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Integral with sine in degrees and the cosine module
The solution to this type integrals required for the solution of tasks related to the Lorentz factor. Next, we consider the integral (1) within (\(0..T\)), where: \(T \le 2\pi\) \[I(T) = \int \limits_0^{T} \sin(x)^n |\cos(x)| \,\Bbb{d}x, \quad n=0,1,2,3,\ldots \qquad (1)\] Without the module in expanded terms, its solution presents no difficulties. The module also imposes certain limitations on each range limit of integration. Think of the integral as \[I(T) = \int \limits_0^{T} \sin(x)^n {|\cos(x)| \over \cos(x)} \,\Bbb{d}\sin(x) = \int \limits_0^{T} A \qquad (2)\] Then in each band the integral is taken in different ways: \[I(T) = \begin{cases} \int \limits_0^{T} A, & \mbox{if } T \le \pi/2 \\ \int \limits_0^{T} A - 2 \int \limits_{\pi/2}^{T} A, & \mbox{if } T \gt \pi/2 \mbox{ and } T \le 3\pi/2 \\ \int \limits_0^{T} A - 2 \int \limits_{\pi/2}^{T} A + \int \limits_{3\pi/2}^{T} A, & \mbox{otherwise} \end{cases} \qquad (3)\] In each of the ranges are the solution to this classic integral: \[\int \limits_{T_1}^{T_2} A = {\sin(T_2)^{n+1} - \sin(T_1)^{n+1} \over n+1} \qquad (4)\] Hence, reducing and converting some of the functions, we obtain the final result for each range is: \[I(T) = \frac{1}{n+1} \begin{cases} \sin(T)^{n+1}, & \mbox{if } T \le \pi/2 \\ 2 - \sin(T)^{n+1}, & \mbox{if } T \gt \pi/2 \mbox{ and } T \le 3\pi/2 \\ \sin(T)^{n+1} + 2 \left(1 - (-1)^{n+1} \right), & \mbox{otherwise} \end{cases} \qquad (5)\] this is an Interesting and quite simple may be the solution of the integral (1) for the full period (\(0..2\pi\)): \[I(2\pi) = 2 \frac{1 - (-1)^{n+1}}{n+1} \qquad (6)\] For multiple periods, the solution to (6), simply multiply by their number. Interestingly, the sum of the series with an infinite number of \(n\), made up of squares such decisions will be like this: \[\sum \limits_{n=0}^{\infty} I(2\pi)^2 = 4 \sum \limits_{n=0}^{\infty} \left(\frac{1 - (-1)^{n+1}}{n+1} \right)^2 = 2\pi^2 \qquad (7)\]