This paper is devoted to the study of the integral of the form \(\int \limits \sin(x)^n |\cos(x)| \,\Bbb{d}x\), where the cosine is taken by absolute value, which gives rise to interesting symmetry and features in calculations. The methods of analytical and numerical integration are studied in detail depending on the range of the variable x, emphasizing the use of classical methods of analysis and integral calculus.

The solution of this type of integrals is required to solve problems related to the Lorentz factor. Next, we will consider the integral (1) within the limits (\(0..T\)), where: \(T \le 2\pi\) \[I(T) = \int \limits_0^{T} \sin(x)^n |\cos(x)| \,\Bbb{d}x, \quad n=0,1,2,3,\ldots \tag{1}\] Without the modulus in the integrand, its solution does not present any difficulties. The modulus imposes certain restrictions on each range of the integration limit. Let us represent the integral as follows: \[I(T) = \int \limits_0^{T} \sin(x)^n {|\cos(x)| \over \cos(x)} \,\Bbb{d}\sin(x) = \int \limits_0^{T} A \tag{2}\] Then in each of the ranges the integral will be taken differently: \[I(T) = \begin{cases} \int \limits_0^{T} A, & \mbox{if } T \le \pi/2 \\ \int \limits_0^{T} A - 2 \int \limits_{\pi/2}^{T} A, & \mbox{if } T \gt \pi/2 \mbox{ and } T \le 3\pi/2 \\ \int \limits_0^{T} A - 2 \int \limits_{\pi/2}^{T} A + \int \limits_{3\pi/2}^{T} A, & \mbox{otherwise} \end{cases} \tag{3}\] In each of the ranges, the solution to such an integral is classical: \[\int \limits_{T_1}^{T_2} A = {\sin(T_2)^{n+1} - \sin(T_1)^{n+1} \over n+1} \tag{4}\] Hence, reducing and transforming some functions, we obtain the final result for each range: \[I(T) = \frac{1}{n+1} \begin{cases} \sin(T)^{n+1}, & \mbox{if } T \le \pi/2 \\ 2 - \sin(T)^{n+1}, & \mbox{if } T \gt \pi/2 \mbox{ and } T \le 3\pi/2 \\ \sin(T)^{n+1} + 2 \left(1 - (-1)^{n+1} \right), & T < 2\pi \end{cases} \tag{5}\] The solution to integral (1) for the full period (\(0..2\pi\) can be interesting and quite simple: \[I(2\pi) = 2 \frac{1 - (-1)^{n+1}}{n+1} \tag{6}\] For several periods, the solution to (6) simply needs to be multiplied by their number. Interestingly, the sum of a series with an infinite number of \(n\) composed of the squares of such solutions will be: \[\sum \limits_{n=0}^{\infty} I(2\pi)^2 = 4 \sum \limits_{n=0}^{\infty} \left(\frac{1 - (-1)^{n+1}}{n+1} \right)^2 = 2\pi^2 \tag{7}\] This solution assumes one full period, but what if there are more such periods?

In this case, the solution is a little more complicated, and formula (5) becomes: \[I(T,N) = I(T_{mod}) + 2N \frac{1 - (-1)^{n+1}}{n+1} \\ T_{mod} = \mbox{mod} (T, 2\pi) \tag{8}\] Here \(N\) is the number of full periods from \(T\), that is: \[ N = \mbox{floor} \left( {T \over 2\pi} \right) \tag{9}\] Whence, by the way, it follows that \(T_{mod}\) can be written as so: \[ T_{mod} = T - 2\pi N \tag{10}\] Continuing this logic, we obtain the transformed formula (6): \[I(N) = 2 (N+1) \frac{1 - (-1)^{n+1}}{n+1} \tag{11}\] From expression (7), we also obtain a formula for an integer number of periods from the sum of integrals over all powers of \(n\): \[\sum \limits_{n=0}^{\infty} I(N)^2 = 2\pi^2 (N+1)^2 \tag{12}\]

In the next section, we will complicate the problem and instead of sine, we will put an exponential with an integer power and an imaginary unit.