`int(cos2x)/((e^(-x)+cosx)sqrt(1+sin2x))dx,x in(0,(pi)/(2)) ` is equal to

Options

A. `log_(e)|1+e^(x)sinx|+c`
B. `log_(e)|e^(x)+cosx|+c`
C. `log_(e)|1+e^(x)cosx|-x+c`
D. `log_(e)|1+e^(x)cosx|+c`

Explanation

To solve the integral

\[ I = \int \frac{\cos(2x)}{(e^{-x} + \cos x) \sqrt{1 + \sin(2x)}} \, dx \]

we will follow these steps:

### Step 1: Simplify the integrand
We start by rewriting the expression in the denominator. Notice that:

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

Since \(x\) is in the interval \((0, \frac{\pi}{2})\), both \(\sin x\) and \(\cos x\) are non-negative, so we can drop the absolute value:

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

Thus, the integral becomes:

\[ I = \int \frac{\cos(2x)}{(e^{-x} + \cos x)(\sin x + \cos x)} \, dx \]

### Step 2: Rewrite \(\cos(2x)\)
Using the double angle formula, we have:

\[ \cos(2x) = \cos^2 x - \sin^2 x \]

This allows us to express the integral as:

\[ I = \int \frac{\cos^2 x - \sin^2 x}{(e^{-x} + \cos x)(\sin x + \cos x)} \, dx \]

### Step 3: Substitute \(e^{-x}\)
We can rewrite \(e^{-x}\) as \(\frac{1}{e^x}\) to simplify the expression:

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

Multiplying the numerator and denominator by \(e^x\) gives:

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

### Step 4: Use substitution
Let \(t = 1 + e^x \cos x\). Then, we need to find \(dt\):

\[ dt = e^x \cos x \, dx - e^x \sin x \, dx \]

This implies:

\[ dt = e^x (\cos x - \sin x) \, dx \]

Rearranging gives:

\[ dx = \frac{dt}{e^x (\cos x - \sin x)} \]

### Step 5: Substitute back into the integral
Now we can substitute \(dx\) back into the integral:

\[ I = \int \frac{e^x(\cos^2 x - \sin^2 x)}{t(\sin x + \cos x)} \cdot \frac{dt}{e^x (\cos x - \sin x)} \]

This simplifies to:

\[ I = \int \frac{\cos^2 x - \sin^2 x}{t(\sin x + \cos x)(\cos x - \sin x)} \, dt \]

### Step 6: Solve the integral
Now we can integrate:

\[ I = \ln |t| + C = \ln |1 + e^x \cos x| + C \]

### Final Result
Thus, the final answer for the integral is:

\[ I = \ln(1 + e^x \cos x) + C \]