Problem 28 - Calculus Volume 3

Question

In the following exercises, evaluate the iterated integrals by choosing the order of integration. $$ \int_{0}^{1} \int_{0}^{1 / 2}(\arcsin x+\arcsin y) d y d x $$

Step-by-Step Solution

Answer

The iterated integral evaluates to \( \frac{5\pi}{24} - \frac{1}{2}(\sqrt{3} - 1) \).

Step 1: Understand the Integral

The given integral is an iterated integral of the form \( \int_{0}^{1} \int_{0}^{1/2} (\arcsin x + \arcsin y) dy \, dx \). Here, we are integrating with respect to \( y \) first, and then with respect to \( x \).

Step 2: Split the Integral

The integral inside can be split into two separate integrals due to the distributive property of integrals: \( \int_{0}^{1} \left( \int_{0}^{1/2} \arcsin x \, dy + \int_{0}^{1/2} \arcsin y \, dy \right) dx \).

Step 3: Integrate with Respect to y for \( \arcsin x \)

For the first part, \( \int_{0}^{1/2} \arcsin x \, dy = \arcsin x \times \frac{1}{2} \), since \( \arcsin x \) is treated as a constant with respect to \( y \).

Step 4: Integrate with Respect to y for \( \arcsin y \)

For the second part, we find \( \int_{0}^{1/2} \arcsin y \, dy \) by using integration by parts. Let \( u = \arcsin y \), then \( du = \frac{1}{\sqrt{1-y^2}}dy \), and let \( dv = dy \), then \( v = y \).The integration by parts formula gives us:\[ \int u \, dv = uv - \int v \, du \]Thus, it becomes:\[ y \arcsin y \bigg|_{0}^{1/2} - \int_{0}^{1/2} \frac{y}{\sqrt{1-y^2}} \, dy \]Calculating: \[ y \arcsin y \bigg|_{0}^{1/2} = \frac{1}{2} \arcsin \frac{1}{2} - 0 = \frac{1}{2} \times \frac{\pi}{6} = \frac{\pi}{12} \]For the remaining integral, use substitution with \( y = \sin \theta \), then \( dy = \cos \theta d\theta \). This simplifies to an integral of a trigonometric identity, and finally results in \( -\sqrt{1-y^2} \bigg|_{0}^{1/2} = -\left( \sqrt{\frac{3}{4}} - 1 \right) = -\frac{1}{2}(\sqrt{3} - 1)\).

Step 5: Combine and Integrate with Respect to x

Now that we have \( \frac{1}{2}\arcsin x \) and the value of the \( y \)-integral \( \frac{\pi}{12} - \frac{1}{2}(\sqrt{3} - 1) \), the integral becomes:\[ \int_{0}^{1} \left( \frac{1}{2}\arcsin x + \frac{\pi}{12} - \frac{1}{2}(\sqrt{3} - 1) \right) \, dx \]This further simplifies into:\[ \int_{0}^{1} \frac{1}{2} \arcsin x \, dx + \left( \frac{\pi}{12} - \frac{1}{2}(\sqrt{3} - 1) \right) \int_{0}^{1} dx \]The first term involves integrating \( \arcsin x \), which gives \( x \arcsin x + \sqrt{1-x^2} \bigg|_{0}^{1} = \frac{\pi}{4} + 0 \), and the second integral is simply the constant evaluated: \( \left( \frac{\pi}{12} - \frac{1}{2}(\sqrt{3} - 1) \right) \), times 1, since integration over 0 to 1 of a constant results in the constant.

Step 6: Calculate the Final Result

Gather all the partial results: - \( \frac{\pi}{8} \) from integrating \( \frac{1}{2} \arcsin x \).- Adding the evaluated constant, the total is:\[ \frac{\pi}{8} + \frac{\pi}{12} - \frac{1}{2}(\sqrt{3} - 1) \]Converting fractions, this simplifies to \( \frac{3\pi}{24} = \frac{\pi}{8} \).Hence the integral evaluates to \( \frac{5\pi}{24} - \frac{1}{2}(\sqrt{3} - 1) \).

Key Concepts

Understanding the Arcsine Function

The arcsine function, denoted as \( \arcsin x \), is the inverse of the sine function that restricts its output range so that any angle \( \theta \) given by \( \theta = \arcsin x \) satisfies \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \).
This means when you take the sine of \( \theta \), you get back the original input \( x \) that lies between -1 and 1.

• It turns input values (like length of the opposite side over the hypotenuse in a right triangle) to angles in radians.
• The arcsine function is particularly useful in trigonometry and calculus for converting between linear and angular measurements.

When working with iterated integrals involving \( \arcsin \), as seen in our problem, it can be treated as a constant if it's not the variable of integration. This aspect simplifies the integration process significantly.Remember: Arcsine is not defined for values outside the interval [-1, 1].

Understanding the Order of Integration

The order of integration refers to the sequence in which multiple integrals are evaluated in iterated integrals. In our exercise, we initially integrate with respect to \( y \), and then with respect to \( x \).

• This order is specified by the integral limits and affects how the problem is solved, especially when splitting integrals or when dealing with functions involving both variables.
• Changing the order of integration might sometimes simplify the calculation by aligning with properties of the specific functions.
• Evaluate integrals by first ensuring that the bounds are correctly established, reflected in the given function within the iterated integral.

In this exercise, sticking to the given order of integration showcases the process of breaking down complex multivariable integrals while using recognizable identities and methods to achieve a solution.

Applying Integration by Parts

Integration by parts is a method derived from the product rule of differentiation. It is crucial in solving integrals where the standard integration methods are cumbersome or inapplicable. The formula is given by:\[\int u \, dv = uv - \int v \, du\]

• Choose \( u \) and \( dv \) wisely: \( u \) should simplify upon differentiation, and \( dv \) should be easily integrable.
• In the exercise, \( u = \arcsin y \) was chosen. This selection works well as \( du \) becomes simpler, and \( dv = dy \) easily integrates to \( v = y \).
• Integration by parts may require additional steps like substitution to simplify remaining integrals, as seen with transforming the trigonometric integral.

When applied correctly, integration by parts not only reveals the solution but also uncovers connections between functions. Remember to evaluate boundary terms carefully, as these contribute to the final result.