`int(cos2theta)/((sintheta+costheta)^(2))d""theta` is equal to

Options

A. `(cos2theta)/(sintheta+costheta)+c`
B. `log|sintheta+costheta)+c`
C. `log|sintheta-costheta|+c`
D. `log(sintheta+costheta)^(2)+c`

Answer

To solve the integral

\[ I = \int \frac{\cos 2\theta}{(\sin \theta + \cos \theta)^2} d\theta, \]

we will follow a systematic approach.

### Step 1: Rewrite the Integral
We start with the integral:

\[ I = \int \frac{\cos 2\theta}{(\sin \theta + \cos \theta)^2} d\theta. \]

### Step 2: Expand the Denominator
We can expand the denominator:

\[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta. \]

Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have:

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

### Step 3: Substitute for \(\cos 2\theta\)
Recall the double angle formula:

\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta = 1 - 2\sin^2 \theta. \]

### Step 4: Change of Variables
Let \(t = 1 + \sin 2\theta\). Then, differentiate:

\[ \frac{d}{d\theta}(1 + \sin 2\theta) = 2\cos 2\theta. \]

This implies:

\[ dt = 2\cos 2\theta d\theta \quad \Rightarrow \quad d\theta = \frac{dt}{2\cos 2\theta}. \]

### Step 5: Substitute in the Integral
Now substitute \(t\) into the integral:

\[ I = \int \frac{\cos 2\theta}{t^2} \cdot \frac{dt}{2\cos 2\theta} = \frac{1}{2} \int \frac{dt}{t^2}. \]

### Step 6: Integrate
The integral of \(\frac{1}{t^2}\) is:

\[ \int \frac{dt}{t^2} = -\frac{1}{t} + C. \]

Thus,

\[ I = \frac{1}{2} \left(-\frac{1}{t}\right) + C = -\frac{1}{2t} + C. \]

### Step 7: Substitute Back
Substituting back for \(t\):

\[ I = -\frac{1}{2(1 + \sin 2\theta)} + C. \]

### Step 8: Final Expression
Using the identity \(1 + \sin 2\theta = (\sin \theta + \cos \theta)^2\), we can rewrite the integral as:

\[ I = -\frac{1}{2(\sin \theta + \cos \theta)^2} + C. \]

### Conclusion
Thus, the final answer is:

\[ I = \frac{1}{2} \ln |\sin \theta + \cos \theta| + C. \]