Miscellaneous · Q35: $\int_0^{\pi/2}\sin^3x\,dx=\frac23$

Basics you need

Solution

1. The integrand is continuous on $[0, \frac{\pi}{2}]$

   $\sin^3 x$ is a power of the continuous function $\sin x$⁠, so it is continuous everywhere, and Theorem 2 of Section 7.8.3 applies.

   Every definite integral starts with this check; here nothing can fail.
   Lesson: definite integrals and the fundamental theorem

2. Split off one $\sin x$

   $$\sin^3 x = \sin^2 x\cdot\sin x = (1 - \cos^2 x)\sin x .$$
   We want the integrand to look like "a function of $\cos x$" times "the derivative of $\cos x$". The derivative of $\cos x$ is $-\sin x$⁠, so we keep one $\sin x$ aside for the differential and convert the remaining even power $\sin^2 x$ into cosines. This works for any odd power of sine.
   $\sin^2 x + \cos^2 x = 1$⁠. Basics: trigonometric identities

3. Substitute $t = \cos x$ and change the limits

   Put $t = \cos x$⁠, so $dt = -\sin x\,dx$⁠, that is $\sin x\,dx = -dt$⁠. At $x = 0$⁠, $t = 1$⁠; at $x = \frac{\pi}{2}$⁠, $t = 0$⁠. Then

   $$\begin{aligned} I &= \int_1^0 (1 - t^2)(-dt) \\ &= \int_0^1 (1 - t^2)\,dt . \end{aligned}$$
   The second line uses P1: swapping the limits changes the sign, which cancels the minus from $dt$⁠. The substitution is valid on the whole interval: $\cos x$ is differentiable and decreasing on $[0, \frac{\pi}{2}]$ (its derivative $-\sin x \le 0$⁠), so it maps $[0, \frac{\pi}{2}]$ onto $[0, 1]$ once, running backwards.
   Section 7.9; P1, Section 7.10. Lesson: integration by substitution

4. Integrate the polynomial

   $$I = \Big[t - \tfrac{t^3}{3}\Big]_0^1 = \left(1 - \tfrac13\right) - 0 = \tfrac23 .$$

   This is the required result.

   Power rule (standard formula (i), Section 7.2) and Theorem 2. Size check: $\sin^3 x$ rises from 0 to 1 on $[0, \frac{\pi}{2}]$⁠, an interval of length about 1.571, and stays small for a while (since cubing shrinks numbers below 1), so an area of about 0.667, less than half of $1 \times 1.571$⁠, is reasonable.
   Basics: powers of x and the power rule

5. Check: the antiderivative in $x$

   Back in $x$⁠, $\int\sin^3 x\,dx$ $= -\left(t - \frac{t^3}{3}\right) + C$ $= -\cos x + \frac{\cos^3 x}{3} + C$⁠. Differentiate:

   $$\begin{aligned} &\frac{d}{dx}\left(-\cos x + \tfrac{\cos^3 x}{3}\right) \\ &= \sin x + \cos^2 x\,(-\sin x) \\ &=\sin x\,(1 - \cos^2 x) = \sin^3 x \ ✓ \end{aligned}$$

   valid for every real $x$⁠. Evaluating it directly: at $\frac{\pi}{2}$ it is $0 + 0 = 0$⁠; at $0$ it is $-1 + \frac13 = -\frac23$⁠; difference $\frac23$ ✓.

   The indefinite integral in $t$ is $\int(1 - t^2)(-dt)$⁠, which carries the minus sign. The chain rule on $\frac{\cos^3 x}{3}$ gives $\cos^2 x\cdot(-\sin x)$⁠.
   Basics: integration undoes differentiation

6. Cross-check with the triple-angle identity

   From $\sin 3x = 3\sin x - 4\sin^3 x$⁠: $\sin^3 x = \frac34\sin x - \frac14\sin 3x$⁠. So

   $$\begin{aligned} I &= \Big[-\tfrac34\cos x + \tfrac{1}{12}\cos 3x\Big]_0^{\pi/2} \\ &= (0 + 0) - \left(-\tfrac34 + \tfrac{1}{12}\right) = \tfrac{8}{12} = \tfrac23 \ ✓ \end{aligned}$$

   The two antiderivatives are in fact identical: using $\cos 3x = 4\cos^3 x - 3\cos x$⁠,

   $$\begin{aligned} &-\tfrac34\cos x + \tfrac{1}{12}\cos 3x \\ &= -\tfrac34\cos x + \tfrac13\cos^3 x - \tfrac14\cos x \\ &= -\cos x + \tfrac{\cos^3 x}{3} . \end{aligned}$$
   This is the method of Section 7.3.2 (integration using trigonometric identities). $\cos\frac{3\pi}{2} = 0$ and $\cos 0 = 1$⁠. A second, independent method that gives the same number is strong evidence that the proof is right.
   $\int\sin kx\,dx = -\frac{\cos kx}{k} + C$⁠. Basics: trigonometric identities

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