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 find the range of \( y = f(x) \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int e^x \frac{1 + \sin x}{1 + \cos x} \, dx \]

### Step 2: Simplify the Fraction
We can rewrite \( \frac{1 + \sin x}{1 + \cos x} \) using trigonometric identities. We know:
\[ \sin x = 2 \tan \frac{x}{2} \cdot \frac{1}{1 + \tan^2 \frac{x}{2}} \quad \text{and} \quad \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \]
Thus, we can express the integral in terms of \( \tan \frac{x}{2} \).

### Step 3: Substitute and Simplify
Substituting these identities into the integral, we have:
\[ I = \int e^x \frac{1 + 2 \tan \frac{x}{2}}{1 + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} \, dx \]
After simplification, we will find that:
\[ I = \int e^x \left( \tan \frac{x}{2} + \text{constant} \right) \, dx \]

### Step 4: Integrate
Using integration by parts or recognizing the standard integral form, we can derive:
\[ I = e^x f(x) + C \]
where \( f(x) = \tan \frac{x}{2} \).

### Step 5: Determine the Range of \( f(x) \)
The function \( f(x) = \tan \frac{x}{2} \) can take any real value. As \( x \) approaches \( 90^\circ \) (or \( \frac{\pi}{2} \) radians), \( f(x) \) approaches \( +\infty \), and as \( x \) approaches \( -90^\circ \) (or \( -\frac{\pi}{2} \) radians), \( f(x) \) approaches \( -\infty \).

### Conclusion
Thus, the range of \( y = f(x) \) is:
\[ (-\infty, +\infty) \]