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

Answer

To evaluate the integral \( I = \int e^{2x} \frac{1 + \sin 2x}{1 + \cos 2x} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We can rewrite the integrand:
\[ \frac{1 + \sin 2x}{1 + \cos 2x} = \frac{1 + 2 \sin x \cos x}{1 + 2 \cos^2 x - 1} = \frac{1 + 2 \sin x \cos x}{2 \cos^2 x} \]
This simplifies to:
\[ \frac{1}{2 \cos^2 x} + \frac{\sin 2x}{2 \cos^2 x} = \frac{1}{2} \sec^2 x + \tan x \]
Thus, we can rewrite the integral as:
\[ I = \int e^{2x} \left( \frac{1}{2} \sec^2 x + \tan x \right) \, dx \]

### Step 2: Separate the integral
Now we can separate the integral:
\[ I = \frac{1}{2} \int e^{2x} \sec^2 x \, dx + \int e^{2x} \tan x \, dx \]

### Step 3: Evaluate the first integral using integration by parts
Let’s evaluate \( \int e^{2x} \sec^2 x \, dx \) using integration by parts. Let:
- \( u = \tan x \) (first function)
- \( dv = e^{2x} \, dx \) (second function)

Then, we have:
- \( du = \sec^2 x \, dx \)
- \( v = \frac{1}{2} e^{2x} \)

Now applying integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Thus,
\[ \int e^{2x} \sec^2 x \, dx = \tan x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \sec^2 x \, dx \]

### Step 4: Solve for the integral
Let \( J = \int e^{2x} \sec^2 x \, dx \). Then we have:
\[ J = \frac{1}{2} e^{2x} \tan x - \frac{1}{2} J \]
Adding \( \frac{1}{2} J \) to both sides:
\[ \frac{3}{2} J = \frac{1}{2} e^{2x} \tan x \]
Thus,
\[ J = \frac{1}{3} e^{2x} \tan x \]

### Step 5: Evaluate the second integral
Now we evaluate \( \int e^{2x} \tan x \, dx \) using integration by parts again:
Let:
- \( u = \tan x \)
- \( dv = e^{2x} \, dx \)

Then,
\[ \int e^{2x} \tan x \, dx = \tan x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \sec^2 x \, dx \]
We already know \( \int e^{2x} \sec^2 x \, dx = J \), so:
\[ \int e^{2x} \tan x \, dx = \frac{1}{2} e^{2x} \tan x - \frac{1}{2} J \]

### Step 6: Combine results
Now substituting back into our expression for \( I \):
\[ I = \frac{1}{2} J + \left( \frac{1}{2} e^{2x} \tan x - \frac{1}{2} J \right) \]
This simplifies to:
\[ I = \frac{1}{2} e^{2x} \tan x \]

### Final Result
Thus, the final result is:
\[ I = \frac{1}{3} e^{2x} \tan x + C \]