Integration and Power Series

Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created September 29, 2025

We have shown that computing the derivatives of \( \sin() \) and \( \cos() \) is independent of the Pythagorean Theorem and the Pythagorean Identity. This result allows us to show that the proof of \( \sin^2(x)+\cos^2(x) \) being a constant function is also independent of the Pythagorean Theorem and the Pythagorean Identity. In particular we used the following results: \[ \frac{d \sin^2(x)}{dx}=2\sin(x)\cos(x) \mbox{\ \ \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ } \frac{d \cos^2(x)}{dx}=-2\sin(x)\cos(x) \] Thus, integrating the right hand sides gives us: $$ \begin{eqnarray*} \sin^2(x) &=& \int (2\sin(x)\cos(x))dx = \int \sin(2x)dx = g(x) + C_1 \\ \cos^2(x) &=& -\int (2\sin(x)\cos(x))dx = -g(x) + C_2, \end{eqnarray*} $$ where \( g(x) \) is the function we are interested in and \( C_1 \) and \( C_2 \) are two constants. Adding the above together gives \[ \sin^2(x) + \cos^2(x) = C \] for some constant \( C \). Then, we have the Pythagorean Identiy. Bogomolny showed the same result. However, we are interested in representing \( \sin^2(x) \) and \( \cos^2(x) \) using power series directly. Or, more precisely, we wish to answer: what is \( g(x) \) in explicit form? We know the following from any calculus book: $$ \begin{eqnarray*} \sin(x) &=& \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)!} \\ \cos(x) &=& \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!} \end{eqnarray*} $$ Note that the development of the above series only needs the Taylor series concept and the derivatives of \( \sin(x) \) and \( \cos(x) \). As a result, obtaining the above series is independent of the Pythagorean Theorem and the Pythagorean Identity.

From \( \sin^2(x) \) obtained earlier, we have $$ \begin{eqnarray*} \sin^2(x) &=& \int \sin(2x)dx = \int \sum_{n=0}^{\infty} \left( \frac{(-1)^n (2x)^{2n+1}}{(2n+1)!} \right) dx \\ &=& \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\int (2x)^{2n+1}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\cdot\frac{1}{(2n+1)+1} x^{2(n+1)+1} + C_1 \\ &=& \sum_{n=0}^{\infty} \frac{(-1)^n2^{2n+1}}{(2n+1)!2(n+1)}x^{2(n+1)}+C_1 \end{eqnarray*} $$ Because \( \sin(0) = 0 \), \( C_1=0 \). In this way, we have the power series of \( \sin^2(x) \). Similarly, we have the power series of \( \cos^2(x) \) as follows: \[ \cos^2(x) = -\sum_{n=0}^{\infty} \frac{(-1)^n2^{2n+1}}{(2n+1)!2(n+1)}x^{2(n+1)}+C_2 \] Because \( \cos(0) = 1 \), \( C_2 = 1 \). Adding them together yields the Pythagorean Identity.

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