Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{6}{\sqrt{4-4 x^{2}}} d x$$

Short answer

Based on the step-by-step solution, create a short answer for the integral of (6/(sqrt(4-4x^2)))dx: The integral of (6/(sqrt(4-4x^2)))dx can be solved using the substitution method. We first make the substitution x = (sin(u))/2 and dx = (1/2)cos(u)du. After replacing x and dx in the integral and simplifying, we obtain the integral of 3du. Integrating this, we get 3u + C. Replacing u back with our original substitution (u = arcsin(2x)), we arrive at the final answer of 3arcsin(2x) + C. The correctness of our solution is confirmed by differentiating the result and obtaining the original integrand.

Step-by-step solution

Choose an appropriate substitution

We have the integral: $$\int \frac{6}{\sqrt{4 - 4x^2}} dx$$ Let's try the following substitution: $$x = \frac{\sin(u)}{2}$$ Then, by differentiating we get: $$dx = \frac{1}{2}\cos(u) du$$

Replace x and dx in our integral

Replace the x and dx in our integral using the substitution and its derivative: $$\int \frac{6}{\sqrt{4 - 4\left(\frac{\sin(u)}{2}\right)^2}} \cdot \frac{1}{2}\cos(u) du$$

Simplify the integral

Simplify the integral to obtain a standard trigonometric integral: $$\int \frac{6}{\sqrt{4 - \sin^2(u)}} \cdot \frac{1}{2}\cos(u) du$$ Notice that \(4 - \sin^2(u) = (2\cos(u))^2\). Then we simplify the integral: $$\int \frac{6}{2\cos(u)} \cdot \frac{1}{2}\cos(u) du$$ $$\int 3 \frac{\cos(u)}{\cos(u)}du = \int 3 du$$

Integrate and solve for x

Integrate the simplified integral and solve for x: $$ \int 3 du = 3u + C$$ Now replace u back with our original substitution, \(u = \arcsin{2x}\): $$ 3\arcsin{2x} + C $$

Check the answer using differentiation

Differentiate the result with respect to x to check our answer: $$\frac{d}{dx} (3\arcsin{2x} + C) = \frac{d}{dx} (\arcsin{2x})$$ Using the chain rule: $$3 \cdot \frac{1}{\sqrt{1 - (2x)^2}} \cdot 2 = \frac{6}{\sqrt{4 - 4x^2}}$$ Our differentiation result matches the integrand; thus, our solution is correct. The indefinite integral is: $$\int \frac{6}{\sqrt{4 - 4x^2}} dx = 3\arcsin{2x} + C$$

Key concepts

Substitution Method

The substitution method, often referred to as u-substitution, is a technique utilized in calculus to simplify integration problems. Essentially, it involves replacing a portion of the integrand—the function inside the integral—with a single variable (typically denoted as 'u'). This replacement makes the integral easier to evaluate.

For example, in our exercise with the integral \[ \int \frac{6}{\sqrt{4 - 4x^2}} dx \], a substitution of \( x = \frac{\sin(u)}{2} \) is used. Following this, the differential \(dx\) is expressed in terms of \(u\), as \(dx = \frac{1}{2}\cos(u) du\). This technique simplifies the original problem by converting it into a basic trigonometric integral. In practice, this strategy is pivotal for working with integrals that involve complex or composite functions.

Trigonometric Integration

Trigonometric integration refers to the process of evaluating integrals that involve trigonometric functions. In our given problem, after substitution, we are faced with a trigonometric integral that simplifies further due to the specific trigonometric identity involved.

By recognizing that \(4 - \sin^2(u) = (2\cos(u))^2\) we simplify the integrand, which initially may seem daunting, to a much more straightforward expression. The ability to identify and use such identities is crucial in evaluating trigonometric integrals. Furthermore, such integrals often appear in the analysis of oscillatory systems, such as waves or harmonic motion, making the method highly valuable in physics and engineering disciplines.

Integration by Parts

Integration by parts is another useful calculus technique for finding indefinite integrals. According to the formula \[ \int u dv = uv - \int v du \] where \(u\) and \(dv\) are differentiable functions of \(x\), it recasts the integral into another form that can ideally be integrated more easily.

While it isn't directly used in the given example, it's important to know that integration by parts is particularly helpful for integrals involving the product of algebraic expressions and transcendental functions (such as exponential, logarithmic, or trigonometric functions). Recognizing when to use substitution versus integration by parts comes with practice and understanding of different integral forms.

Anti-Differentiation

Anti-differentiation, or finding an indefinite integral, is the inverse process of differentiation. What this means is that we are looking for a function whose derivative is the given integrand. The process of anti-differentiation is pivotal in calculus because it allows us to uncover the original function from its rate of change.

In the context of our exercise, the anti-derivative of \(3 \) with respect to \(u\) was found to be \(3u + C\), where \(C\) represents the constant of integration. It's critical to understand that there are multiple functions with the same derivative, hence the addition of the constant \(C\). Anti-differentiation applies across various domains of science and engineering where it's essential to determine the quantity from its rate.

Calculus

Calculus is a branch of mathematics focused on studying rates of change and accumulation. The indefinite integrals we have been discussing are a fundamental component of integral calculus, which, alongside differential calculus, forms the two main branches of calculus.

Calculus is not just an academic exercise but is also a powerful tool used to solve real-world problems in various fields such as physics, engineering, economics, and biology. The concepts and techniques like the substitution method or anti-differentiation are essential for analyzing and modeling dynamic systems. Understanding the foundational principles of integral and differential calculus is key to success in any technical field that involves mathematical analysis.