If \(\int[(\cos 9 x+\cos 6 x) /(2 \cos 5 x-1)] d x=k_{1} \sin 4 x+k_{2} \sin x+c\) then \(4 \mathrm{k}_{1}+\mathrm{k}_{2}=\) (a) 1 (b) 2 (c) 4 (d) 5

To rewrite the numerator in the expression as a product, we need to use the product-to-sum formula: \[\cos a + \cos b = 2 \cos(\frac{a+b}{2}) \cos(\frac{a-b}{2})\] So for our numerator: \[\cos 6x + \cos 9x = 2 \cos(\frac{6x+9x}{2}) \cos(\frac{6x-9x}{2})\]

Applying the product-to-sum formula to our numerator, we get: \[\cos 6x + \cos 9x = 2 \cos(\frac{15x}{2}) \cos(-\frac{3x}{2})\] Notice that the cosine function is even, thus \(\cos(-\frac{3x}{2}) = \cos(\frac{3x}{2})\). Therefore, substituting this into our previous expression: \[\cos 6x + \cos 9x = 2 \cos(\frac{15x}{2}) \cos(\frac{3x}{2})\] Now the given expression becomes: \[\int\frac{2 \cos(\frac{15x}{2}) \cos(\frac{3x}{2})}{2\cos 5x - 1}dx\]

Notice that the function inside the integral can be simplified further: \[\int\frac{2 \cos(\frac{15x}{2}) \cos(\frac{3x}{2})}{2\cos 5x - 1}dx = \int\frac{\cos(\frac{15x}{2}) \cos(\frac{3x}{2})}{\cos 5x - \frac{1}{2}}dx\] This integral is solved using the trick of making the denominator equal to the derivative of the numerator: \[\int\frac{\cos(\frac{15x}{2}) \cos(\frac{3x}{2})}{\cos 5x - \frac{1}{2}}dx = k_1 \sin 4x + k_2 \sin x + c\] Originally, we're tasked to find the value of \(4k_1 + k_2\). To do that, we can take the derivative of the expression on the right-hand side and set it equal to the integrand on the left-hand side: \[\frac{\cos(\frac{15x}{2}) \cos(\frac{3x}{2})}{\cos 5x - \frac{1}{2}} = 4k_1 \cos 4x + k_2 \cos x\] We need to find the values of \(k_1\) and \(k_2\) to get to our final solution. In order to find them, we'll equate coefficients.

Equating the coefficients, we get the following system of linear equations: \[\begin{cases} 4k_1 + k_2 = 1 \\ 15k_1 - 3k_2 = 0 \end{cases}\] Solve this system of linear equations. Multiply the first equation by 3 to eliminate \(k_2\): \[\begin{cases} 12k_1 + 3k_2 = 3\\ 15k_1 - 3k_2 = 0 \end{cases}\] Add the two modified equations to isolate \(k_1\): \[\begin{cases} 27k_1 = 3 \\ \end{cases}\] Divide by 27: \[k_1 = \frac{1}{9}\] Now, substitute \(k_1\) back into the first equation to solve for \(k_2\): \[4 (\frac{1}{9}) + k_2 = 1\] \[k_2 = 1 - \frac{4}{9}\] \[k_2 = \frac{5}{9}\]

Now we have the values for \(k_1\) and \(k_2\). We can calculate \(4k_1 + k_2\) as follows: \[ 4k_1 + k_2 = 4\left(\frac{1}{9}\right) + \frac{5}{9} = \frac{4+5}{9} = \frac{9}{9} = 1 \] The value of \(4k_1 + k_2\) is 1. This corresponds to option (a).