Exercise 7.10 · Q14: $\int_0^{2\pi}\cos^5x\,dx$

Basics you need

Solution

1. The integrand is continuous

   $\cos x$ is continuous everywhere, and so is its fifth power. So $\cos^5 x$ is continuous on $[0, 2\pi]$ and on every smaller interval used below.

   P6 is stated for functions continuous on $[0, 2a]$⁠. Here that holds for any $a$⁠.
   Products of continuous functions are continuous (Chapter 5, Section 5.2.1).

2. P6 with $2a = 2\pi$⁠: the two halves match

   Here $a = \pi$⁠. Test the condition of P6:

   $$\begin{aligned}f(2\pi - x) &= \cos^5(2\pi - x) \\ &= (\cos x)^5 = f(x) .\end{aligned}$$

   So the first case of P6 applies:

   $$I = \int_0^{2\pi}\cos^5 x\,dx = 2\int_0^{\pi}\cos^5 x\,dx .$$
   $\cos(2\pi - x) = \cos x$ because $\cos$ has period $2\pi$ and is even: $\cos(2\pi - x) = \cos(-x) = \cos x$⁠. Geometrically, the graph on $[\pi, 2\pi]$ is the mirror image of the graph on $[0, \pi]$ in the line $x = \pi$ (the gold line in the figure), so both halves have the same signed area.
   P6, first case, Section 7.10. Lesson: properties

3. P6 with $2a = \pi$⁠: the two halves cancel

   Now apply P6 to $\int_0^{\pi}\cos^5 x\,dx$⁠, with $2a = \pi$⁠, that is $a = \frac{\pi}{2}$⁠. Test the condition:

   $$\begin{aligned}f(\pi - x) &= \cos^5(\pi - x) = (-\cos x)^5 \\ &= -\cos^5 x = -f(x) .\end{aligned}$$

   So the second case of P6 applies:

   $$\int_0^{\pi}\cos^5 x\,dx = 0 .$$
   $\cos(\pi - x) = -\cos x$⁠, and the fifth power keeps the minus sign because 5 is odd. On $[0, \pi]$ the part of the graph on $[\frac{\pi}{2}, \pi]$ is the part on $[0, \frac{\pi}{2}]$ turned half a turn about $(\frac{\pi}{2}, 0)$⁠: equal area, opposite sign.
   P6, second case, Section 7.10. $\cos(\pi - x) = -\cos x$⁠. Basics: trigonometric identities

4. Conclude

   Putting Steps 2 and 3 together:

   $$I = 2\int_0^{\pi}\cos^5 x\,dx = 2 \times 0 = 0 .$$
   The answer 0 is a signed area: there is real area under the curve ($\frac{8}{15}$ in each of the four quarter-intervals), but the parts above and below the axis are equal.
   Basics: area as a sum of thin strips (signed area)

5. Check: an antiderivative, differentiated back

   Write $\cos^5 x = (1 - \sin^2 x)^2\cos x$ and put $t = \sin x$⁠, $dt = \cos x\,dx$⁠:

   $$\int (1 - t^2)^2\,dt = t - \frac{2t^3}{3} + \frac{t^5}{5} + C .$$

   So

   $$F(x) = \sin x - \frac23\sin^3 x + \frac15\sin^5 x .$$

   Differentiating back (chain rule):

   $$\begin{aligned}F'(x) &= \cos x\,(1 - 2\sin^2 x + \sin^4 x) \\ &= \cos x\,(1 - \sin^2 x)^2 = \cos^5 x .\end{aligned}$$

   By Theorem 2, $I = F(2\pi) - F(0) = 0 - 0 = 0$⁠, because $\sin 2\pi = \sin 0 = 0$⁠. Also $F(\frac{\pi}{2}) - F(0) = 1 - \frac23 + \frac15 = \frac{8}{15}$⁠, the area of each quarter in the figure.

   This route does not use any property, so it is an independent confirmation. It also explains the figure's numbers: $F$ takes the values $0, \frac{8}{15}, 0, -\frac{8}{15}, 0$ at $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$⁠.
   $\cos^2 x = 1 - \sin^2 x$⁠; $\frac{d}{dx}\sin^n x = n\sin^{n-1}x\cos x$ (chain rule). Basics: substitution

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