91Ó°ÊÓ

The given integral is: $$\int_{0}^{\pi / 2} \ln (1+\cos \theta \cos x) \frac{d x}{\cos x}$$

Let us make a substitution to simplify the integral. Let \(u = \cos x\). Note that \(du = -\sin x dx\). We also need to change the limits of integration based on \(u\): when \(x = 0\), \(u = \cos{0} = 1\), and when \(x = \frac{\pi}{2}\), \(u = \cos{\frac{\pi}{2}} = 0\). So, the new integral will be: $$\int_{1}^{0} \ln (1+\cos \theta u) \frac{-du}{u}$$

In order to make the limits of integration go from low to high (from \(0\) to \(1\)), we can swap the limits and change the sign of the integral: $$\int_{0}^{1} \ln (1+\cos \theta u) \frac{du}{u}$$

Now, we can integrate \(\ln (1 + \cos \theta u)\) with respect to \(u\). This integration requires some manipulation: $$\int_{0}^{1} \ln (1+\cos \theta u) \frac{du}{u} = \int_{0}^{1} \frac{\ln (1 + u(\cos \theta - 1) + u)}{u}du$$

We can use integration by parts in this case, defining the two functions \(f = \ln (1 + u(\cos \theta - 1) + u)\) and \(g' = \frac{1}{u}\): $$f'(u) = \frac{\cos \theta - 1}{1+u(\cos \theta - 1) + u}$$ $$g(u) = \ln u$$ Applying the integration by parts formula, we get: $$\int_{0}^{1} \ln (1+u(\cos \theta - 1) + u)\frac{du}{u} = f(1)g(1)-f(0)g(0) - \int_{0}^{1} f'(u)g(u)du$$

Now, we can simplify the integral: $$= 0-0 - \int_{0}^{1} \frac{\cos \theta - 1}{1+u(\cos \theta - 1) + u}\ln u du$$ Since the first terms are zero, we have: $$\int_{0}^{1} \ln (1+u(\cos \theta - 1) + u)\frac{du}{u} = -\int_{0}^{1} \frac{\cos \theta - 1}{1+u(\cos \theta - 1) + u}\ln u du$$ At this point, further evaluation of the integral using elementary functions is not possible. The result can be represented in terms of special functions, but it is beyond the scope of high school calculus. Nevertheless, we have simplified the original integral quite a bit, making it easier for further analysis using numerical methods or special functions. Thus, the final simplified form of the integral is: $$\int_{0}^{\pi / 2} \ln (1+\cos \theta \cos x) \frac{d x}{\cos x}= -\int_{0}^{1} \frac{\cos \theta - 1}{1+u(\cos \theta - 1) + u}\ln u \,du$$