`int_(-pi//2)^(pi//2)(sin^(2n-1)x)/(1+cos^(2n)x)dx=`

Options

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

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^{2n-1}(x)}{1 + \cos^{2n}(x)} \, dx, \]

we will first determine whether the integrand is an odd or even function.

### Step 1: Determine if the function is odd or even

Let \( f(x) = \frac{\sin^{2n-1}(x)}{1 + \cos^{2n}(x)} \).

To check if \( f(x) \) is odd, we need to evaluate \( f(-x) \):

\[ f(-x) = \frac{\sin^{2n-1}(-x)}{1 + \cos^{2n}(-x)}. \]

Using the properties of sine and cosine:

- \( \sin(-x) = -\sin(x) \) (sine is an odd function)
- \( \cos(-x) = \cos(x) \) (cosine is an even function)

Substituting these properties into \( f(-x) \):

\[ f(-x) = \frac{(-\sin(x))^{2n-1}}{1 + \cos^{2n}(x)} = \frac{-(\sin^{2n-1}(x))}{1 + \cos^{2n}(x)} = -f(x). \]

Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 2: Evaluate the integral

For odd functions, the integral over symmetric limits around zero is zero:

\[ \int_{-a}^{a} f(x) \, dx = 0. \]

In our case, the limits are from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), which are symmetric around zero. Therefore, we can conclude:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \, dx = 0. \]

### Final Result

Thus, the value of the integral is:

\[ \boxed{0}. \]

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