Exercise 7.3 · Q19: $\int\frac{dx}{\sin x\cos^3x}$

Basics you need

Solution

1. Where is the integrand defined?

   We need $\sin x \ne 0$ and $\cos x \ne 0$⁠, that is,

   $$x \ne \frac{n\pi}{2}, \quad n \text{ an integer}.$$

   On each interval between consecutive multiples of $\frac{\pi}{2}$⁠, $\tan x$ is defined and non-zero.

   The non-zero $\tan x$ matters: after the substitution we divide by $t = \tan x$⁠, and $\log|\tan x|$ appears in the answer. Both need $\tan x \ne 0$⁠, which holds exactly because $\sin x \ne 0$⁠.

2. Divide top and bottom by $\cos^4 x$

   $$\begin{aligned} \frac{1}{\sin x\cos^3 x} &= \frac{\dfrac{1}{\cos^4 x}}{\dfrac{\sin x\cos^3 x}{\cos^4 x}} \\ &= \frac{\sec^4 x}{\dfrac{\sin x}{\cos x}} = \frac{\sec^4 x}{\tan x}. \end{aligned}$$
   Dividing numerator and denominator by the same non-zero quantity does not change a fraction. We choose $\cos^4 x$ because the denominator has four trigonometric factors in total: one $\sin x$ pairs with one $\cos x$ to make $\tan x$⁠, and the other three $\cos x$ cancel against $\cos^4 x$⁠, leaving everything in $\tan x$ and $\sec x$⁠.
   $\frac{\sin x}{\cos x} = \tan x$⁠, $\frac{1}{\cos x} = \sec x$⁠. Basics: trigonometric identities for integrals

3. Split $\sec^4 x$ and substitute $t = \tan x$

   Write $\sec^4 x = \sec^2 x\cdot\sec^2 x$⁠, which is $(1 + \tan^2 x)\sec^2 x$⁠. Then

   $$\int\frac{dx}{\sin x\cos^3 x} = \int\frac{1 + \tan^2 x}{\tan x}\,\sec^2 x\,dx.$$

   Put $t = \tan x$⁠, so $dt = \sec^2 x\,dx$⁠:

   $$\int\frac{dx}{\sin x\cos^3 x} = \int\frac{1 + t^2}{t}\,dt.$$
   One factor $\sec^2 x$ is kept with $dx$ because together they form $dt$⁠. The other factor is rewritten with the identity $\sec^2 x = 1 + \tan^2 x$ so that it too is a function of $\tan x$⁠. Now the integrand is entirely in $t$⁠.
   $1 + \tan^2 x = \sec^2 x$⁠; change of variable, Section 7.3.1. Lesson: integration by substitution

4. Integrate in $t$

   $$\begin{aligned} \int\frac{1 + t^2}{t}\,dt &= \int\left(\frac1t + t\right)dt \\ &= \log|t| + \frac{t^2}{2} + C. \end{aligned}$$
   Splitting the fraction gives two standard integrals: formula (xii) for $\frac1t$ (valid since $t \ne 0$⁠) and the power rule (formula (i)) for $t$⁠. The modulus in $\log|t|$ is needed because $\tan x$ is negative on some intervals of the domain.
   Standard formulae (i) and (xii), Section 7.2. Basics: logarithms and 1/x

5. Return to $x$

   $$\int\frac{dx}{\sin x\cos^3 x} = \log|\tan x| + \frac12\tan^2 x + C.$$
   Replace $t$ by $\tan x$⁠. This is the textbook's answer exactly.

6. Check: differentiate the answer back

   By the chain rule, $\frac{d}{dx}\log|\tan x| = \frac{\sec^2 x}{\tan x}$ and $\frac{d}{dx}\tfrac12\tan^2 x = \tan x\sec^2 x$⁠. With $F(x) = \log|\tan x| + \frac12\tan^2 x$⁠,

   $$\begin{aligned} F'(x) &= \frac{\sec^2 x}{\tan x} + \tan x\sec^2 x \\ &= \frac{\sec^2 x\,(1 + \tan^2 x)}{\tan x} = \frac{\sec^4 x}{\tan x} \\ &= \frac{1}{\cos^4 x}\cdot\frac{\cos x}{\sin x} = \frac{1}{\sin x\cos^3 x}. \end{aligned}$$

   This is the integrand, so the answer is correct on every interval avoiding the multiples of $\frac{\pi}{2}$⁠.

   $\frac{d}{du}\log|u| = \frac1u$ (list in Section 7.2, (xii)); chain rule (Theorem 4, Section 5.3.1); Property (I), Section 7.2.1. Basics: integration undoes differentiation

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