Question

Evaluate \(\int_{0}^{\pi / 2} \ln (1+\cos \theta \cos x) \frac{d x}{\cos x}\)

Short answer

Question: Evaluate the definite integral \(\int_{0}^{\pi / 2} \ln (1+\cos \theta \cos x) \frac{d x}{\cos x}\). Answer: The simplified form of the integral is: \(-\int_{0}^{1} \frac{\cos \theta - 1}{1+u(\cos \theta - 1) + u}\ln u \,du\). Further evaluation using elementary functions is not possible, and the result can be represented in terms of special functions or numerical methods.

Step-by-step solution

Identify the integral

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

Change of variables

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: $$\int_{1}^{0} \ln (1+\cos \theta u) \frac{-du}{u}$$

Reverse the limits of integration

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: $$\int_{0}^{1} \ln (1+\cos \theta u) \frac{du}{u}$$

Break down the integral

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

Use integration by parts

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: $$\int_{0}^{1} \ln (1+u(\cos \theta - 1) + u)\frac{du}{u} = f(1)g(1)-f(0)g(0) - \int_{0}^{1} f'(u)g(u)du$$

Evaluate integrals and simplify

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: $$\int_{0}^{\pi / 2} \ln (1+\cos \theta \cos x) \frac{d x}{\cos x}= -\int_{0}^{1} \frac{\cos \theta - 1}{1+u(\cos \theta - 1) + u}\ln u \,du$$