Exercise 7.8 · Q8: $\int_{\pi/6}^{\pi/4}\operatorname{cosec}x\,dx$

Basics you need

Solution

1. Check continuity on $\left[\frac{\pi}{6}, \frac{\pi}{4}\right]$

   $\operatorname{cosec}x = \frac{1}{\sin x}$ is continuous wherever $\sin x \ne 0$⁠. On $\left[\frac{\pi}{6}, \frac{\pi}{4}\right]$⁠, $\sin x$ increases from $\frac12$ to $\frac{1}{\sqrt2}$⁠, so $\sin x \ge \frac12 > 0$⁠. The integrand is continuous on the whole interval.

   $\operatorname{cosec}x$ breaks down only at multiples of $\pi$ (where $\sin x = 0$⁠); none lies in $\left[\frac{\pi}{6}, \frac{\pi}{4}\right]$⁠. So Theorem 2 applies (Remark (iv) after Theorem 2).
   Quotient of continuous functions: Chapter 5, Theorem 1, Section 5.2.1. Lesson: the fundamental theorem

2. Antiderivative: the book's trick

   Multiply top and bottom by $\operatorname{cosec}x + \cot x$ (which is positive on our interval):

   $$\int\operatorname{cosec}x\,dx = \int\frac{\operatorname{cosec}x\left(\operatorname{cosec}x + \cot x\right)}{\operatorname{cosec}x + \cot x}\,dx.$$

   Let $t = \operatorname{cosec}x + \cot x$⁠. Then

   $$\begin{aligned}dt &= \left(-\operatorname{cosec}x\cot x - \operatorname{cosec}^2 x\right)dx \\ &= -\operatorname{cosec}x\left(\operatorname{cosec}x + \cot x\right)dx,\end{aligned}$$

   which is minus the numerator. So

   $$\int\operatorname{cosec}x\,dx = -\int\frac{dt}{t} = -\log|\operatorname{cosec}x + \cot x| + C.$$

   Now use the identity $\operatorname{cosec}^2 x - \cot^2 x = 1$⁠:

   $$(\operatorname{cosec}x + \cot x)(\operatorname{cosec}x - \cot x) = 1,$$

   so the two brackets are reciprocals of each other, and

   $$\begin{aligned}-\log|\operatorname{cosec}x + \cot x| &= \log\frac{1}{|\operatorname{cosec}x + \cot x|} \\ &= \log|\operatorname{cosec}x - \cot x|.\end{aligned}$$

   Take

   $$F(x) = \log\left|\operatorname{cosec}x - \cot x\right|.$$
   The multiplier is chosen so that the numerator becomes (minus) the derivative of the denominator; then the integral is a logarithm. The final form is result (iv) of Section 7.3.1: $\int\operatorname{cosec}x\,dx = \log|\operatorname{cosec}x - \cot x| + C$⁠.
   $\frac{d}{dx}\operatorname{cosec}x = -\operatorname{cosec}x\cot x$⁠, $\frac{d}{dx}\cot x = -\operatorname{cosec}^2 x$⁠. Lesson: integration by substitution

3. Check: differentiate $F$ back

   On our interval $\operatorname{cosec}x - \cot x = \frac{1 - \cos x}{\sin x} > 0$⁠, so $F(x) = \log(\operatorname{cosec}x - \cot x)$ and, by the chain rule,

   $$\begin{aligned}F'(x) &= \frac{-\operatorname{cosec}x\cot x + \operatorname{cosec}^2 x}{\operatorname{cosec}x - \cot x} \\ &= \frac{\operatorname{cosec}x\left(\operatorname{cosec}x - \cot x\right)}{\operatorname{cosec}x - \cot x} = \operatorname{cosec}x.\end{aligned}$$
   Factor $\operatorname{cosec}x$ out of the numerator; the bracket cancels with the denominator, leaving exactly the integrand.
   $\frac{d}{dx}\log u = \frac{u'}{u}$ for $u > 0$⁠. Basics: the chain rule

4. Values at the two limits

   At $x = \frac{\pi}{4}$⁠: $\operatorname{cosec}\frac{\pi}{4} = \sqrt2$ and $\cot\frac{\pi}{4} = 1$⁠, so

   $$F\left(\tfrac{\pi}{4}\right) = \log\left(\sqrt2 - 1\right).$$

   At $x = \frac{\pi}{6}$⁠: $\operatorname{cosec}\frac{\pi}{6} = \frac{1}{1/2} = 2$ and $\cot\frac{\pi}{6} = \frac{\sqrt3/2}{1/2} = \sqrt3$⁠, so

   $$F\left(\tfrac{\pi}{6}\right) = \log\left(2 - \sqrt3\right).$$
   Both $\sqrt2 - 1 \approx 0.414$ and $2 - \sqrt3 \approx 0.268$ are positive, so the bars are not needed. Both logarithms are negative (their arguments are less than 1); that is fine, only their difference matters.
   $\operatorname{cosec} = \frac{1}{\sin}$⁠, $\cot = \frac{\cos}{\sin}$⁠. Basics: trigonometric values

5. Subtract: upper minus lower

   $$\begin{aligned}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\operatorname{cosec}x\,dx &= \log\left(\sqrt2 - 1\right) - \log\left(2 - \sqrt3\right) \\ &= \log\frac{\sqrt2 - 1}{2 - \sqrt3}.\end{aligned}$$

   In decimals: $\log(\sqrt2 - 1) \approx -0.8813736$ and $\log(2 - \sqrt3) \approx -1.3169579$⁠, so the integral is

   $$-0.8813736 - (-1.3169579) \approx 0.4355843.$$
   $\log a - \log b = \log\frac{a}{b}$ gives the book's form. Multiplying top and bottom by $2 + \sqrt3$ (and using $(2 - \sqrt3)(2 + \sqrt3) = 1$⁠) gives an equivalent form: $\frac{\sqrt2 - 1}{2 - \sqrt3}$ equals $(\sqrt2 - 1)(2 + \sqrt3) \approx 1.5458661$⁠, and $\log 1.5458661 \approx 0.4355843$⁠.
   Laws of logarithms. Basics: logarithms

6. Cross-check with $\tan\frac{x}{2}$ and the graph

   By the half-angle formulas,

   $$\begin{aligned}\operatorname{cosec}x - \cot x &= \frac{1 - \cos x}{\sin x} \\ &= \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \tan\frac{x}{2}.\end{aligned}$$

   So $F(x) = \log\tan\frac{x}{2}$⁠, and the two values above are $\tan\frac{\pi}{8} = \sqrt2 - 1$ and $\tan\frac{\pi}{12} = 2 - \sqrt3$⁠, as expected.

   Size: $\operatorname{cosec}x$ falls from 2 to $\sqrt2$ over a base of width $\frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12}$⁠, so

   $$\sqrt2\cdot\frac{\pi}{12} \approx 0.370 < 0.436 < 2\cdot\frac{\pi}{12} \approx 0.524.$$
   Two independent facts (the well-known values of $\tan\frac{\pi}{8}$ and $\tan\frac{\pi}{12}$⁠, and the rectangle bounds) agree with the answer.
   $1 - \cos x = 2\sin^2\frac{x}{2}$⁠, $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$⁠. Basics: trigonometric identities

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