Question

Verify each of the following answers for an indefinite integral by one or more of the methods suggested above. \(\int \frac{K d r}{\sqrt{1-K^{2} r^{2}}}=\sin ^{-1} K r \quad\) or \(\quad-\cos ^{-1} K r \quad\) or \(\quad \tan ^{-1} \frac{K r}{\sqrt{1-K^{2} r^{2}}}.\) Hints: Sketch a right triangle with acute angles \(u\) and \(v\) and label the sides so that \(\sin u=K r .\) Also note that \(u+v=\pi / 2 ;\) then if \(u\) is an indefinite integral, so is \(-v\) since they differ by a constant of integration.

Short answer

\(\sin^{-1} Kr\) is verified by substitution. \(-\cos^{-1} Kr\) and \(\tan^{-1} \frac{Kr}{\sqrt{1-K^2 r^2}}\) solutions are verified using trigonometric identities.

Step-by-step solution

Identify the integral and proposed solutions

Identify the integral given in the problem: \[\int \frac{K \, dr}{\sqrt{1-K^2 r^2}}\]and the proposed solutions: \[\sin^{-1} K r, \; -\cos^{-1} K r, \; \tan^{-1} \frac{K r}{\sqrt{1-K^2 r^2}}.\]

Use trigonometric substitution

Let \(\theta = \sin^{-1}(Kr)\), which implies \(Kr = \sin(\theta)\). Use the identity \(1 - \sin^2(\theta) = \cos^2(\theta)\) to rewrite the integral in terms of \(\theta\). Thus, \(\sqrt{1-K^2 r^2} = \cos(\theta)\).

Rewrite the integral using substitution

Substitute \(r = \frac{\sin(\theta)}{K}\) and \(dr = \frac{\cos(\theta)}{K} \, d\theta\) in the integral: \[\int \frac{K \cdot \frac{\cos(\theta)}{K} \, d\theta}{\cos(\theta)} = \int d\theta = \theta = \sin^{-1}(Kr).\]

Verify the other forms

For the given solutions, note that \(u + v = \frac{\pi}{2}\). If \(u = \sin^{-1}(Kr)\), then \(v = \cos^{-1}(Kr)\), so: \[\sin^{-1}(Kr) + \cos^{-1}(Kr) = \frac{\pi}{2} \Rightarrow \cos^{-1}(Kr) = \frac{\pi}{2} - \sin^{-1}(Kr)\] and thus, \(\int \frac{K \, dr}{\sqrt{1-K^2 r^2}} = -\cos^{-1}(Kr)\) is valid up to a constant.Also, using the identity \(\tan^{-1}x = \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\), we get \(\tan^{-1}\left(\frac{Kr}{\sqrt{1-K^2 r^2}} \right) = \sin^{-1}Kr\).

Key concepts

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals using trigonometric identities. It is especially useful when dealing with roots of quadratic expressions.

For the integral \(\frac{K dr}{\frac{1-K^{2} r^{2}}}\), we let \(\theta = \frac{K r}\). This makes \(\text{sin}\theta=K r\), which simplifies finding the integral. With trigonometric substitution, we often convert roots and other complex parts of the integrand into trigonometric functions.

In this problem, the substitution of \(\text{sin}\theta = Kr\) transforms our integral. We use the identity \(1-\text{sin}^2(\theta) = \text{cos}^2(\theta)\) to convert all expressions into trigonometric forms. This lets us simplify the integral to an easier form. This substitution helps reduce the complexity and makes solving integrals more manageable.

Substituting back once the integral is solved is necessary to revert to the original variable.

Inverse Trigonometric Functions

Inverse trigonometric functions reverse the roles of the angles and the trigonometric ratios. In this context, they help verify our integral results.

For example, \(\text{sin}^{-1}x\) gives us the angle whose sine is \(x\). For the integral solution, \(\theta = \frac{K r}\), recognizing \(\text{sin}^{-1}K r\) as an angle is key.

In this solution, we verify \(\text{sin}^{-1}K r \) by integrating through trigonometric substitution. We also check if other forms like \(-\text{cos}^{-1}K r\) and \(\tan^{-1}\frac{K r}{\frac{1-K^{2}} \}\) equal \(\text{sin}^{-1}K r \) by using known trigonometric identities.

Understanding the relationships among these inverse functions helps confirm that the integrals are consistent and correct even if they look different initially. Each form provides the same integral result but represents different inverse trigonometric relationships.

Constant of Integration

When dealing with indefinite integrals, we always include the constant of integration, denoted as \(+C\). This constant represents an infinite family of functions.

Indefinite integrals differ by a constant since the derivative of a constant is zero. In the given problem, the solutions \(\text{sin}^{-1}K r\),\(-\text{cos}^{-1}K r\), and \(\text{tan}^{-1}\frac{K r}{\frac{1-K^{2}} \}\) really differ only by a constant.

For example, \(\text{sin}^{-1}K r + \text{cos}^{-1}K r = \frac{\frac{1+} {2}}\). This confirms that the functions are equivalent except for a constant shift.

Always adding \(+C\) ensures our integral reflects all possible solutions. Without it, we might miss out on the full range of potential solutions represented by the integral.
Understanding the concept of the constant of integration is crucial for accurately solving indefinite integrals and accounting for all possible outcomes.