`int (cos 9x +cos 6x )/( 2 cos 5x-1)dx =A sin 4x +B sin+C,` then `A+B` is equal to:
(Where C is constant of integration)

Options

A. `1/2`
B. `3/4`
C. `2`
D. `5/4`

Answer

To solve the integral

\[ I = \int \frac{\cos 9x + \cos 6x}{2 \cos 5x - 1} \, dx, \]

we will follow these steps:

### Step 1: Rewrite the Integral
First, we rewrite the integral by multiplying and dividing the numerator and denominator by \(\sin 5x\):

\[ I = \int \frac{\sin 5x (\cos 9x + \cos 6x)}{\sin 5x (2 \cos 5x - 1)} \, dx. \]

### Step 2: Simplify the Denominator
The denominator simplifies as follows:

\[ \sin 5x (2 \cos 5x - 1) = 2 \sin 5x \cos 5x - \sin 5x = \sin 10x - \sin 5x. \]

### Step 3: Apply Trigonometric Identities
Next, we apply the identity for the sum of cosines:

\[ \cos a + \cos b = 2 \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right). \]

For \(a = 9x\) and \(b = 6x\):

\[ \cos 9x + \cos 6x = 2 \cos\left(\frac{15x}{2}\right) \cos\left(\frac{3x}{2}\right). \]

### Step 4: Substitute Back into the Integral
Substituting this back into the integral gives:

\[ I = \int \frac{2 \sin 5x \cos\left(\frac{15x}{2}\right) \cos\left(\frac{3x}{2}\right)}{\sin 10x - \sin 5x} \, dx. \]

### Step 5: Simplify Further
Using the identity for the difference of sines:

\[ \sin a - \sin b = 2 \sin\left(\frac{a-b}{2}\right) \cos\left(\frac{a+b}{2}\right), \]

we can rewrite the denominator:

\[ \sin 10x - \sin 5x = 2 \sin\left(\frac{5x}{2}\right) \cos\left(\frac{15x}{2}\right). \]

### Step 6: Cancel Common Terms
Now, we can cancel \(\cos\left(\frac{15x}{2}\right)\) from the numerator and denominator:

\[ I = \int \frac{2 \sin 5x \cos\left(\frac{3x}{2}\right)}{2 \sin\left(\frac{5x}{2}\right)} \, dx. \]

### Step 7: Further Simplification
This simplifies to:

\[ I = \int \frac{\sin 5x \cos\left(\frac{3x}{2}\right)}{\sin\left(\frac{5x}{2}\right)} \, dx. \]

### Step 8: Use Trigonometric Identities Again
Using the identity \(2 \sin A \cos B = \sin(A + B) + \sin(A - B)\):

\[ \sin 5x = 2 \sin\left(\frac{5x}{2}\right) \cos\left(\frac{5x}{2}\right), \]

we can rewrite the integral as:

\[ I = \int \cos\left(\frac{3x}{2}\right) \, dx. \]

### Step 9: Integrate
The integral of \(\cos\left(\frac{3x}{2}\right)\) is:

\[ \int \cos\left(\frac{3x}{2}\right) \, dx = \frac{2}{3} \sin\left(\frac{3x}{2}\right) + C. \]

### Step 10: Compare with Given Form
Now, we compare this result with the form \(A \sin 4x + B \sin x + C\).

Since we have:

\[ I = \frac{2}{3} \sin\left(\frac{3x}{2}\right) + C, \]

we can see that \(A = 0\) and \(B = \frac{2}{3}\).

### Final Calculation
Thus, \(A + B = 0 + \frac{2}{3} = \frac{2}{3}\).

### Conclusion
The value of \(A + B\) is:

\[ \boxed{\frac{2}{3}}. \]