The value of the integral `int_(-pi//2)^(pi//2) sqrt(cosx-cos^(2)x)dx` is

Options

A. 0
B. `2//3`
C. `4//3`
D. none of these

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\cos x - \cos^2 x} \, dx, \]

we can simplify the expression under the square root and then evaluate the integral step by step.

### Step 1: Simplify the integrand

We start with the expression under the square root:

\[ \sqrt{\cos x - \cos^2 x} = \sqrt{\cos x (1 - \cos x)}. \]

### Step 2: Use symmetry of the integral

Since the function \(\sqrt{\cos x (1 - \cos x)}\) is even (i.e., \(f(-x) = f(x)\)), we can simplify the integral:

\[ I = 2 \int_0^{\frac{\pi}{2}} \sqrt{\cos x (1 - \cos x)} \, dx. \]

### Step 3: Substitute \(u = \cos x\)

Let \(u = \cos x\), then \(du = -\sin x \, dx\) or \(dx = -\frac{du}{\sqrt{1 - u^2}}\). The limits change as follows:
- When \(x = 0\), \(u = 1\).
- When \(x = \frac{\pi}{2}\), \(u = 0\).

Thus, the integral becomes:

\[ I = 2 \int_1^0 \sqrt{u (1 - u)} \left(-\frac{du}{\sqrt{1 - u^2}}\right) = 2 \int_0^1 \sqrt{u (1 - u)} \frac{du}{\sqrt{1 - u^2}}. \]

### Step 4: Change the order of the integral

Now we have:

\[ I = 2 \int_0^1 \sqrt{u (1 - u)} \frac{du}{\sqrt{1 - u^2}}. \]

### Step 5: Use the identity for \(\sqrt{u(1-u)}\)

Using the identity \(\sqrt{u(1-u)} = \frac{1}{2} \sin(2\theta)\) where \(u = \sin^2 \theta\), we can rewrite the integral in terms of \(\theta\):

\[ I = 2 \cdot \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(2\theta) \, d\theta. \]

### Step 6: Evaluate the integral

The integral of \(\sin(2\theta)\) is:

\[ \int \sin(2\theta) \, d\theta = -\frac{1}{2} \cos(2\theta). \]

Evaluating from \(0\) to \(\frac{\pi}{2}\):

\[ \left[-\frac{1}{2} \cos(2\theta)\right]_0^{\frac{\pi}{2}} = -\frac{1}{2} \left(\cos(\pi) - \cos(0)\right) = -\frac{1}{2} (-1 - 1) = 1. \]

### Step 7: Final result

Thus, we find that:

\[ I = 2 \cdot \frac{1}{2} \cdot 1 = 1. \]

Therefore, the value of the integral is:

\[ \boxed{1}. \]