`int_(0)^(pi//2) (sin x cos x) / ((1 + 2 sin x ) ( 1+ sin x))dx = `

Options

A. `1/2 log (3/4)`
B. `1/2 log (4/3)`
C. `1/2 log (3/2)`
D. `1/2 log (2/3)`

Answer

To solve the integral

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

we can follow these steps:

### Step 1: Simplify the Integral
We know that \(\sin x \cos x = \frac{1}{2} \sin(2x)\). Thus, we can rewrite the integral as:

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

### Step 2: Use Substitution
Let \(t = \sin x\). Then, \(dt = \cos x \, dx\) and when \(x = 0\), \(t = 0\), and when \(x = \frac{\pi}{2}\), \(t = 1\). The integral becomes:

\[ I = \frac{1}{2} \int_{0}^{1} \frac{t}{(1 + 2t)(1 + t)} \, dt. \]

### Step 3: Partial Fraction Decomposition
We can decompose the integrand using partial fractions:

\[ \frac{t}{(1 + 2t)(1 + t)} = \frac{A}{1 + 2t} + \frac{B}{1 + t}. \]

Multiplying through by the denominator \((1 + 2t)(1 + t)\) gives:

\[ t = A(1 + t) + B(1 + 2t). \]

Expanding and combining like terms, we get:

\[ t = A + At + B + 2Bt \implies t = (A + B) + (A + 2B)t. \]

### Step 4: Solve for Coefficients
Equating coefficients, we have:

1. \(A + B = 0\)
2. \(A + 2B = 1\)

From the first equation, \(B = -A\). Substituting into the second equation:

\[ A - 2A = 1 \implies -A = 1 \implies A = -1, \quad B = 1. \]

Thus, we can rewrite the integral as:

\[ I = \frac{1}{2} \int_{0}^{1} \left( \frac{-1}{1 + 2t} + \frac{1}{1 + t} \right) dt. \]

### Step 5: Integrate Each Term
Now we can integrate each term separately:

\[ I = \frac{1}{2} \left( -\int_{0}^{1} \frac{1}{1 + 2t} dt + \int_{0}^{1} \frac{1}{1 + t} dt \right). \]

Calculating the first integral:

\[ \int \frac{1}{1 + 2t} dt = \frac{1}{2} \ln(1 + 2t) + C. \]

Evaluating from \(0\) to \(1\):

\[ \left[ \frac{1}{2} \ln(1 + 2t) \right]_{0}^{1} = \frac{1}{2} \ln(3) - \frac{1}{2} \ln(1) = \frac{1}{2} \ln(3). \]

Calculating the second integral:

\[ \int \frac{1}{1 + t} dt = \ln(1 + t) + C. \]

Evaluating from \(0\) to \(1\):

\[ \left[ \ln(1 + t) \right]_{0}^{1} = \ln(2) - \ln(1) = \ln(2). \]

### Step 6: Combine Results
Putting it all together:

\[ I = \frac{1}{2} \left( -\frac{1}{2} \ln(3) + \ln(2) \right) = \frac{1}{2} \left( \ln(2) - \frac{1}{2} \ln(3) \right). \]

### Step 7: Final Simplification
This can be simplified to:

\[ I = \frac{1}{2} \ln\left(\frac{2}{\sqrt{3}}\right). \]

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

\[ I = \frac{1}{2} \ln\left(\frac{2}{\sqrt{3}}\right). \]