Find: `int e^x sqrt(1+sin2x)/(1+ cos 2x) dx.`

Answer

To solve the integral

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

we will simplify the integrand step by step.

### Step 1: Simplify the expression under the square root

We know that

\[ \sin 2x = 2 \sin x \cos x \quad \text{and} \quad \cos 2x = 2 \cos^2 x - 1. \]

Thus, we can rewrite \(1 + \cos 2x\) as:

\[ 1 + \cos 2x = 1 + (2 \cos^2 x - 1) = 2 \cos^2 x. \]

Now, let's simplify \(1 + \sin 2x\):

\[ 1 + \sin 2x = 1 + 2 \sin x \cos x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2. \]

### Step 2: Substitute back into the integral

Now substituting these back into the integral, we have:

\[ \sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2} = \sin x + \cos x. \]

Thus, the integral becomes:

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

### Step 3: Rewrite the integrand

This can be rewritten as:

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

### Step 4: Use integration by parts

We can use the formula for the integral of the product of an exponential function and a function of \(x\):

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

Let \(f(x) = \sec x\). Then \(f'(x) = \sec x \tan x\).

Applying this formula, we have:

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

### Step 5: Combine the results

Thus, the integral becomes:

\[ \frac{1}{2} \left( e^x \sec x + C \right), \]

where \(C\) is the constant of integration.

### Final Answer

Therefore, the final result is:

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