`int(3 cos x + 2 sin x)/(4 sin x + 5 cos x) dx = ..........`

Answer

To solve the integral

\[ \int \frac{3 \cos x + 2 \sin x}{4 \sin x + 5 \cos x} \, dx, \]

we can use the method of substitution and partial fractions.

### Step 1: Rewrite the Numerator
We want to express the numerator \(3 \cos x + 2 \sin x\) in terms of the derivative of the denominator \(4 \sin x + 5 \cos x\).

Let \(u = 4 \sin x + 5 \cos x\). Then, we find the derivative:

\[ \frac{du}{dx} = 4 \cos x - 5 \sin x. \]

### Step 2: Express the Numerator in Terms of \(u\)
We can express \(3 \cos x + 2 \sin x\) in terms of \(u\). We need to find constants \(A\) and \(B\) such that:

\[ 3 \cos x + 2 \sin x = A(4 \cos x - 5 \sin x) + B(4 \sin x + 5 \cos x). \]

### Step 3: Set Up the System of Equations
By comparing coefficients, we can set up the following equations:

1. For \(\cos x\):
\[ 3 = 5B + 4A \quad \text{(Equation 1)} \]

2. For \(\sin x\):
\[ 2 = 4B - 5A \quad \text{(Equation 2)} \]

### Step 4: Solve the System of Equations
From Equation 1:
\[ 4A + 5B = 3. \]

From Equation 2:
\[ -5A + 4B = 2. \]

Now, we can solve these equations simultaneously.

Multiply Equation 1 by 5:
\[ 20A + 25B = 15. \]

Multiply Equation 2 by 4:
\[ -20A + 16B = 8. \]

Now, add these two equations:
\[ (20A - 20A) + (25B + 16B) = 15 + 8, \]
which simplifies to:
\[ 41B = 23 \implies B = \frac{23}{41}. \]

Substituting \(B\) back into Equation 1:
\[ 4A + 5\left(\frac{23}{41}\right) = 3, \]
which gives:
\[ 4A + \frac{115}{41} = 3. \]
Multiplying through by 41 to eliminate the fraction:
\[ 164A + 115 = 123 \implies 164A = 8 \implies A = \frac{2}{41}. \]

### Step 5: Substitute Back into the Integral
Now we substitute \(A\) and \(B\) back into the integral:

\[ \int \frac{3 \cos x + 2 \sin x}{4 \sin x + 5 \cos x} \, dx = \int \left(\frac{2}{41}(4 \sin x + 5 \cos x) + \frac{23}{41}(4 \cos x - 5 \sin x)\right) \frac{1}{4 \sin x + 5 \cos x} \, dx. \]

### Step 6: Integrate
This simplifies to:

\[ \frac{2}{41} \int dx + \frac{23}{41} \int \frac{4 \cos x - 5 \sin x}{4 \sin x + 5 \cos x} \, dx. \]

The first integral is straightforward:

\[ \frac{2}{41} x + C_1. \]

The second integral can be solved using the substitution \(u = 4 \sin x + 5 \cos x\):

\[ \int \frac{du}{u} = \ln |u| + C_2 = \ln |4 \sin x + 5 \cos x| + C_2. \]

### Final Answer
Combining everything, we get:

\[ \int \frac{3 \cos x + 2 \sin x}{4 \sin x + 5 \cos x} \, dx = \frac{2}{41} x + \frac{23}{41} \ln |4 \sin x + 5 \cos x| + C. \]