`int_(0)^(pi//2)(sin x cos x dx)/(cos^(2)x+3cosx+2)=`

Options

A. `log(8/9)`
B. `log(9/8)`
C. `log(8xx9)`
D. `log(3/8)`

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 follow these steps:

### Step 1: Substitution
Let \( t = \cos x \). Then, the differential \( dt = -\sin x \, dx \) or \( \sin x \, dx = -dt \).

### Step 2: Change the limits of integration
When \( x = 0 \), \( t = \cos(0) = 1 \).
When \( x = \frac{\pi}{2} \), \( t = \cos\left(\frac{\pi}{2}\right) = 0 \).

### Step 3: Rewrite the integral
Substituting \( t \) into the integral, we have:

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

### Step 4: Factor the denominator
The quadratic \( t^2 + 3t + 2 \) can be factored as:

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

### Step 5: Partial fraction decomposition
We can express

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

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

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

### Step 6: Solve for A and B
Expanding and rearranging gives:

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

By comparing coefficients, 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, we find:

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

Thus, we can rewrite the integral as:

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

### Step 7: Integrate
Now we can integrate term by term:

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

Calculating these integrals:

1. \( \int -\frac{1}{t + 1} dt = -\ln(t + 1) \)
2. \( \int \frac{1}{t + 2} dt = \ln(t + 2) \)

Thus,

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

Calculating the limits:

- For \( -\ln(t + 1) \):

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

- For \( 2\ln(t + 2) \):

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

### Step 8: Combine results
Putting it all together:

\[ I = -\ln(2) + 2\ln\left(\frac{3}{2}\right) = -\ln(2) + \ln\left(\frac{9}{4}\right) = \ln\left(\frac{9}{4}\right) - \ln(2) = \ln\left(\frac{9}{8}\right). \]

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

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