Exercise 7.10 · Q1: $\int_0^{\pi/2}\cos^2x\,dx$

Basics you need

Solution

1. The integrand is continuous; label the integral (1)

   $\cos x$ is continuous for every real $x$⁠, so its square $\cos^2 x$ is continuous on $\left[0, \frac{\pi}{2}\right]$⁠. Write

   $$I = \int_0^{\frac{\pi}{2}}\cos^2 x\,dx \qquad (1)$$
   The properties P0 to P7 and Theorem 2 are stated for functions continuous on the closed interval. A continuous integrand means the integral is a definite finite number, so we may later add it to itself and divide by 2.
   A product (here $\cos x \cdot \cos x$⁠) of continuous functions is continuous.

2. Apply P4 with $a = \frac{\pi}{2}$

   Here $f(x) = \cos^2 x$ and $a = \frac{\pi}{2}$⁠. Work out $f(a - x)$⁠:

   $$f\left(\tfrac{\pi}{2} - x\right) = \cos^2\left(\tfrac{\pi}{2} - x\right) = \sin^2 x .$$

   So P4 gives

   $$I = \int_0^{\frac{\pi}{2}}\sin^2 x\,dx \qquad (2)$$
   P4 has no special condition: it holds for every function continuous on $[0, a]$⁠. The only work is to compute $f(a - x)$ and see that it is simpler or more useful. Here it is the "partner" $\sin^2 x$⁠, whose sum with $\cos^2 x$ is the constant 1. (Proof of P4 in the book: put $t = a - x$⁠, so $dt = -dx$⁠, and the limits 0 and $a$ swap.)
   $\cos\left(\frac{\pi}{2} - x\right) = \sin x$⁠; squaring both sides gives $\cos^2\left(\frac{\pi}{2} - x\right) = \sin^2 x$⁠. Basics: trigonometric identities

3. Add (1) and (2)

   $$\begin{aligned}2I &= \int_0^{\frac{\pi}{2}}\left(\cos^2 x + \sin^2 x\right)dx \\ &= \int_0^{\frac{\pi}{2}}1\,dx = \big[x\big]_0^{\frac{\pi}{2}} = \frac{\pi}{2}.\end{aligned}$$
   Both forms are integrals over the same interval, so they can be added under one integral sign. The identity $\sin^2 x + \cos^2 x = 1$ replaces a function we would otherwise have to integrate by a constant, whose integral over an interval is the interval's length.
   $\int 1\,dx = x + C$ (the line $\int dx = x + C$ under standard formula (i), Section 7.2), then Theorem 2, Section 7.8.3. Lesson: the fundamental theorem

4. Solve for $I$

   $$I = \frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4} \approx 0.7853982 .$$
   $I$ is a finite number (Step 1), so dividing $2I = \frac{\pi}{2}$ by 2 is allowed. The answer is positive, as it must be: $\cos^2 x \ge 0$⁠, so the region lies above the axis.
   Dividing both sides of an equation by the same non-zero number keeps it true.

5. The picture: two mirror images fill a rectangle

   This is P4 seen as a picture: replacing $x$ by $\frac{\pi}{2} - x$ reflects a graph in the vertical line through the middle of the interval, and a reflected region has the same area.
   Mirror images have equal areas. Basics: symmetry and mirror images

6. A second route: the double-angle formula

   Without P4: $\cos^2 x = \frac{1 + \cos 2x}{2}$⁠, and an antiderivative is $\frac{x}{2} + \frac{\sin 2x}{4}$ (check: its derivative is $\frac12 + \frac{\cos 2x}{2}$⁠). So

   $$\begin{aligned}I &= \left[\frac{x}{2} + \frac{\sin 2x}{4}\right]_0^{\frac{\pi}{2}} \\ &= \left(\frac{\pi}{4} + \frac{\sin\pi}{4}\right) - 0 = \frac{\pi}{4}.\end{aligned}$$
   Both methods must agree, which is a strong check on the answer. The exercise asks for the properties, so the P4 route (Steps 2 to 4) is the expected one; it also works for integrands such as those in Questions 2 to 4, which have no simple antiderivative.
   $\cos 2x = 2\cos^2 x - 1$⁠, so $\cos^2 x = \frac{1 + \cos 2x}{2}$ (Section 7.3.2 uses this identity); $\sin\pi = 0$⁠. Basics: trigonometric identities

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