Exercise 7.3 · Q15: $\int\tan^3 2x\,\sec 2x\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   $\tan 2x$ and $\sec 2x$ both need $\cos 2x \ne 0$⁠, that is, $2x \ne \frac{\pi}{2} + n\pi$⁠, so

   $$x \ne \frac{\pi}{4} + \frac{n\pi}{2}, \quad n \text{ an integer}.$$

   We work on any interval avoiding these points.

   Both factors have $\cos 2x$ in a denominator. Knowing the domain also guarantees that the identity in Step 2 is valid at every point we use.

2. Split off $\sec 2x\tan 2x$ and use $\tan^2 = \sec^2 - 1$

   $$\begin{aligned} \tan^3 2x\,\sec 2x &= \tan^2 2x\,(\sec 2x\tan 2x) \\ &= (\sec^2 2x - 1)(\sec 2x\tan 2x). \end{aligned}$$
   We want the integrand to read "a function of $\sec 2x$" times "the derivative of $\sec 2x$". The derivative of $\sec 2x$ contains $\sec 2x\tan 2x$⁠, so we set that product aside. What remains, $\tan^2 2x$⁠, is an even power of tangent, and the identity $\tan^2\theta = \sec^2\theta - 1$ (with $\theta = 2x$⁠) rewrites it in secants.
   $1 + \tan^2\theta = \sec^2\theta$⁠. Basics: trigonometric identities for integrals

3. Substitute $t = \sec 2x$

   Let $t = \sec 2x$⁠. By the chain rule,

   $$\begin{aligned} dt &= 2\sec 2x\tan 2x\,dx, \\ \text{so}\quad \sec 2x\tan 2x\,dx &= \frac12\,dt. \end{aligned}$$

   Therefore

   $$\int \tan^3 2x\,\sec 2x\,dx = \frac12\int (t^2 - 1)\,dt.$$
   The bracket $\sec^2 2x - 1$ becomes $t^2 - 1$⁠, and the set-aside factor with $dx$ becomes $\frac12\,dt$⁠. The $\frac12$ comes from the inner function $2x$⁠: forgetting it is the most common error here.
   Change of variable, Section 7.3.1; $\frac{d}{dx}\sec x = \sec x\tan x$ (list in Section 7.2, (vi)). Lesson: integration by substitution

4. Integrate the polynomial in $t$

   $$\begin{aligned} \frac12\int (t^2 - 1)\,dt &= \frac12\left(\frac{t^3}{3} - t\right) + C \\ &= \frac{t^3}{6} - \frac{t}{2} + C. \end{aligned}$$
   Each term uses the power rule (formula (i)): $\int t^2\,dt = \frac{t^3}{3}$ and $\int 1\,dt = t$⁠. Property (V) lets us integrate term by term and keep the constant $\frac12$ outside.
   Standard formula (i), Section 7.2; Property (V), Section 7.2.1. Basics: powers and the power rule

5. Return to $x$

   $$\int \tan^3 2x\,\sec 2x\,dx = \frac16\sec^3 2x - \frac12\sec 2x + C.$$
   Replace $t$ by $\sec 2x$⁠. This is exactly the textbook's form.

6. Check: differentiate the answer back

   By the chain rule (outer function $u^3$⁠, inner function $u = \sec 2x$⁠):

   $$\begin{aligned} \frac{d}{dx}\sec^3 2x &= 3\sec^2 2x\cdot 2\sec 2x\tan 2x \\ &= 6\sec^3 2x\tan 2x. \end{aligned}$$

   So, with $F(x) = \frac16\sec^3 2x - \frac12\sec 2x$⁠,

   $$\begin{aligned} F'(x) &= \sec^3 2x\tan 2x - \sec 2x\tan 2x \\ &= \sec 2x\tan 2x\,(\sec^2 2x - 1) \\ &= \sec 2x\tan 2x\cdot\tan^2 2x \\ &= \tan^3 2x\,\sec 2x. \end{aligned}$$

   This is the integrand, so the answer is correct on every interval where $\cos 2x \ne 0$⁠.

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

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