Exercise 7.10 · Q11: $\int_{-\pi/2}^{\pi/2}\sin^2x\,dx$

Basics you need

Solution

1. The integrand is continuous

   $\sin x$ is continuous for every real $x$⁠, so $\sin^2 x$ is continuous on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. Label

   $$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2 x\,dx .$$
   P7 and P4 are applied to functions continuous on the closed interval, so that every integral below is a finite number.
   A product of continuous functions ($\sin x \cdot \sin x$⁠) is continuous.

2. $\sin^2 x$ is an even function

   Let $f(x) = \sin^2 x$⁠. For every real $x$⁠:

   $$\begin{aligned}f(-x) &= \big(\sin(-x)\big)^2 = (-\sin x)^2 \\ &= \sin^2 x = f(x) .\end{aligned}$$
   This is the condition of P7(i), checked for an arbitrary $x$⁠, not just at a few points. $\sin$ itself is odd, and squaring removes the minus sign. (The interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ has the form $[-a, a]$ with $a = \frac{\pi}{2}$⁠, which is the other requirement of P7.)
   $\sin(-x) = -\sin x$ (sine is odd); $(-u)^2 = u^2$⁠. Basics: symmetry and mirror images

3. Apply P7(i) with $a = \frac{\pi}{2}$

   $$I = 2\int_0^{\frac{\pi}{2}}\sin^2 x\,dx .$$
   For an even function the part over $\left[-\frac{\pi}{2}, 0\right]$ is the mirror image of the part over $\left[0, \frac{\pi}{2}\right]$⁠, so the two halves are equal and the whole is twice one half. (The book's proof: split at 0 with P2, substitute $t = -x$ in the left half.)
   P7(i), Section 7.10. Lesson: properties of definite integrals

4. Evaluate the half with P4

   Let $J = \int_0^{\pi/2}\sin^2 x\,dx$⁠. By P4 with $a = \frac{\pi}{2}$⁠, since $\sin^2\left(\frac{\pi}{2} - x\right) = \cos^2 x$⁠,

   $$J = \int_0^{\frac{\pi}{2}}\cos^2 x\,dx .$$

   Adding the two forms of $J$⁠:

   $$\begin{aligned}2J &= \int_0^{\frac{\pi}{2}}\left(\sin^2 x + \cos^2 x\right)dx = \frac{\pi}{2}, \\ J &= \frac{\pi}{4}.\end{aligned}$$
   This is exactly the argument of Question 1: on $\left[0, \frac{\pi}{2}\right]$ the integrals of $\sin^2 x$ and $\cos^2 x$ are equal, and together they integrate the constant 1 over a length $\frac{\pi}{2}$⁠.
   P4, Section 7.10; $\sin^2 x + \cos^2 x = 1$⁠. Exercise 7.10 Q1

5. Double

   $$I = 2J = 2\cdot\frac{\pi}{4} = \frac{\pi}{2} \approx 1.5707963 .$$
   Sense check: $\sin^2 x$ lies between 0 and 1 and has average value $\frac12$ over this interval (it equals $\frac12$ on average because $\sin^2$ and $\cos^2$ share the total 1), and the interval length is $\pi$⁠; $\frac12 \cdot \pi = \frac{\pi}{2}$⁠.
   Mirror images have equal areas. Basics: symmetry and mirror images

6. A second route: the double-angle formula

   $\sin^2 x = \frac{1 - \cos 2x}{2}$⁠, with antiderivative $\frac{x}{2} - \frac{\sin 2x}{4}$ (its derivative is $\frac12 - \frac{\cos 2x}{2}$⁠). So

   $$\begin{aligned}I &= \left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\ &= \left(\frac{\pi}{4} - 0\right) - \left(-\frac{\pi}{4} - 0\right) = \frac{\pi}{2},\end{aligned}$$

   using $\sin\pi = 0$ and $\sin(-\pi) = 0$⁠.

   An independent method gives the same value, which confirms Steps 2 to 5.
   $\cos 2x = 1 - 2\sin^2 x$ (Section 7.3.2 uses such identities); Theorem 2, Section 7.8.3. Basics: trigonometric identities

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