`int_0^(pi//2)(sinxcosx)/(cos^2x+3cosx+2)dx=`

Options

A. `log (9//8)`
B. `log (8//9)`
C. `log(3//4)`
D. None of these

Answer

To solve the integral

\[ I = \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\cos^2 x + 3 \cos x + 2} \, dx, \]

we will use substitution and partial fractions.

### Step 1: Substitution
Let \( t = \cos x \). Then, \( dt = -\sin x \, dx \) or \( \sin x \, dx = -dt \). The limits change as follows:
- When \( x = 0 \), \( t = \cos(0) = 1 \).
- When \( x = \frac{\pi}{2} \), \( t = \cos\left(\frac{\pi}{2}\right) = 0 \).

Thus, the integral becomes:

\[ I = \int_1^0 \frac{-t}{t^2 + 3t + 2} \, dt = \int_0^1 \frac{t}{t^2 + 3t + 2} \, dt. \]

### Step 2: Factor the Denominator
We need to factor the denominator \( t^2 + 3t + 2 \):

\[ t^2 + 3t + 2 = (t + 1)(t + 2). \]

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

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

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

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

Expanding and rearranging gives:

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

Setting coefficients equal, we have:

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

From the first equation, \( B = 1 - A \). Substituting into the second equation:

\[ 2A + (1 - A) = 0 \implies A + 1 = 0 \implies A = -1. \]

Then, substituting back to find \( B \):

\[ B = 1 - (-1) = 2. \]

Thus, we have:

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

### Step 4: Integrate
Now we can integrate:

\[ I = \int_0^1 \left( \frac{-1}{t + 1} + \frac{2}{t + 2} \right) dt. \]

This can be split into two integrals:

\[ I = -\int_0^1 \frac{1}{t + 1} \, dt + 2 \int_0^1 \frac{1}{t + 2} \, dt. \]

Calculating each integral:

1. For the first integral:

\[ -\int_0^1 \frac{1}{t + 1} \, dt = -[\log(t + 1)]_0^1 = -(\log(2) - \log(1)) = -\log(2). \]

2. For the second integral:

\[ 2 \int_0^1 \frac{1}{t + 2} \, dt = 2[\log(t + 2)]_0^1 = 2(\log(3) - \log(2)) = 2\log\left(\frac{3}{2}\right). \]

### Step 5: Combine Results
Combining these results gives:

\[ I = -\log(2) + 2\log\left(\frac{3}{2}\right) = -\log(2) + \log(3^2) - \log(2^2) = -\log(2) + \log(9) - \log(4). \]

This simplifies to:

\[ I = \log\left(\frac{9}{4}\right) - \log(2) = \log\left(\frac{9}{8}\right). \]

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

\[ I = \log\left(\frac{9}{8}\right). \]