Exercise 7.3 · Q17: $\int\frac{\sin^3x+\cos^3x}{\sin^2x\cos^2x}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   The denominator $\sin^2 x\cos^2 x$ must be non-zero, so $\sin x \ne 0$ and $\cos x \ne 0$⁠, that is,

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

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

   $\sin x = 0$ at multiples of $\pi$ and $\cos x = 0$ at odd multiples of $\frac{\pi}{2}$⁠; together these are all multiples of $\frac{\pi}{2}$⁠. On the remaining intervals every cancellation below is legitimate.

2. Split the fraction and cancel

   $$\begin{aligned} f(x) &= \frac{\sin^3 x}{\sin^2 x\cos^2 x} + \frac{\cos^3 x}{\sin^2 x\cos^2 x} \\ &= \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x}. \end{aligned}$$
   In the first fraction $\sin^2 x$ cancels from $\sin^3 x$⁠, leaving $\sin x$ over $\cos^2 x$⁠; in the second, $\cos^2 x$ cancels from $\cos^3 x$⁠. We may cancel because $\sin x$ and $\cos x$ are non-zero on our domain (Step 1).
   $\frac{A + B}{D} = \frac AD + \frac BD$⁠. Basics: splitting fractions

3. Recognise the two standard integrands

   $$\begin{aligned} \frac{\sin x}{\cos^2 x} &= \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x\tan x, \\ \frac{\cos x}{\sin^2 x} &= \frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} = \operatorname{cosec} x\cot x. \end{aligned}$$
   Writing $\cos^2 x$ as $\cos x\cdot\cos x$ and giving one factor to $\sin x$ produces $\tan x$⁠; the other factor gives $\sec x$⁠. The same trick with $\sin^2 x$ gives $\cot x$ and $\operatorname{cosec} x$⁠. These products are exactly the derivatives of $\sec x$ and of $-\operatorname{cosec} x$⁠.
   Definitions of $\sec$⁠, $\operatorname{cosec}$⁠, $\tan$⁠, $\cot$⁠. Basics: trigonometric identities for integrals

4. Integrate

   $$\begin{aligned} \int f(x)\,dx &= \int\sec x\tan x\,dx \\ &\quad + \int\operatorname{cosec} x\cot x\,dx \\ &= \sec x + (-\operatorname{cosec} x) + C \\ &= \sec x - \operatorname{cosec} x + C. \end{aligned}$$
   Property (III): the integral of a sum is the sum of the integrals. Formula (vi) gives $\sec x$⁠; formula (vii) gives $-\operatorname{cosec} x$⁠, because $\frac{d}{dx}\operatorname{cosec} x = -\operatorname{cosec} x\cot x$⁠.
   Standard formulae (vi) and (vii), Section 7.2; Property (III), Section 7.2.1. Lesson: integration as anti-differentiation

5. Check: differentiate the answer back

   Let $F(x) = \sec x - \operatorname{cosec} x$⁠. Then

   $$\begin{aligned} F'(x) &= \sec x\tan x + \operatorname{cosec} x\cot x \\ &= \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} \\ &= \frac{\sin^3 x + \cos^3 x}{\sin^2 x\cos^2 x}. \end{aligned}$$

   The last line puts the two fractions over the common denominator $\sin^2 x\cos^2 x$ (multiply the first by $\frac{\sin^2 x}{\sin^2 x}$ and the second by $\frac{\cos^2 x}{\cos^2 x}$⁠). This is the integrand, so the answer is correct on every interval avoiding the multiples of $\frac{\pi}{2}$⁠.

   List in Section 7.2, (vi) and (vii); Property (I), Section 7.2.1. Basics: integration undoes differentiation

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