`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 use a substitution method.

### Step 1: Substitution
Let \( u = \sin \theta + \cos \theta \). Then, we differentiate \( u \) with respect to \( \theta \):

\[ \frac{du}{d\theta} = \cos \theta - \sin \theta \implies du = (\cos \theta - \sin \theta) d\theta. \]

### Step 2: Express \( d\theta \) in terms of \( du \)
From the above, we can express \( d\theta \):

\[ d\theta = \frac{du}{\cos \theta - \sin \theta}. \]

### Step 3: Rewrite \( \cos 2\theta \)
We know that

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

### Step 4: Substitute in the integral
Now substitute \( \cos 2\theta \) and \( d\theta \) into the integral:

\[ I = \int \frac{u(\cos \theta - \sin \theta)}{u^2} \cdot \frac{du}{\cos \theta - \sin \theta}. \]

The \( \cos \theta - \sin \theta \) terms cancel out:

\[ I = \int \frac{u}{u^2} \, du = \int \frac{1}{u} \, du. \]

### Step 5: Integrate
Now, we can integrate:

\[ I = \ln |u| + C = \ln |\sin \theta + \cos \theta| + C. \]

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

\[ \int \frac{\cos 2\theta}{(\sin \theta + \cos \theta)^2} \, d\theta = \ln |\sin \theta + \cos \theta| + C. \]

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