Exercise 7.3 · Q23 (MCQ): $\int\frac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   The denominator $\sin^2 x\cos^2 x$ must be non-zero:

   $$x \ne \frac{n\pi}{2}, \quad n \text{ an integer}.$$
   $\sin x = 0$ at multiples of $\pi$⁠, $\cos x = 0$ at odd multiples of $\frac{\pi}{2}$⁠. We work on an interval between consecutive multiples of $\frac{\pi}{2}$⁠, where every cancellation below is valid.

2. Split the fraction and cancel

   $$\begin{aligned} f(x) &= \frac{\sin^2 x}{\sin^2 x\cos^2 x} - \frac{\cos^2 x}{\sin^2 x\cos^2 x} \\ &= \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} \\ &= \sec^2 x - \operatorname{cosec}^2 x. \end{aligned}$$
   In the first fraction $\sin^2 x$ cancels, in the second $\cos^2 x$ cancels (both non-zero on our domain). The result is a difference of two functions whose antiderivatives are in the standard list.
   $\sec x = \frac{1}{\cos x}$⁠, $\operatorname{cosec} x = \frac{1}{\sin x}$⁠. Basics: splitting fractions

3. Integrate

   $$\begin{aligned} \int f(x)\,dx &= \int\sec^2 x\,dx - \int\operatorname{cosec}^2 x\,dx \\ &= \tan x - (-\cot x) + C \\ &= \tan x + \cot x + C. \end{aligned}$$
   Property (V) splits the integral. Formula (iv) gives $\tan x$⁠; formula (v) gives $\int\operatorname{cosec}^2 x\,dx = -\cot x$⁠, and subtracting it produces $+\cot x$⁠. The two minus signs are where most errors happen.
   Standard formulae (iv) and (v), Section 7.2; Property (V), Section 7.2.1. Lesson: integration as anti-differentiation

4. Check: differentiate the answer back

   $$\begin{aligned} \frac{d}{dx}(\tan x + \cot x) &= \sec^2 x - \operatorname{cosec}^2 x \\ &= \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} \\ &= \frac{\sin^2 x - \cos^2 x}{\sin^2 x\cos^2 x}. \end{aligned}$$

   The last line uses the common denominator $\sin^2 x\cos^2 x$⁠. This is the integrand. ✓

   List in Section 7.2, (iv) and (v): $\frac{d}{dx}\tan x = \sec^2 x$⁠, $\frac{d}{dx}\cot x = -\operatorname{cosec}^2 x$⁠; Property (I), Section 7.2.1. Basics: integration undoes differentiation

5. A second route, through double angles

   $\sin^2 x - \cos^2 x = -\cos 2x$ and $\sin x\cos x = \frac12\sin 2x$⁠, so

   $$f(x) = \frac{-\cos 2x}{\frac14\sin^2 2x} = \frac{-4\cos 2x}{\sin^2 2x}.$$

   Put $t = \sin 2x$⁠, $dt = 2\cos 2x\,dx$⁠:

   $$\begin{aligned} \int f(x)\,dx &= -2\int\frac{dt}{t^2} = \frac{2}{t} + C \\ &= \frac{2}{\sin 2x} + C. \end{aligned}$$

   This is the same function as $\tan x + \cot x$⁠:

   $$\begin{aligned} \tan x + \cot x &= \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} \\ &= \frac{1}{\sin x\cos x} = \frac{2}{\sin 2x}. \end{aligned}$$
   Two independent methods agreeing is strong evidence. Here they agree exactly, not only up to a constant.
   $\cos 2x = \cos^2 x - \sin^2 x$⁠, $\sin 2x = 2\sin x\cos x$⁠; substitution, Section 7.3.1. Lesson: integration by substitution

6. Examine every option

   Test each option's derivative at $x = \frac{\pi}{6}$⁠, where $\sin^2 x = \frac14$⁠, $\cos^2 x = \frac34$⁠, so the integrand is $\frac{-1/2}{3/16} = -\frac83$⁠. Also $\sec^2\frac{\pi}{6} = \frac43$⁠, $\operatorname{cosec}^2\frac{\pi}{6} = 4$⁠, $\sec\frac{\pi}{6}\tan\frac{\pi}{6} = \frac23$⁠, $\operatorname{cosec}\frac{\pi}{6}\cot\frac{\pi}{6} = 2\sqrt3$⁠.

   • (A) The derivative of $\tan x + \cot x$ is $\sec^2 x - \operatorname{cosec}^2 x$⁠, which equals the integrand everywhere (Step 4). At $\frac{\pi}{6}$⁠: $\frac43 - 4 = -\frac83$⁠. Correct.
   • (B) The derivative of $\tan x + \operatorname{cosec} x$ is $\sec^2 x - \operatorname{cosec} x\cot x$⁠. At $\frac{\pi}{6}$⁠: $\frac43 - 2\sqrt3 \approx -2.130768$⁠, not $-\frac83$⁠. Wrong.
   • (C) The derivative of $-\tan x + \cot x$ is $-\sec^2 x - \operatorname{cosec}^2 x$⁠. At $\frac{\pi}{6}$⁠: $-\frac43 - 4 = -\frac{16}{3}$⁠. Wrong (the sign of the $\tan$ term is reversed).
   • (D) The derivative of $\tan x + \sec x$ is $\sec^2 x + \sec x\tan x$⁠. At $\frac{\pi}{6}$⁠: $\frac43 + \frac23 = 2$⁠. Wrong.
   For (B), (C), (D) one point of disagreement is enough to show the derivative is not the integrand, so the option is not an antiderivative. For (A) we have the general identity from Step 4, not just one point.

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