Exercise 7.2 · Q37: $\int\frac{x^3\sin(\tan^{-1}x^4)}{1+x^8}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   $\tan^{-1}$ accepts every real number, $\sin$ too, and $1 + x^8 \ge 1 > 0$⁠. So the integrand is defined for every real $x$⁠.

   No points need to be excluded, so the answer will hold on the whole real line.

2. Differentiate $\tan^{-1}(x^4)$

   Chain rule, with outer function $\tan^{-1}u$ and inner function $u = x^4$⁠:

   $$\begin{aligned} \frac{d}{dx}\tan^{-1}(x^4) &= \frac{1}{1 + (x^4)^2}\cdot 4x^3 \\ &= \frac{4x^3}{1 + x^8}. \end{aligned}$$
   The integrand contains $\tan^{-1}(x^4)$ inside a sine, and also $\frac{x^3}{1 + x^8}$⁠. This computation shows that the second factor is a quarter of the derivative of the first. That match is what makes the substitution work (Section 7.3.1).
   $(x^4)^2 = x^8$⁠; standard formula (x), Section 7.2; chain rule, Theorem 4, Section 5.3.1.

3. Substitute $t = \tan^{-1}(x^4)$

   Put $t = \tan^{-1}(x^4)$⁠. By Step 2,

   $$dt = \frac{4x^3}{1 + x^8}\,dx, \quad\text{so}\quad \frac{x^3\,dx}{1 + x^8} = \tfrac14\,dt.$$

   Then

   $$\int\sin\left(\tan^{-1}x^4\right)\cdot\frac{x^3\,dx}{1 + x^8} = \tfrac14\int\sin t\,dt.$$

   Since $x^4 \ge 0$⁠, the new variable satisfies $0 \le t < \frac{\pi}{2}$⁠.

   We substitute for the whole inner function $\tan^{-1}(x^4)$⁠, just as the book does for $\sin(\tan^{-1}x)$ in Example 5(iv). One could also go in two stages ($u = x^4$⁠, then $t = \tan^{-1}u$⁠); the result is the same.
   Lesson: integration by substitution

4. Integrate and return to $x$

   $$\tfrac14\int\sin t\,dt = -\tfrac14\cos t + C = -\tfrac14\cos\left(\tan^{-1}x^4\right) + C.$$
   Standard formula (iii), then $t$ replaced by $\tan^{-1}(x^4)$⁠.
   Standard formula (iii), Section 7.2: $\int\sin t\,dt = -\cos t + C$⁠.

5. Check: differentiate the answer back

   Let $F(x) = -\frac14\cos(\tan^{-1}x^4)$⁠. Chain rule, outer $-\frac14\cos t$⁠, inner $t = \tan^{-1}(x^4)$ with $t' = \frac{4x^3}{1 + x^8}$ (Step 2):

   $$\begin{aligned} F'(x) &= -\tfrac14\cdot\left(-\sin(\tan^{-1}x^4)\right)\cdot\frac{4x^3}{1 + x^8} \\ &= \frac{x^3\sin(\tan^{-1}x^4)}{1 + x^8}. \end{aligned}$$

   This is the integrand. ✓

   Property (I), Section 7.2.1. The $\frac14$ and the $4$ cancel, and the two minus signs cancel.
   Basics: integration undoes differentiation

6. An algebraic form of the answer

   Let $\theta = \tan^{-1}s$ with $s = x^4$⁠. Then $\tan\theta = s$ and $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$⁠, where $\cos\theta > 0$⁠. From $\sec^2\theta = 1 + \tan^2\theta = 1 + s^2$⁠:

   $$\cos\theta = \frac{1}{\sqrt{1 + s^2}},$$

   taking the positive root because $\cos\theta > 0$⁠.

   With $s = x^4$⁠, $\cos(\tan^{-1}x^4) = \frac{1}{\sqrt{1 + x^8}}$⁠, so the answer can also be written

   $$-\frac{1}{4\sqrt{1 + x^8}} + C.$$

   Differentiating this form (chain rule, inner $1 + x^8$⁠):

   $$\begin{aligned} \frac{d}{dx}\left(-\tfrac14(1 + x^8)^{-1/2}\right) &= \tfrac18(1 + x^8)^{-3/2}\cdot 8x^7 \\ &= \frac{x^7}{(1 + x^8)^{3/2}}. \end{aligned}$$

   And in the same way $\sin\theta = \cos\theta\tan\theta = \frac{s}{\sqrt{1 + s^2}}$⁠, so the integrand itself is

   $$\frac{x^3}{1 + x^8}\cdot\frac{x^4}{\sqrt{1 + x^8}} = \frac{x^7}{(1 + x^8)^{3/2}}.$$

   The two agree. ✓

   This is the same function as $-\frac14\cos(\tan^{-1}x^4)$⁠, not just the same family, so no constant is involved. The principal range of $\tan^{-1}$ (Chapter 2) is what fixes the sign of the root.
   $\sec^2\theta = 1 + \tan^2\theta$⁠. Basics: trigonometric identities

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