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The Maclaurin Series is a type of Taylor series expansion of a function at zero. It's a potent tool for approximating functions using polynomial terms. The general formula for the Maclaurin series of a function \( f(x) \) is:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \)

For the sine function, the Maclaurin series expansion is:

• \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \, \cdots \)

When dealing with transformations, such as \( \sin(2x) \), we adjust for the factor, leading to:

• \( \sin(2x) = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \, \cdots \)
• After simplification, \( \sin(2x) = 2x - \frac{4x^3}{3} + \frac{16x^5}{15} - \, \cdots \)

Knowing this is crucial for further operations, like integration, without starting from scratch each time.

Trigonometric identities are equations involving trigonometric functions that hold true for every value of the occurring variables. One of the foundational identities is \( 2 \sin(x) \cos(x) = \sin(2x) \). This identity allows us to transform products of sine and cosine into a more manageable single sine function with double the angle. To solve our problem of finding the power series for \( \sin^2(x) \), the identity simplifies the complexity significantly. By rewriting \( \sin^2(x) \) in terms of \( \sin(2x) \), it becomes easier to handle within the realm of series. We express \( \sin^2(x) \) as:

• \( \sin^2(x) = \left(\frac{1}{2} \sin(2x)\right)^2 = \frac{1}{4} \sin^2(2x) \)

This conversion is pivotal, setting the stage for further exploration through integration and series expansion.

Integrating a power series term by term is a systematic approach to obtaining the integral of a function expressed by a series. This method is particularly useful in calculus due to its straightforward property: we integrate each term of the series individually.Given the series representation of \( \sin(2x) \) as:

• \( 2x - \frac{4x^3}{3} + \frac{16x^5}{15} - \, \cdots \)

We integrate each term separately:

• \( \int 2x \, dx = x^2 \)
• \( \int \left(-\frac{4x^3}{3}\right) dx = -\frac{x^4}{3} \)
• \( \int \frac{16x^5}{15} \, dx = \frac{x^6}{9} \)

After integrating, we must remember the context of these operations for \( \sin^2(x) = \frac{1}{4} \int \sin(2x) \, dx \). This involves multiplying the result by \( \frac{1}{4} \), leading us to our final series:

• \( \sin^2(x) = \frac{x^2}{4} - \frac{x^4}{12} + \frac{x^6}{36} - \, \cdots \)

With term-by-term integration, the complex process of series manipulations becomes more approachable.