`int_(0) ^(pi//2) (cos^(2) x)/(sin x + cos x ) dx = `

Options

A. `1/(2sqrt(2)) log ((sqrt(2)-1)/(sqrt(2)+1))`
B. `1/(2sqrt(2)) log ((sqrt(2)+1)/(sqrt(2)-1))`
C. `1/(sqrt(2)) log ((sqrt(2)-1)/(sqrt(2)+1))`
D. `1/(sqrt(2)) log ((sqrt(2)+1)/(sqrt(2)-1))`

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x}{\sin x + \cos x} \, dx, \]

we can use the property of definite integrals:

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx. \]

### Step 1: Apply the property of definite integrals

We will substitute \(x\) with \(\frac{\pi}{2} - x\):

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)} \, dx. \]

Using the trigonometric identities \(\cos\left(\frac{\pi}{2} - x\right) = \sin x\) and \(\sin\left(\frac{\pi}{2} - x\right) = \cos x\), we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{\cos x + \sin x} \, dx. \]

### Step 2: Combine the two integrals

Now we have two expressions for \(I\):

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x}{\sin x + \cos x} \, dx, \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} \, dx. \]

Adding these two equations gives:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x + \sin^2 x}{\sin x + \cos x} \, dx. \]

Since \(\cos^2 x + \sin^2 x = 1\), we can simplify this to:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx. \]

### Step 3: Solve the integral

Now we need to evaluate:

\[ I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx. \]

To evaluate this integral, we can multiply the numerator and denominator by \(\sqrt{2}\):

\[ \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{2}}{\sqrt{2}(\sin x + \cos x)} \, dx. \]

Now, we can rewrite \(\sin x + \cos x\) as:

\[ \sin x + \cos x = \sqrt{2} \left(\sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}}\right) = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right). \]

Thus, we have:

\[ \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)} \, dx. \]

### Step 4: Evaluate the integral

This integral can be solved using the substitution \(u = x + \frac{\pi}{4}\), which gives:

\[ du = dx \quad \text{and when } x = 0, u = \frac{\pi}{4}, \text{ and when } x = \frac{\pi}{2}, u = \frac{3\pi}{4}. \]

Thus, we have:

\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{\sqrt{2} \sin u} \, du = \frac{1}{\sqrt{2}} \left[-\log(\csc u + \cot u)\right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}. \]

Evaluating this gives us:

\[ \frac{1}{\sqrt{2}} \left[-\log(\sqrt{2} + 1) + \log(\sqrt{2} - 1)\right]. \]

### Final Result

Thus,

\[ I = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \left[-\log(\sqrt{2} + 1) + \log(\sqrt{2} - 1)\right]. \]

This simplifies to:

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