Question

Evaluate the following integrals. Assume a and b are real numbers and \(n\) is a positive integer. \(\int x^{n} \sin ^{-1} x d x\) (Hint. integration by parts.)$

Short answer

Question: Find the integral of \(x^n \sin^{-1}(x)\) with respect to \(x\) using integration by parts. Answer: \(\int x^n \sin^{-1}(x) \, dx = \frac{x^{n+1}}{n+1} \sin^{-1}(x) - \int \frac{x^{n+1}}{(n+1)\sqrt{1-x^2}} \, dx\), where further integration depends on the specific value(s) of \(n\).

Step-by-step solution

Identify u and dv

Let's choose \(u\) and \(dv\) as follows: \(u = \sin^{-1}(x)\), and \(dv = x^n \, dx\)

Compute du and v

Now, we need to find \(du\) and \(v\). 1. Differentiate \(u\) with respect to \(x\) to get \(du\): \(du = \frac{1}{\sqrt{1-x^2}} \, dx\) 2. Integrate \(dv\) with respect to \(x\) to get \(v\): \(v = \int x^n \, dx = \frac{x^{n+1}}{n+1}\)

Apply integration by parts formula

Now let's apply the integration by parts formula to the given integral: \(\int x^n \sin^{-1}(x) \, dx = \int u \, dv = uv - \int v \, du\) Substituting the values of \(u\), \(v\), and \(du\), we get: \(\int x^n \sin^{-1}(x) \, dx = \frac{x^{n+1}}{n+1} \sin^{-1}(x) - \int \frac{x^{n+1}}{n+1} \frac{1}{\sqrt{1-x^2}} \, dx\) This problem can be solved further in general depending on the integer value of \(n\). The integral of \(\frac{x^{n+1}}{\sqrt{1-x^2}} \, dx\) may be solvable using additional integration techniques or may require a reduction formula, but it will depend on the specific value(s) of \(n\).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions play a vital role in calculus. They help us find angles when the ratios of lengths within a right-angled triangle are known. In our exercise, we deal with the inverse sine function, denoted as \(\sin^{-1}(x)\). This function is significant because it serves as a bridge between algebraic expressions and trigonometric concepts. It returns the angle whose sine is the given number \(x\), and its range is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). When working with inverse trigonometric functions in calculus, it's important to remember their derivatives. The derivative of \(\sin^{-1}(x)\) is \(\frac{1}{\sqrt{1-x^2}}\). This derivative is critical when applying techniques like integration by parts, as it helps us simplify and solve the problem at hand. Inverse trigonometric functions can also occur in a variety of integration scenarios, often requiring specific methods to integrate them accurately, such as substitution or integration by parts.

Definite Integrals

Definite integrals are a central concept in calculus, providing a powerful tool for calculating the area under a curve. They offer a way to accumulate quantities that can be represented by a function, such as the distance traveled over time or the total accumulation of a substance. This ability to encapsulate continuous sums makes definite integrals highly valuable in various fields, including physics, engineering, and economics.In our example, although the problem is not explicitly about definite integrals, the techniques applied are essential for evaluating both definite and indefinite integrals. In practice, a definite integral has the form \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the limits of integration. It represents the net area between the function \(f(x)\) and the x-axis, from \(x = a\) to \(x = b\). While working with definite integrals, one must consider properties such as the sum rule and the constant multiple rule, which simplify the integration process. Additionally, fundamental theorems of calculus link the concept of differentiability with integrability, proving indispensable in solving calculus problems.

Calculus Problem Solving

Effective problem solving in calculus involves blending creativity and strategic thinking with various integration and differentiation techniques. To solve the problem \(\int x^n \sin^{-1}(x) \, dx\), we employ integration by parts. This technique is especially useful when dealing with products of functions that aren't easily integrable on their own. By carefully choosing parts for \(u\) and \(dv\), we turn a complex integral into simpler parts. The integration by parts formula, \(\int u \, dv = uv - \int v \, du\), is the cornerstone of our solution. Prior to using this technique, identifying the correct \(u\) and \(dv\) based on ease of differentiation and integration can facilitate the process. In our problem, \(u\) was chosen as \(\sin^{-1}(x)\) because it simplifies the derivative. Moreover, it's imperative to recognize when additional techniques or refinements are necessary, as calculus problems can demand a layered approach. Subsequent substitutions, reductions, or numerical methods may be needed to fully solve or approximate the remaining integral, depending on the specific problem constraints.