Exercise 7.8 · Q7: $\int_0^{\pi/4}\tan x\,dx$

Basics you need

Solution

1. Check continuity on $\left[0, \frac{\pi}{4}\right]$

   $\tan x = \frac{\sin x}{\cos x}$ is continuous wherever $\cos x \ne 0$⁠. On $\left[0, \frac{\pi}{4}\right]$⁠, $\cos x$ decreases from 1 to $\frac{1}{\sqrt2}$⁠, so $\cos x \ge \frac{1}{\sqrt2} > 0$⁠. Hence $\tan x$ is continuous on the whole interval.

   $\tan x$ breaks down only where $\cos x = 0$⁠, at $x = \pm\frac{\pi}{2}, \pm\frac{3\pi}{2}, \ldots$⁠. None of these lies in $\left[0, \frac{\pi}{4}\right]$⁠, so Theorem 2 can be used (Remark (iv) after Theorem 2).
   A quotient of continuous functions is continuous where the denominator is not 0 (Chapter 5, Theorem 1). Lesson: the fundamental theorem

2. Antiderivative by the substitution $t = \cos x$

   Write $\tan x = \frac{\sin x}{\cos x}$ and let $t = \cos x$⁠. Then $dt = -\sin x\,dx$⁠, so $\sin x\,dx = -dt$⁠:

   $$\begin{aligned}\int\tan x\,dx &= \int\frac{\sin x\,dx}{\cos x} = -\int\frac{dt}{t} \\ &= -\log|t| + C = -\log|\cos x| + C.\end{aligned}$$

   Since $-\log|\cos x| = \log\frac{1}{|\cos x|}$⁠, which is $\log|\sec x|$⁠, take

   $$F(x) = \log|\sec x| = \log\sec x \quad\text{on } \left[0, \tfrac{\pi}{4}\right].$$
   The numerator $\sin x$ is (minus) the derivative of the denominator $\cos x$⁠, which is the signal for $t = \cos x$⁠. On our interval $\sec x = \frac{1}{\cos x} \ge 1 > 0$⁠, so the absolute value bars can be dropped. This is result (i) of Section 7.3.1: $\int\tan x\,dx = \log|\sec x| + C$⁠.
   Section 7.3.1, result (i); standard formula (xii). Lesson: integration by substitution

3. Check: differentiate $F$ back

   Chain rule (outer $\log$⁠, inner $\sec x$⁠):

   $$F'(x) = \frac{1}{\sec x}\cdot\sec x\tan x = \tan x.$$
   $\frac{d}{dx}\sec x = \sec x\tan x$⁠, and the $\sec x$ cancels, leaving exactly the integrand.
   $\frac{d}{dx}\log u = \frac{1}{u}\cdot\frac{du}{dx}$ for $u > 0$⁠. Basics: the chain rule

4. Evaluate $F\left(\frac{\pi}{4}\right) - F(0)$

   $\sec\frac{\pi}{4} = \frac{1}{\cos\frac{\pi}{4}} = \frac{1}{1/\sqrt2} = \sqrt2$ and $\sec 0 = \frac{1}{\cos 0} = 1$⁠. So

   $$\begin{aligned}\int_0^{\frac{\pi}{4}}\tan x\,dx &= \log\sqrt2 - \log 1 = \log 2^{\frac12} - 0 \\ &= \tfrac12\log 2 \approx 0.3465736.\end{aligned}$$
   Upper minus lower. $\log 1 = 0$⁠, and $\log\sqrt2 = \log 2^{1/2} = \frac12\log 2$ by the power law of logarithms. This is the book's form.
   $\log a^k = k\log a$⁠. Basics: logarithms

5. Does the size make sense?

   On $\left[0, \frac{\pi}{4}\right]$ the curve $y = \tan x$ lies above the line $y = x$ and below the chord $y = \frac{4}{\pi}x$ joining $(0, 0)$ to $\left(\frac{\pi}{4}, 1\right)$⁠. So the area under it lies between the two triangle areas:

   $$\frac{\pi^2}{32} \approx 0.308 < 0.347 < \frac{\pi}{8} \approx 0.393.$$
   $\tan x \ge x$ because $\tan x - x$ is 0 at $x = 0$ and has derivative $\sec^2 x - 1 = \tan^2 x \ge 0$⁠. The curve is below the chord because it bends upwards: its second derivative $2\sec^2 x\tan x$ is $\ge 0$ there. The triangles under $y = x$ and under the chord have areas $\frac12\left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{32}$ and $\frac12\cdot\frac{\pi}{4}\cdot 1 = \frac{\pi}{8}$⁠.
   Definite integral as area, Section 7.8.1. Basics: area as a sum of thin strips

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