Let g (x) be a differentiable function satisfying `(d)/(dx){g(x)}=g(x) and g (0)=1` , then `intg(x)((2-sin2x)/(1-cos2x))dx` is equal to

Options

A. `g(x)cotx+C`
B. `-g(x)cotx+C`
C. `(g(x))/(1-cos2x)+C`
D. none of these

Explanation

To solve the problem, we need to find the value of the integral:

\[ \int g(x) \frac{2 - \sin(2x)}{1 - \cos(2x)} \, dx \]

Given that \( g'(x) = g(x) \) and \( g(0) = 1 \), we can start by solving for \( g(x) \).

### Step 1: Solve for \( g(x) \)

Since \( g'(x) = g(x) \), we can rewrite this as:

\[ \frac{g'(x)}{g(x)} = 1 \]

Integrating both sides with respect to \( x \):

\[ \int \frac{g'(x)}{g(x)} \, dx = \int 1 \, dx \]

This gives us:

\[ \ln |g(x)| = x + C \]

Exponentiating both sides, we have:

\[ g(x) = e^{x+C} = e^C e^x \]

Let \( e^C = k \), then:

\[ g(x) = k e^x \]

Using the initial condition \( g(0) = 1 \):

\[ g(0) = k e^0 = k = 1 \]

Thus, we find:

\[ g(x) = e^x \]

### Step 2: Substitute \( g(x) \) into the integral

Now we substitute \( g(x) \) into the integral:

\[ \int g(x) \frac{2 - \sin(2x)}{1 - \cos(2x)} \, dx = \int e^x \frac{2 - \sin(2x)}{1 - \cos(2x)} \, dx \]

### Step 3: Simplify the integrand

We know that:

\[ 1 - \cos(2x) = 2\sin^2(x) \]

Thus, we can rewrite the integral as:

\[ \int e^x \frac{2 - \sin(2x)}{2\sin^2(x)} \, dx \]

This simplifies to:

\[ \int e^x \left(\frac{1}{\sin^2(x)} - \frac{\sin(2x)}{2\sin^2(x)}\right) \, dx \]

### Step 4: Further simplification

Since \( \sin(2x) = 2\sin(x)\cos(x) \), we can rewrite:

\[ \frac{\sin(2x)}{2\sin^2(x)} = \frac{2\sin(x)\cos(x)}{2\sin^2(x)} = \frac{\cos(x)}{\sin(x)} = \cot(x) \]

Thus, the integral becomes:

\[ \int e^x \left(\csc^2(x) - \cot(x)\right) \, dx \]

### Step 5: Split the integral

Now we can split the integral:

\[ \int e^x \csc^2(x) \, dx - \int e^x \cot(x) \, dx \]

### Step 6: Use integration by parts

For the integral \( \int e^x \cot(x) \, dx \), we can use integration by parts. Let:

- \( u = \cot(x) \) and \( dv = e^x \, dx \)
- Then \( du = -\csc^2(x) \, dx \) and \( v = e^x \)

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

This gives us:

\[ \int e^x \cot(x) \, dx = e^x \cot(x) - \int e^x (-\csc^2(x)) \, dx \]

### Step 7: Combine results

Putting everything together, we find:

\[ \int e^x \left(\csc^2(x) - \cot(x)\right) \, dx = e^x \cot(x) + C \]

### Final Result

Thus, the final result for the integral is:

\[ -e^x \cot(x) + C \]