`int(sinx-cosx)/(sqrt(1-sin2x))e^(sinx)cosx dx` is equal to

Options

A. `e^(sinx)+C`
B. `e^(sinx-cosx)+C`
C. `e^(sinx+cosx)+C`
D. `e^(cosx-sinx)+C`

Answer

To solve the integral

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

we will follow these steps:

### Step 1: Simplify the denominator
We know that

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

Thus, we can rewrite the denominator:

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

Using the identity \(\sin^2 x + \cos^2 x = 1\), we can express this as:

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

### Step 2: Substitute the simplified denominator
Now, substituting this back into the integral, we have:

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

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 (or we can adjust the limits accordingly). Thus, we can simplify:

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

### Step 3: Use substitution
Next, we can use the substitution:

\[ u = e^{\sin x}, \quad \text{then} \quad du = e^{\sin x} \cos x \, dx. \]

This means that:

\[ dx = \frac{du}{e^{\sin x} \cos x} = \frac{du}{u \cos x}. \]

### Step 4: Substitute in the integral
Now, substituting \(u\) into the integral, we have:

\[ I = \int e^{\sin x \cos x} \frac{du}{u \cos x}. \]

However, we notice that we can directly integrate \(e^{\sin x \cos x}\) without further substitutions.

### Step 5: Integrate
The integral simplifies to:

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

This integral can be directly evaluated as:

\[ I = e^{\sin x} + C, \]

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

### Final Result
Thus, the final result is:

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

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