If `int (e ^(x) (1+ sin x )dx)/( 1+ cos x ) = e ^(x) f (x) + C,` then f (x) is equal to

Options

A. `sin ""x/2`
B. `cos""x/2`
C. `tan ""x/2`
D. `log "" x/2`

Answer

To solve the integral

\[ \int \frac{e^x (1 + \sin x)}{1 + \cos x} \, dx, \]

we will follow a systematic approach.

### Step 1: Simplify the integrand

We can rewrite the integrand by using the identity for \(1 + \cos x\) and \(1 + \sin x\):

\[ 1 + \sin x = 1 + 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) + 1. \]

The term \(1 + \cos x\) can be expressed as:

\[ 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right). \]

Thus, we rewrite the integral as:

\[ \int \frac{e^x (1 + \sin x)}{1 + \cos x} \, dx = \int \frac{e^x (1 + \sin x)}{2 \cos^2\left(\frac{x}{2}\right)} \, dx. \]

### Step 2: Split the integral

Now we can split the integral:

\[ \int \frac{e^x (1 + \sin x)}{2 \cos^2\left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int e^x \sec^2\left(\frac{x}{2}\right) (1 + \sin x) \, dx. \]

### Step 3: Use substitution

Let’s use the substitution \(u = \frac{x}{2}\), which gives \(dx = 2 \, du\). The integral becomes:

\[ \frac{1}{2} \int e^{2u} \sec^2(u) (1 + 2 \sin u) \cdot 2 \, du = \int e^{2u} \sec^2(u) (1 + 2 \sin u) \, du. \]

### Step 4: Apply integration by parts

We can apply integration by parts here. We know that:

\[ \int e^{ax} f'(x) \, dx = e^{ax} f(x) - \int e^{ax} f(x) \, dx. \]

In our case, we can let \(f(u) = \tan(u)\) since \(\sec^2(u)\) is the derivative of \(\tan(u)\).

### Step 5: Solve the integral

Thus, we have:

\[ \int e^{2u} \sec^2(u) \, du = e^{2u} \tan(u) - \int e^{2u} \tan(u) \, du. \]

After simplifying, we find that:

\[ \int e^{2u} \tan(u) \, du = e^{2u} \tan(u) - \int e^{2u} \sec^2(u) \, du. \]

### Step 6: Back substitute

Now, we can back substitute \(u = \frac{x}{2}\) to express everything in terms of \(x\):

\[ f(x) = \tan\left(\frac{x}{2}\right). \]

### Conclusion

Thus, we conclude that:

\[ f(x) = \tan\left(\frac{x}{2}\right). \]