`int_(0)^(pi//2) (sin^(2)x)/(sin x+cos x) dx` is equal to

Options

A. `pi//2`
B. `sqrt2 log (sqrt2+1)`
C. `(1)/(sqrt2) log (sqrt2+1)`
D. None of these

Answer

To solve the integral

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

we will use a symmetry property of definite integrals.

### Step 1: Set up the integral
Let

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

### Step 2: Use the property of definite integrals
We can use the property that

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

In our case, we will substitute \( x \) with \( \frac{\pi}{2} - x \):

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

### Step 3: Simplify the integral
Using the identities \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \), we have:

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

### Step 4: Add the two integrals
Now we have two expressions for \( I \):

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

Adding these two equations gives:

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

### Step 5: Use the Pythagorean identity
Using the identity \( \sin^2 x + \cos^2 x = 1 \):

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

### Step 6: Solve the integral
Now we need to evaluate

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

We can factor out \( \sqrt{2} \):

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

This can be rewritten as:

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

Thus,

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

### Step 7: Evaluate the integral of secant
The integral of \( \sec x \) is known:

\[ \int \sec x \, dx = \ln |\sec x + \tan x| + C. \]

Applying the limits:

\[ \int_{0}^{\frac{\pi}{2}} \sec\left(x + \frac{\pi}{4}\right) \, dx = \left[ \ln |\sec\left(x + \frac{\pi}{4}\right) + \tan\left(x + \frac{\pi}{4}\right)| \right]_{0}^{\frac{\pi}{2}}. \]

### Step 8: Calculate the limits
Calculating at the limits gives:

At \( x = \frac{\pi}{2} \):

\[ \sec\left(\frac{\pi}{2} + \frac{\pi}{4}\right) + \tan\left(\frac{\pi}{2} + \frac{\pi}{4}\right) \to \infty. \]

At \( x = 0 \):

\[ \sec\left(0 + \frac{\pi}{4}\right) + \tan\left(0 + \frac{\pi}{4}\right) = \sqrt{2} + 1. \]

Thus, the integral diverges at the upper limit.

### Final Calculation
Returning to our expression for \( 2I \):

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

Thus,

\[ I = \frac{1}{2\sqrt{2}} \ln(\sqrt{2} + 1). \]

### Final Answer
Therefore, the value of the integral is:

\[ I = \frac{1}{\sqrt{2}} \ln(\sqrt{2} + 1). \]

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