The integral `I=inte^(x)((1+sinx)/(1+cosx))dx=e^(x)f(x)+C`
(where, C is the constant of integration).
Then, the range of `y=f(x)` (for all x in the domain of `f(x)`) is

Options

A. `[-1, 1]`
B. `(-oo, oo)`
C. `(-1, 1)`
D. `[0, oo)`

Answer

To solve the integral \( I = \int e^x \frac{1 + \sin x}{1 + \cos x} \, dx \) and express it in the form \( I = e^x f(x) + C \), we will follow these steps:

### Step 1: Simplify the integrand
We start with the integrand \( \frac{1 + \sin x}{1 + \cos x} \). We can rewrite \( \cos x \) in terms of \( \tan \frac{x}{2} \):

\[ \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \]

This gives us:

\[ 1 + \cos x = 1 + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \frac{(1 + \tan^2 \frac{x}{2}) + (1 - \tan^2 \frac{x}{2})}{1 + \tan^2 \frac{x}{2}} = \frac{2}{1 + \tan^2 \frac{x}{2}} \]

Now, we can express \( 1 + \sin x \):

\[ 1 + \sin x = 1 + \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \frac{(1 + \tan^2 \frac{x}{2}) + 2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \frac{1 + 2 \tan \frac{x}{2} + \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \frac{(1 + \tan \frac{x}{2})^2}{1 + \tan^2 \frac{x}{2}} \]

Thus, we have:

\[ \frac{1 + \sin x}{1 + \cos x} = \frac{(1 + \tan \frac{x}{2})^2}{2} \]

### Step 2: Substitute into the integral
Now we can substitute this back into the integral:

\[ I = \int e^x \frac{(1 + \tan \frac{x}{2})^2}{2} \, dx \]

### Step 3: Use integration by parts
We can use integration by parts where we let:

\[ u = \frac{(1 + \tan \frac{x}{2})^2}{2}, \quad dv = e^x \, dx \]

Then we find \( du \) and \( v \):

\[ du = (1 + \tan \frac{x}{2}) \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} \, dx, \quad v = e^x \]

### Step 4: Apply integration by parts
Using integration by parts:

\[ I = e^x \cdot \frac{(1 + \tan \frac{x}{2})^2}{2} - \int e^x \cdot \left( (1 + \tan \frac{x}{2}) \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} \right) \, dx \]

### Step 5: Solve for \( f(x) \)
From the integration, we find that:

\[ I = e^x f(x) + C \]

where \( f(x) = \frac{(1 + \tan \frac{x}{2})^2}{2} \).

### Step 6: Determine the range of \( f(x) \)
To find the range of \( f(x) \):

1. As \( x \) varies over all real numbers, \( \tan \frac{x}{2} \) will take all real values from \( -\infty \) to \( +\infty \).
2. The expression \( (1 + \tan \frac{x}{2})^2 \) will be non-negative and can take any value from \( 0 \) to \( +\infty \).
3. Therefore, \( f(x) = \frac{(1 + \tan \frac{x}{2})^2}{2} \) will also take values from \( 0 \) to \( +\infty \).

### Final Answer
The range of \( y = f(x) \) is \( [0, +\infty) \).

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