Exercise 7.2 · Q34: $\int\frac{\sqrt{\tan x}}{\sin x\cos x}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   • $\sqrt{\tan x}$ needs $\tan x$ defined and $\tan x \ge 0$⁠.
   • The denominator needs $\sin x\cos x \ne 0$⁠, so $\sin x \ne 0$ and $\cos x \ne 0$⁠; then $\tan x \ne 0$⁠.

   Together: $\tan x > 0$⁠, that is,

   $$n\pi < x < n\pi + \tfrac{\pi}{2}, \quad n \text{ an integer}.$$

   On these intervals $\sin x$ and $\cos x$ have the same sign, so $\sin x\cos x > 0$ as well.

   The integrand only exists on these intervals, so the answer is claimed only there (Note, Section 7.2). Knowing $\tan x > 0$ also lets us simplify $\frac{\sqrt{\tan x}}{\tan x} = \frac{1}{\sqrt{\tan x}}$ in Step 2.
   $\tan x > 0$ in the first and third quadrants. Basics: trigonometric facts

2. Rewrite the denominator using $\tan x$

   Since $\tan x\cos^2 x = \frac{\sin x}{\cos x}\cos^2 x = \sin x\cos x$⁠,

   $$\begin{aligned} \frac{\sqrt{\tan x}}{\sin x\cos x} &= \frac{\sqrt{\tan x}}{\tan x\,\cos^2 x} \\ &= \frac{\sqrt{\tan x}}{\tan x}\cdot\sec^2 x = \frac{\sec^2 x}{\sqrt{\tan x}}. \end{aligned}$$
   We want the integrand as "a function of $\tan x$" times "the derivative of $\tan x$". The derivative of $\tan x$ is $\sec^2 x = \frac{1}{\cos^2 x}$⁠, so we pull a factor $\cos^2 x$ out of the denominator. The last equality uses $\frac{\sqrt u}{u} = \frac{1}{\sqrt u}$⁠, valid since $u = \tan x > 0$⁠.
   $\frac{1}{\cos^2 x} = \sec^2 x$⁠. Basics: trigonometric identities

3. Substitute $t = \tan x$

   Put $t = \tan x$⁠, so $dt = \sec^2 x\,dx$⁠. On the domain, $t > 0$⁠. Then

   $$\int\frac{\sec^2 x}{\sqrt{\tan x}}\,dx = \int\frac{dt}{\sqrt t} = \int t^{-1/2}\,dt.$$
   After Step 2, the integrand is exactly "a function of $\tan x$" times $\sec^2 x$⁠, the pattern substitution handles (Section 7.3.1).
   Standard formula (iv), Section 7.2: $\frac{d}{dx}\tan x = \sec^2 x$⁠. Lesson: integration by substitution

4. Integrate and return to $x$

   Power rule with $n = -\frac12$⁠, $n + 1 = \frac12$⁠:

   $$\begin{aligned} \int t^{-1/2}\,dt &= \frac{t^{1/2}}{1/2} + C = 2\sqrt t + C \\ &= 2\sqrt{\tan x} + C. \end{aligned}$$
   $n = -\frac12 \ne -1$⁠, so the power rule applies; dividing by $\frac12$ doubles.
   Standard formula (i), Section 7.2. Basics: power rule

5. Check: differentiate the answer back

   Let $F(x) = 2(\tan x)^{1/2}$⁠. Chain rule, outer $2u^{1/2}$⁠, inner $u = \tan x$⁠:

   $$\begin{aligned} F'(x) &= 2\cdot\tfrac12(\tan x)^{-1/2}\sec^2 x \\ &= \frac{\sec^2 x}{\sqrt{\tan x}} = \frac{\tan x}{\sqrt{\tan x}}\cdot\frac{\sec^2 x}{\tan x} \\ &= \frac{\sqrt{\tan x}}{\sin x\cos x}. \end{aligned}$$

   This is the integrand. ✓ (The last step uses $\frac{\tan x}{\sqrt{\tan x}} = \sqrt{\tan x}$ and $\tan x\cos^2 x = \sin x\cos x$⁠.)

   Differentiating back confirms the antiderivative (Property (I), Section 7.2.1) and reverses the rewriting of Step 2.
   Chain rule, Theorem 4, Section 5.3.1. Basics: integration undoes differentiation

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