Exercise 7.2 · Q39 (MCQ): $\int\frac{dx}{\sin^2x\cos^2x}$

Basics you need

Solution

1. Where is the integrand defined?

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

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

   So we work on any interval between consecutive multiples of $\frac{\pi}{2}$⁠, such as $\left(0, \frac{\pi}{2}\right)$⁠.

   At these points the denominator is zero. They are also exactly the points where $\tan x$ or $\cot x$ fails to exist, so the answer in Step 3 lives on the same intervals.

2. Split the integrand using $1 = \sin^2 x + \cos^2 x$

   $$\begin{aligned} \frac{1}{\sin^2 x\cos^2 x} &= \frac{\sin^2 x + \cos^2 x}{\sin^2 x\cos^2 x} \\ &= \frac{\sin^2 x}{\sin^2 x\cos^2 x} + \frac{\cos^2 x}{\sin^2 x\cos^2 x} \\ &= \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \\ &= \sec^2 x + \operatorname{cosec}^2 x. \end{aligned}$$
   A product in the denominator is hard to integrate, but a sum of simple fractions is easy. Writing the numerator 1 as $\sin^2 x + \cos^2 x$ lets each term cancel one factor of the denominator. Each cancellation is valid because $\sin x \ne 0$ and $\cos x \ne 0$ on the domain.
   $\sin^2 x + \cos^2 x = 1$⁠. Basics: trigonometric identities

3. Integrate each piece

   $$\begin{aligned} \int\frac{dx}{\sin^2 x\cos^2 x} &= \int\sec^2 x\,dx \\ &\quad + \int\operatorname{cosec}^2 x\,dx \\ &= \tan x - \cot x + C. \end{aligned}$$
   The integral of a sum is the sum of the integrals (Property (III), Section 7.2.1). Then standard formula (iv) gives $\tan x$⁠, and standard formula (v) gives $-\cot x$⁠: the minus sign is because $\frac{d}{dx}\cot x = -\operatorname{cosec}^2 x$⁠.
   Standard formulae (iv) and (v), Section 7.2. Lesson: integration as anti-differentiation

4. Check: differentiate the answer back

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

   This is the integrand. ✓

   Property (I), Section 7.2.1. The second line reverses Step 2 by putting the two fractions over the common denominator $\sin^2 x\cos^2 x$⁠.
   Basics: integration undoes differentiation

5. A second route, by substitution

   Since $\sin x\cos x = \frac12\sin 2x$⁠, the integrand is $\frac{4}{\sin^2 2x} = 4\operatorname{cosec}^2 2x$⁠. Put $u = 2x$⁠, $dx = \frac12\,du$⁠:

   $$\int 4\operatorname{cosec}^2 2x\,dx = 2\int\operatorname{cosec}^2 u\,du = -2\cot 2x + C.$$

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

   $$\begin{aligned} \tan x - \cot x &= \frac{\sin^2 x - \cos^2 x}{\sin x\cos x} \\ &= \frac{-\cos 2x}{\frac12\sin 2x} = -2\cot 2x. \end{aligned}$$
   The two routes agree exactly (the difference is the constant 0), which confirms option (B) independently. This route also shows why option (D), which contains $\cot 2x$⁠, can look tempting; but (D) has the wrong coefficient and an extra $\tan x$⁠.
   $\sin 2x = 2\sin x\cos x$⁠, $\cos 2x = \cos^2 x - \sin^2 x$⁠. Lesson: integration by substitution

6. Examine every option

   An option is right exactly when its derivative is the integrand. Test at $x = \frac{\pi}{6}$⁠, where $\sin^2 x = \frac14$⁠, $\cos^2 x = \frac34$⁠, so the integrand is $\frac{1}{3/16} = \frac{16}{3}$⁠. Also $\sec^2\frac{\pi}{6} = \frac43$⁠, $\operatorname{cosec}^2\frac{\pi}{6} = 4$⁠, $\operatorname{cosec}^2\frac{\pi}{3} = \frac43$⁠.

   • (A) $\frac{d}{dx}(\tan x + \cot x) = \sec^2 x - \operatorname{cosec}^2 x$⁠. At $\frac{\pi}{6}$⁠: $\frac43 - 4 = -\frac83 \ne \frac{16}{3}$⁠. Wrong (the sign of the $\cot$ term is wrong).
   • (B) Derivative $\sec^2 x + \operatorname{cosec}^2 x$⁠, equal to the integrand for every $x$ in the domain (Step 4). At $\frac{\pi}{6}$⁠: $\frac43 + 4 = \frac{16}{3}$⁠. ✓ Correct.
   • (C) $\tan x\cot x = 1$ wherever both exist, so this option is the constant $1 + C$⁠; its derivative is $0$⁠, never $\frac{1}{\sin^2 x\cos^2 x}$ (which is always positive). Wrong.
   • (D) $\frac{d}{dx}(\tan x - \cot 2x) = \sec^2 x + 2\operatorname{cosec}^2 2x$⁠. At $\frac{\pi}{6}$⁠: $\frac43 + 2\cdot\frac43 = 4 \ne \frac{16}{3}$⁠. Wrong.
   A single point where the derivatives differ is enough to reject an option. (Do not test (D) at $x = \frac{\pi}{4}$⁠: there both sides happen to equal 4, a coincidence that proves nothing.)

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