`int(cos2x)/((sinx+cosx)^(2))dx` is equal to

Options

A. `(-1)/(sinx+cos)+C`
B. `log(sinx+cosx)+C`
C. `log(sinx-cosx)+C`
D. `log(sinx+cosx)^(2)+C`

Answer

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

### Step 1: Rewrite the integrand
Using the identity for \( \cos(2x) \):
\[ \cos(2x) = \cos^2(x) - \sin^2(x) \]
we can express the integral as:
\[ \int \frac{\cos^2(x) - \sin^2(x)}{(\sin x + \cos x)^2} \, dx \]

### Step 2: Factor the numerator
The expression \( \cos^2(x) - \sin^2(x) \) can be factored as:
\[ \cos^2(x) - \sin^2(x) = (\cos x + \sin x)(\cos x - \sin x) \]
Thus, we can rewrite the integral:
\[ \int \frac{(\cos x + \sin x)(\cos x - \sin x)}{(\sin x + \cos x)^2} \, dx \]

### Step 3: Simplify the integrand
Notice that \( \frac{(\cos x + \sin x)}{(\sin x + \cos x)^2} \) simplifies to:
\[ \frac{\cos x - \sin x}{\sin x + \cos x} \]
This gives us:
\[ \int \frac{\cos x - \sin x}{\sin x + \cos x} \, dx \]

### Step 4: Use substitution
Let:
\[ t = \sin x + \cos x \]
Then, differentiate \( t \) with respect to \( x \):
\[ dt = (\cos x - \sin x) \, dx \]
This allows us to rewrite the integral as:
\[ \int \frac{dt}{t} \]

### Step 5: Integrate
The integral \( \int \frac{dt}{t} \) is:
\[ \ln |t| + C \]
Substituting back for \( t \):
\[ \ln |\sin x + \cos x| + C \]

### Final Answer
Thus, the solution to the integral is:
\[ \int \frac{\cos(2x)}{(\sin x + \cos x)^2} \, dx = \ln |\sin x + \cos x| + C \]

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