`int_(-pi//2)^(pi//2) sqrt((1-cos2x)/2)dx = `

Options

A. 0
B. 2
C. `1/2`
D. `(-1)/2`

Answer

To solve the definite integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\frac{1 - \cos 2x}{2}} \, dx, \]

we will follow these steps:

### Step 1: Simplify the integrand
Using the trigonometric identity \( \cos 2x = 1 - 2\sin^2 x \), we can rewrite \( 1 - \cos 2x \) as follows:

\[ 1 - \cos 2x = 2\sin^2 x. \]

Thus, we have:

\[ \sqrt{\frac{1 - \cos 2x}{2}} = \sqrt{\frac{2\sin^2 x}{2}} = \sqrt{\sin^2 x} = |\sin x|. \]

### Step 2: Determine the behavior of \(|\sin x|\) over the interval
On the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\), \(\sin x\) is non-negative, so \(|\sin x| = \sin x\). Therefore, we can rewrite the integral:

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

### Step 3: Evaluate the integral
Now we can evaluate the integral:

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

Since \(\sin x\) is an odd function, we can use the property of definite integrals for odd functions:

\[ \int_{-a}^{a} f(x) \, dx = 0 \quad \text{if } f(x) \text{ is odd.} \]

In this case, since \(\sin(-x) = -\sin x\), we conclude that:

\[ I = 0. \]

### Final Answer
Thus, the value of the definite integral is:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\frac{1 - \cos 2x}{2}} \, dx = 0. \]