The integral `I=intsin (2theta)[(1+cos^(2)theta)/(2sin^(2)theta)]d theta` simplifies to (where, c is the integration constant)

Options

A. `ln|sin theta|+cos theta+c`
B. `2ln|sin theta|-(sin^(2)theta)/(2)+c`
C. `ln|sin theta|-sin^(2)theta+c`
D. `ln|cos theta|+cos^(2)theta+c`

Answer

To solve the integral

\[ I = \int \sin(2\theta) \frac{1 + \cos^2 \theta}{2 \sin^2 \theta} d\theta, \]

we can follow these steps:

### Step 1: Rewrite the integral
We can express \(\sin(2\theta)\) in terms of \(\sin\theta\) and \(\cos\theta\):
\[ \sin(2\theta) = 2 \sin\theta \cos\theta. \]
Thus, we can rewrite the integral as:
\[ I = \int 2 \sin\theta \cos\theta \frac{1 + \cos^2 \theta}{2 \sin^2 \theta} d\theta = \int \frac{(1 + \cos^2 \theta) \cos\theta}{\sin\theta} d\theta. \]

### Step 2: Simplify the integral
Now, we can split the integral:
\[ I = \int \frac{\cos\theta}{\sin\theta} d\theta + \int \frac{\cos^3\theta}{\sin\theta} d\theta. \]
The first integral can be simplified as:
\[ \int \frac{\cos\theta}{\sin\theta} d\theta = \ln|\sin\theta| + C_1. \]

### Step 3: Solve the second integral
For the second integral, we can use the identity \(\cos^2\theta = 1 - \sin^2\theta\):
\[ \int \frac{\cos^3\theta}{\sin\theta} d\theta = \int \frac{\cos^2\theta \cos\theta}{\sin\theta} d\theta = \int \frac{(1 - \sin^2\theta) \cos\theta}{\sin\theta} d\theta. \]
Let \(u = \sin\theta\), then \(du = \cos\theta d\theta\):
\[ \int \frac{(1 - u^2)}{u} du = \int \left(\frac{1}{u} - u\right) du = \ln|u| - \frac{u^2}{2} + C_2 = \ln|\sin\theta| - \frac{\sin^2\theta}{2} + C_2. \]

### Step 4: Combine the results
Now, combining both parts:
\[ I = \ln|\sin\theta| + \left(\ln|\sin\theta| - \frac{\sin^2\theta}{2}\right) + C, \]
which simplifies to:
\[ I = 2\ln|\sin\theta| - \frac{\sin^2\theta}{2} + C. \]

### Final Result
Thus, the final result for the integral is:
\[ I = 2 \ln|\sin\theta| - \frac{\sin^2\theta}{2} + C. \]