Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\int_{0}^{u} \frac{\sin x d x}{\sqrt{1-x^{2}}}$$

Integral calculus is a fundamental part of calculus focused on the concept of integration. Integrating a function essentially means finding the area under the curve of that function. This is used in many areas of mathematics and physics to solve problems involving area, volume, displacement, and more.

In this exercise, we are tasked with finding the integral of the function \(\frac{\text{sin}(x)}{\text{sqrt}(1-x^2)}\) from 0 to 'u'. Integral calculus allows us to calculate definite integrals like this one, which represent the accumulation of quantities. For the given function, it is not straightforward to integrate directly, so we employ a technique to simplify our work by using series expansions.

The process involves several steps:

• Define the function to be integrated.
• Use suitable expansions to express the function in a simpler form.
• Finally, integrate term-by-term.

This approach makes use of both integral calculus and series expansion, highlighting how these mathematical tools can work together to simplify complex problems.

Series expansion is a way of expressing a function as a sum of terms from a sequence. This technique is very useful when dealing with complicated functions, allowing us to approximate them using simpler polynomials. In this exercise, we utilize the Maclaurin series expansion, a specific case of the Taylor series expansion, where the series is centered around zero.

To simplify our integral, we need the series expansions of two functions:

• Binomial Series Expansion: Applies to functions of the form \((1 - x^2)^{-1/2}\). Expanding this gives: \( (1 - x^2)^{-1/2} = 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \text{...} \).
• Maclaurin Series for Sine: The sine function can be expanded as: \(\text{sin}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \text{...} \).

Combining these expansions helps us to approximate the integrand. Multiplying these series together and simplifying gives us a new series that is easier to integrate.

The Taylor series is a powerful mathematical tool that represents a function as an infinite sum of terms. Each term is calculated from the derivatives of the function at a single point. The Maclaurin series is a special case of the Taylor series centered at zero.

The general formula for the Taylor series of a function \(f(x)\) around a point \(a\) is:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \text{...} \]

For the Maclaurin series, \(a = 0\), simplifying the formula to:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \text{...} \]

In the given exercise, after defining the function and using suitable expansions, we combine and simplify to obtain a polynomial form. By integrating this polynomial term-by-term, we find the first few terms of the integral’s Maclaurin series:

• \frac{u^2}{2}\li>
\frac{u^4}{12}\li>\frac{u^6}{240}\li>

These terms provide a valuable approximation that simplifies the original complex integral into a more manageable form. You can verify this using computational tools such as WolframAlpha or SymPy.