqrt x = \frac{\pi}{2} - \sin^{-1}\sqrt x$ for $0 \le x \le 1$⁠, the two answers differ by the constant $\frac{\pi}{2}$⁠, and both are correct.

   Antiderivatives on an interval differ by a constant (Remark, Section 7.2); $\sin^{-1}y + \cos^{-1}y = \frac{\pi}{2}$ for $y$ in $[-1, 1]$ (Chapter 2).

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