Q.By using the properties of definite integrals, evaluate the integral $\int_{-\pi/2}^{\pi/2}\sin^2 x\,dx$

Concept: Definite integral of an even function; antiderivative of cos x is sin x.

Integral from -pi/2 to pi/2 of cos x dx = [ sin x ] from -pi/2 to pi/2

= sin(pi/2) - sin(-pi/2)

= 1 - (-1)

= 2 …

The key idea here is symmetry and the absolute value. The function $|x \cos \pi x|$ is even because $|(-x) \cos(-\pi x)| = | -x \cos \pi x| = |x \cos \pi x|$. So we can integrate from $0$ to $2$ and double the result. But the absolute value also means we have to handle where $\cos \pi x$ changes sign — those are the points where the integrand would otherwise be negative, so the absolute value flips the sign there.

The zeros of $\cos \pi x$ occur when $\pi x = \frac{\pi}{2} + n\pi$, i.e., $x = \frac{1}{2} + n$. Over $[0,2]$, these are at $x = \frac{1}{2}$ and $x = \frac{3}{2}$. Between these zeros, $\cos \pi x$ alternates sign, so the absolute value makes each piece positive. We'll integrate piecewise.

1. Use evenness to reduce the interval. Since the integrand is even:

2. Identify the sign changes of $\cos \pi x$ on $[0,2]$.

   $\cos \pi x = 0$ at $x = \frac{1}{2}$ and $x = \frac{3}{2}$.

   • On $[0, \frac{1}{2}]$: $\cos \pi x \ge 0$, so $|x \cos \pi x| = x \cos \pi x$.
   • On $[\frac{1}{2}, \frac{3}{2}]$: $\cos \pi x \le 0$, so $|x \cos \pi x| = -x \cos \pi x$.
   • On $[\frac{3}{2}, 2]$: $\cos \pi x \ge 0$, so $|x \cos \pi x| = x \cos \pi x$.

   Thus:

3. Compute the antiderivative. Use integration by parts: let $u = x$, $dv = \cos \pi x \, dx$, so $du = dx$, $v = \frac{\sin \pi x}{\pi}$. Then:

4. Evaluate each piece.
   • First piece:

 At $x = 1/2$: $\sin(\pi/2) = 1$, $\cos(\pi/2) = 0$, so value = $\frac{(1/2)(1)}{\pi} + 0 = \frac{1}{2\pi}$.  
 At $x = 0$: $\sin 0 = 0$, $\cos 0 = 1$, so value = $0 + \frac{1}{\pi^2}$.  
 Thus first piece = $\frac{1}{2\pi} - \frac{1}{\pi^2}$.

• Second piece (note the minus sign outside):

 At $x = 3/2$: $\sin(3\pi/2) = -1$, $\cos(3\pi/2) = 0$, so value = $\frac{(3/2)(-1)}{\pi} + 0 = -\frac{3}{2\pi}$.  
 At $x = 1/2$: value = $\frac{1}{2\pi}$ (as above).  
 So the bracket $[F]_{1/2}^{3/2} = \left(-\frac{3}{2\pi}\right) - \left(\frac{1}{2\pi}\right) = -\frac{4}{2\pi} = -\frac{2}{\pi}$.  
 Then the second piece = $-(-\frac{2}{\pi}) = \frac{2}{\pi}$.

• Third piece:

 At $x = 2$: $\sin 2\pi = 0$, $\cos 2\pi = 1$, so value = $0 + \frac{1}{\pi^2}$. …

Concept and intuition

The expression under the square root, $\cos x - \cos^3 x$, factors as $\cos x(1 - \cos^2 x) = \cos x \sin^2 x$. Taking the square root gives $\sqrt{\cos x}\,|\sin x|$ — the absolute value is crucial because $\sqrt{\sin^2 x} = |\sin x|$, not $\sin x$. Over $[-\pi/2, \pi/2]$, $\sin x$ changes sign, but the absolute value makes the integrand even and non‑negative. That symmetry lets us double the integral from $0$ to $\pi/2$, where $\sin x \ge 0$, so $|\sin x| = \sin x$. The resulting integral is a standard power‑substitution.

Step‑by‑step solution

1. Simplify the integrand

Hence

2. Exploit symmetry The function $f(x) = \sqrt{\cos x}\,|\sin x|$ is even because $\cos$ is even and $|\sin|$ is even. The interval $[-\pi/2, \pi/2]$ is symmetric about $0$, so

3. Remove the absolute value on $[0, \pi/2]$ For $x \in [0, \pi/2]$, $\sin x \ge 0$, so $|\sin x| = \sin x$. Thus

4. Substitute $u = \cos x$ Then $du = -\sin x\,dx$, so $\sin x\,dx = -du$. When $x=0$, $u=1$; when $x=\pi/2$, $u=0$. The integral becomes $\int_{0}^{\pi/2} \sqrt{\cos x}\,\sin x\,dx = \int_{1}^{0} \sqrt{u}\,(-du) = \int_{0}^{1} u^{1/2}\,du.$ …