Exercise 7.3 · Q16: $\int\tan^4 x\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   $\tan x = \frac{\sin x}{\cos x}$ needs $\cos x \ne 0$⁠, so

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

   We work on any interval between two consecutive such points, for example $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠.

   $\tan x$ blows up at these points, so no antiderivative can cross them. On each interval between them the identity $1 + \tan^2 x = \sec^2 x$ holds at every point.

2. Rewrite $\tan^4 x$ using $\tan^2 x = \sec^2 x - 1$

   Apply the identity to one of the two factors $\tan^2 x$⁠:

   $$\begin{aligned} \tan^4 x &= \tan^2 x\cdot\tan^2 x \\ &= \tan^2 x\,(\sec^2 x - 1) \\ &= \tan^2 x\sec^2 x - \tan^2 x. \end{aligned}$$

   Apply it once more to the last term:

   $$\begin{aligned} \tan^4 x &= \tan^2 x\sec^2 x - (\sec^2 x - 1) \\ &= \tan^2 x\sec^2 x - \sec^2 x + 1. \end{aligned}$$
   We cannot integrate $\tan^4 x$ directly, but we can integrate anything of the form "(function of $\tan x$⁠) $\times \sec^2 x$". Converting one $\tan^2 x$ into $\sec^2 x - 1$ creates exactly such a term, $\tan^2 x\sec^2 x$⁠. The leftover $\tan^2 x$ is converted the same way into $\sec^2 x - 1$⁠, whose pieces are standard integrals.
   $1 + \tan^2 x = \sec^2 x$⁠. Basics: trigonometric identities for integrals

3. Integrate $\tan^2 x\sec^2 x$ by substitution

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

   $$\begin{aligned} \int \tan^2 x\sec^2 x\,dx &= \int t^2\,dt \\ &= \frac{t^3}{3} + C_1 = \frac{\tan^3 x}{3} + C_1. \end{aligned}$$
   The factor $\sec^2 x\,dx$ is precisely $dt$⁠, and $\tan^2 x$ becomes $t^2$⁠. Nothing in $x$ is left over, so the substitution is complete, and the power rule finishes it.
   Change of variable, Section 7.3.1; standard formula (i), Section 7.2. Lesson: integration by substitution

4. Integrate the remaining terms and combine

   By standard formulae (iv) and (i), $\int\sec^2 x\,dx = \tan x + C_2$ and $\int 1\,dx = x + C_3$⁠. By Property (V),

   $$\begin{aligned} \int \tan^4 x\,dx &= \int\tan^2 x\sec^2 x\,dx \\ &\quad - \int\sec^2 x\,dx + \int dx \\ &= \frac{\tan^3 x}{3} - \tan x + x + C, \end{aligned}$$

   where the separate constants are combined into a single $C$⁠.

   The integral of a sum (or difference) is the sum (or difference) of the integrals. Three arbitrary constants added together are just one arbitrary constant.
   Property (V), Section 7.2.1; standard formulae (i) and (iv), Section 7.2. Lesson: integration as anti-differentiation

5. Check: differentiate the answer back

   By the chain rule, $\frac{d}{dx}\tan^3 x = 3\tan^2 x\sec^2 x$⁠. So, with $F(x) = \frac13\tan^3 x - \tan x + x$⁠,

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

   The second and fourth lines use $\sec^2 x - 1 = \tan^2 x$⁠. This is the integrand, so the answer is correct on every interval where $\cos x \ne 0$⁠.

   Chain rule (Theorem 4, Section 5.3.1); Property (I), Section 7.2.1. Basics: integration undoes differentiation

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