`int_(0)^(pi//2) ( sin x cos x )/( 1+ sin x ) dx` is equal to

Options

A. `3 - log 12`
B. `1- log 2`
C. `1+ log 2`
D. none of these

Answer

To solve the integral

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

we will use a substitution method.

### Step 1: Substitution
Let

\[ t = 1 + \sin x. \]

Then, the derivative of \(t\) with respect to \(x\) is

\[ dt = \cos x \, dx. \]

### Step 2: Change of Limits
When \(x = 0\),

\[ t = 1 + \sin(0) = 1. \]

When \(x = \frac{\pi}{2}\),

\[ t = 1 + \sin\left(\frac{\pi}{2}\right) = 2. \]

### Step 3: Express \(\sin x\) in terms of \(t\)
From the substitution \(t = 1 + \sin x\), we can express \(\sin x\) as

\[ \sin x = t - 1. \]

### Step 4: Rewrite the Integral
Now we can rewrite the integral in terms of \(t\):

\[ I = \int_{1}^{2} \frac{(t - 1) \cos x}{t} \, dt. \]

Since \(dt = \cos x \, dx\), we can replace \(\cos x \, dx\) with \(dt\):

\[ I = \int_{1}^{2} \frac{(t - 1)}{t} \, dt. \]

### Step 5: Simplify the Integral
We can simplify the integrand:

\[ \frac{(t - 1)}{t} = 1 - \frac{1}{t}. \]

Thus, the integral becomes:

\[ I = \int_{1}^{2} \left(1 - \frac{1}{t}\right) dt. \]

### Step 6: Integrate
Now we can integrate:

\[ I = \int_{1}^{2} 1 \, dt - \int_{1}^{2} \frac{1}{t} \, dt. \]

Calculating these integrals:

\[ \int_{1}^{2} 1 \, dt = [t]_{1}^{2} = 2 - 1 = 1, \]

and

\[ \int_{1}^{2} \frac{1}{t} \, dt = [\ln t]_{1}^{2} = \ln(2) - \ln(1) = \ln(2). \]

### Step 7: Combine Results
Putting it all together, we have:

\[ I = 1 - \ln(2). \]

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

\[ \boxed{1 - \ln(2)}. \]