`int(cos2x)/(cosx-sinx)dx=`

Options

A. `sinx+cosx+c`
B. `cosx-sinx+c`
C. `sinx-cosx+c`
D. `sinx.cosx+c`

Answer

To solve the integral \( \int \frac{\cos 2x}{\cos x - \sin x} \, dx \), we can follow these steps:

### Step 1: Rewrite \( \cos 2x \)
We use the double angle identity for cosine:
\[ \cos 2x = \cos^2 x - \sin^2 x \]
Thus, we can rewrite the integral as:
\[ \int \frac{\cos^2 x - \sin^2 x}{\cos x - \sin x} \, dx \]

### Step 2: Factor the numerator
Notice that \( \cos^2 x - \sin^2 x \) can be factored using the difference of squares:
\[ \cos^2 x - \sin^2 x = (\cos x + \sin x)(\cos x - \sin x) \]
Now substituting this back into the integral gives:
\[ \int \frac{(\cos x + \sin x)(\cos x - \sin x)}{\cos x - \sin x} \, dx \]

### Step 3: Simplify the integral
The \( \cos x - \sin x \) terms in the numerator and denominator cancel out (assuming \( \cos x - \sin x \neq 0 \)):
\[ \int (\cos x + \sin x) \, dx \]

### Step 4: Integrate
Now we can integrate \( \cos x + \sin x \):
\[ \int (\cos x + \sin x) \, dx = \int \cos x \, dx + \int \sin x \, dx \]
The integrals are:
\[ \int \cos x \, dx = \sin x \]
\[ \int \sin x \, dx = -\cos x \]
So, combining these results, we get:
\[ \sin x - \cos x + C \]
where \( C \) is the constant of integration.

### Final Answer:
Thus, the final result of the integral is:
\[ \int \frac{\cos 2x}{\cos x - \sin x} \, dx = \sin x - \cos x + C \]