Find \( \int \pi 2 \operatorname{cosin} 10(x) \cos (x) d x \) \[ \begin{array}{ll}\text { a. } & 1 \\ \text { b. } & 0 \\ \text { c. } & 13 \pi 1098 \\ \text { d. } & 21 \pi 2048\end{array} \]

To solve the integral \[ \int \pi 2 \cos(10x) \cos(x) \, dx, \] we can use the product-to-sum formulas. The formula states that: \[ \cos(a) \cos(b) = \frac{1}{2}[\cos(a + b) + \cos(a - b)]. \] Applying it here with \(a = 10x\) and \(b = x\): \[ \cos(10x) \cos(x) = \frac{1}{2}[\cos(11x) + \cos(9x)]. \] Substituting this into the integral gives: \[ \int \pi 2 \cos(10x) \cos(x) \, dx = \pi 2 \int \cos(10x) \cos(x) \, dx = \pi 2 \int \frac{1}{2}[\cos(11x) + \cos(9x)] \, dx. \] Simplifying gives: \[ = \pi \int [\cos(11x) + \cos(9x)] \, dx. \] Next, we can integrate each term: \[ \int \cos(11x) \, dx = \frac{1}{11} \sin(11x) + C, \] \[ \int \cos(9x) \, dx = \frac{1}{9} \sin(9x) + C. \] Putting it all together: \[ \int [\cos(11x) + \cos(9x)] \, dx = \frac{1}{11} \sin(11x) + \frac{1}{9} \sin(9x) + C. \] Thus, \[ \pi \left( \frac{1}{11} \sin(11x) + \frac{1}{9} \sin(9x) \right) + C. \] The final expression for the integral is \[ \frac{\pi}{11} \sin(11x) + \frac{\pi}{9} \sin(9x) + C. \] Now let’s evaluate the choices given in the problem. When taking definite bounds for \(x\), the sine functions would evaluate to zero at multiples of \(\pi\), thus the integral evaluates to zero over any complete interval of period. Therefore, the answer is: \[ \boxed{0}. \]