Evaluate`int e^(x) ((1+sinx cos x)/(cos^(2)x))dx`

Answer

To evaluate the integral

\[ \int e^x \frac{1 + \sin x \cos x}{\cos^2 x} \, dx, \]

we can start by rewriting the integrand.

### Step 1: Rewrite the integrand
We can express the integrand as:

\[ \frac{1 + \sin x \cos x}{\cos^2 x} = \frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x} = \sec^2 x + \tan x. \]

Thus, we can rewrite the integral as:

\[ \int e^x \left( \sec^2 x + \tan x \right) \, dx. \]

### Step 2: Split the integral
Now, we can split the integral into two parts:

\[ \int e^x \sec^2 x \, dx + \int e^x \tan x \, dx. \]

### Step 3: Use integration by parts
We will use integration by parts for both integrals. Recall that the integration by parts formula is:

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

#### For the first integral \(\int e^x \sec^2 x \, dx\):
Let \(u = \sec^2 x\) and \(dv = e^x \, dx\). Then, we find:

- \(du = 2 \sec^2 x \tan x \, dx\)
- \(v = e^x\)

Applying integration by parts:

\[ \int e^x \sec^2 x \, dx = e^x \sec^2 x - \int e^x (2 \sec^2 x \tan x) \, dx. \]

#### For the second integral \(\int e^x \tan x \, dx\):
Let \(u = \tan x\) and \(dv = e^x \, dx\). Then, we find:

- \(du = \sec^2 x \, dx\)
- \(v = e^x\)

Applying integration by parts:

\[ \int e^x \tan x \, dx = e^x \tan x - \int e^x \sec^2 x \, dx. \]

### Step 4: Combine results
Now we have:

\[ \int e^x \sec^2 x \, dx + \int e^x \tan x \, dx = e^x \sec^2 x - \int e^x (2 \sec^2 x \tan x) \, dx + e^x \tan x - \int e^x \sec^2 x \, dx. \]

Notice that the two integrals on the left-hand side and right-hand side will cancel each other out, leading to:

\[ \int e^x \sec^2 x \, dx + \int e^x \tan x \, dx = e^x \sec^2 x + e^x \tan x + C. \]

### Final Result
Thus, we can conclude that:

\[ \int e^x \frac{1 + \sin x \cos x}{\cos^2 x} \, dx = e^x + \tan x + C. \]