Example 39: $\int\left(\sqrt{\cot x}+\sqrt{\tan x}\right)dx$

Basics you need

Solution

1. Where is the integrand defined?

   $\sqrt{\tan x}$ needs $\tan x \ge 0$⁠, and $\sqrt{\cot x}$ needs $\cot x = \frac{1}{\tan x}$ to exist and be $\ge 0$⁠, so $\tan x \ne 0$⁠. Together: $\tan x > 0$⁠, which happens exactly on the intervals

   $$k\pi < x < k\pi + \frac{\pi}{2}, \qquad k \in \Z .$$

   We work on one such interval, say $0 < x < \frac{\pi}{2}$⁠; everything below is the same on the others, since $\tan$ and $\cot$ have period $\pi$⁠.

   Square roots of negative numbers are not real, and $\cot x$ is undefined where $\tan x = 0$⁠. On each interval the integrand is continuous, so it has an antiderivative there.
   $\tan x > 0$ in the first and third quadrants.

2. Rewrite with $\sqrt{\tan x}$ only

   Since $\cot x = \frac{1}{\tan x}$ and $\tan x > 0$⁠, $\sqrt{\cot x} = \frac{1}{\sqrt{\tan x}}$⁠. So

   $$\sqrt{\cot x} + \sqrt{\tan x} = \sqrt{\tan x}\,(\cot x + 1) = \sqrt{\tan x} + \frac{1}{\sqrt{\tan x}} .$$
   The textbook writes $\sqrt{\tan x}\,(1 + \cot x)$⁠; the last form is the same thing, and it shows that with $t = \sqrt{\tan x}$ the integrand is simply $t + \frac1t$⁠.
   $\sqrt{\tan x}\cdot\cot x = \frac{\sqrt{\tan x}}{\tan x} = \frac{1}{\sqrt{\tan x}}$⁠.

3. Substitute $\tan x = t^2$ (with $t = \sqrt{\tan x} > 0$⁠)

   Differentiate both sides of $\tan x = t^2$⁠: $\sec^2 x\,dx = 2t\,dt$⁠. Since $\sec^2 x = 1 + \tan^2 x = 1 + t^4$⁠,

   $$dx = \frac{2t\,dt}{1 + t^4} .$$

   So

   $$\begin{aligned}I &= \int\left(t + \frac1t\right)\frac{2t}{1+t^4}\,dt\\ &= \int\frac{2t^2 + 2}{t^4 + 1}\,dt = 2\int\frac{t^2+1}{t^4+1}\,dt .\end{aligned}$$
   The substitution clears both square roots at once. It is legitimate: on $0 < x < \frac{\pi}{2}$⁠, $t = \sqrt{\tan x}$ is a differentiable, increasing function taking every value in $(0, \infty)$⁠. The result is a rational function of $t$⁠, which we know how to attack.
   $\left(t + \frac1t\right)\cdot 2t = 2t^2 + 2$⁠. Lesson: integration by substitution

4. Divide top and bottom by $t^2$

   Since $t > 0$⁠, we may divide numerator and denominator by $t^2$⁠:

   $$I = 2\int\frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}}\,dt = 2\int\frac{1 + \frac{1}{t^2}}{\left(t - \frac1t\right)^2 + 2}\,dt .$$
   This is the clever step. The numerator $1 + \frac{1}{t^2}$ is exactly the derivative of $t - \frac1t$⁠, so if the denominator can be written using $t - \frac1t$ too, a substitution will finish the job. Basic 4 shows it can: $t^2 + \frac{1}{t^2} = \left(t - \frac1t\right)^2 + 2$⁠.
   Basics: completing the square

5. Substitute $y = t - \frac1t$

   Then $dy = \left(1 + \frac{1}{t^2}\right)dt$⁠, and

   $$\begin{aligned}I &= 2\int\frac{dy}{y^2 + (\sqrt2)^2} = 2\cdot\frac{1}{\sqrt2}\tan^{-1}\frac{y}{\sqrt2} + C\\ &= \sqrt2\tan^{-1}\frac{y}{\sqrt2} + C .\end{aligned}$$
   Section 7.4, formula (3) with $a = \sqrt2$⁠. As $t$ runs over $(0, \infty)$⁠, $y = t - \frac1t$ increases from $-\infty$ to $\infty$⁠, so this is a valid one-to-one substitution. Also $2\cdot\frac{1}{\sqrt2} = \sqrt2$⁠.
   $\frac{d}{dt}\left(-\frac1t\right) = \frac{1}{t^2}$⁠.

6. Back to $x$

   $\frac{y}{\sqrt2} = \frac{t - \frac1t}{\sqrt2} = \frac{t^2 - 1}{\sqrt2\,t}$⁠, and with $t^2 = \tan x$⁠, $t = \sqrt{\tan x}$⁠:

   $$I = \sqrt2\tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2\tan x}}\right) + C .$$
   Multiplying top and bottom of $\frac{t - 1/t}{\sqrt2}$ by $t$ clears the inner fraction. Then $\sqrt2\,t = \sqrt2\sqrt{\tan x} = \sqrt{2\tan x}$⁠.

7. Check: $\frac{d}{dx}$ of our answer

   Let $G = \sqrt2\tan^{-1}u$ with $u = \frac{t^2 - 1}{\sqrt2\,t}$⁠. First differentiate with respect to $t$⁠. Here $\frac{du}{dt} = \frac{1}{\sqrt2}\left(1 + \frac{1}{t^2}\right)$⁠, which is $\frac{t^2+1}{\sqrt2\,t^2}$⁠, and

   $$1 + u^2 = 1 + \frac{(t^2-1)^2}{2t^2} = \frac{t^4 + 1}{2t^2} .$$

   So

   $$\frac{dG}{dt} = \sqrt2\cdot\frac{2t^2}{t^4+1}\cdot\frac{t^2+1}{\sqrt2\,t^2} = \frac{2(t^2+1)}{t^4+1} .$$

   Then, with $\frac{dt}{dx} = \frac{1 + t^4}{2t}$ from Step 3,

   $$\frac{dG}{dx} = \frac{2(t^2+1)}{t^4+1}\cdot\frac{t^4+1}{2t} = \frac{t^2 + 1}{t} = t + \frac1t ,$$

   which is $\sqrt{\tan x} + \frac{1}{\sqrt{\tan x}} = \sqrt{\tan x} + \sqrt{\cot x}$⁠, the integrand.

   The chain rule through $t$ keeps the algebra short. $\frac{d}{du}\tan^{-1}u = \frac{1}{1+u^2}$⁠, and $(t^2 - 1)^2 + 2t^2 = t^4 + 1$⁠.

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