`int(cosx-sinx)/(1+2sin x cosx)dx` is equal to

Options

A. `-(1)/(cos x- sinx)+C`
B. `(cos x+sin x)/(cos x-sinx)+C`
C. `-(1)/(sin x+cosx)+C`
D. `(x)/(sinx+cosx)+C`

Explanation

To solve the integral \( \int \frac{\cos x - \sin x}{1 + 2\sin x \cos x} \, dx \), we can follow these steps:

### Step 1: Simplify the Denominator
The denominator can be rewritten using the identity for the square of a sum:
\[ 1 + 2\sin x \cos x = \sin^2 x + \cos^2 x + 2\sin x \cos x = (\sin x + \cos x)^2 \]
Thus, we can rewrite the integral as:
\[ \int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} \, dx \]

### Step 2: Substitute \( u = \sin x + \cos x \)
Let \( u = \sin x + \cos x \). Then, we need to find \( du \):
\[ du = \cos x \, dx - \sin x \, dx = (\cos x - \sin x) \, dx \]
This means \( dx = \frac{du}{\cos x - \sin x} \).

### Step 3: Substitute in the Integral
Now substituting \( u \) and \( dx \) into the integral:
\[ \int \frac{\cos x - \sin x}{u^2} \cdot \frac{du}{\cos x - \sin x} = \int \frac{1}{u^2} \, du \]

### Step 4: Integrate
The integral \( \int \frac{1}{u^2} \, du \) can be solved as:
\[ \int u^{-2} \, du = -\frac{1}{u} + C \]

### Step 5: Substitute Back
Now, substituting back \( u = \sin x + \cos x \):
\[ -\frac{1}{\sin x + \cos x} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{\cos x - \sin x}{1 + 2\sin x \cos x} \, dx = -\frac{1}{\sin x + \cos x} + C \]