`int(sin x - cos x)/(sqrt(1 - sin 2x)) e^("sin x") cos x dx` is equal to

Options

A. `e^(sin x) + C`
B. `e^(sin x - cos x) + c`
C. `e^(sin x + cos x) + c`
D. `e^(cos x - sin x) + c`

Answer

To solve the integral

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

we will follow these steps:

### Step 1: Simplify the Denominator
Recall that \(\sin 2x = 2 \sin x \cos x\). Thus, we can rewrite the denominator:

\[ 1 - \sin 2x = 1 - 2 \sin x \cos x. \]

Now, we can express \(1\) as \(\sin^2 x + \cos^2 x\):

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

### Step 2: Rewrite the Integral
Now substituting this back into the integral gives us:

\[ \int \frac{\sin x - \cos x}{\sqrt{(\sin x - \cos x)^2}} e^{\sin x} \cos x \, dx. \]

Since \(\sqrt{(\sin x - \cos x)^2} = |\sin x - \cos x|\), we can assume \(\sin x - \cos x\) is positive in the interval we are considering, thus simplifying to:

\[ \int \frac{\sin x - \cos x}{\sin x - \cos x} e^{\sin x} \cos x \, dx = \int e^{\sin x} \cos x \, dx. \]

### Step 3: Substitution
Now, we will use the substitution \(t = \sin x\). Then, \(dt = \cos x \, dx\). The integral now becomes:

\[ \int e^t \, dt. \]

### Step 4: Integrate
The integral of \(e^t\) is simply:

\[ e^t + C. \]

### Step 5: Back Substitute
Now, we substitute back \(t = \sin x\):

\[ e^{\sin x} + C. \]

### Final Answer
Thus, the final answer is:

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

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