Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int_{\pi / 6}^{\pi / 2} x \csc ^{2} x d x $$

Step-by-Step Solution

Answer

The evaluated integral is \( \frac{\pi\sqrt{3}}{6} + \ln 2 \).

Step 1: Identify Parts for Integration by Parts

We use the integration by parts formula, which is \( \int u \, dv = uv - \int v \, du \). For the given integral \( \int x \csc^2(x) \, dx \), choose \( u = x \) and \( dv = \csc^2(x) \, dx \).

Step 2: Differentiate and Integrate to Find du and v

Differentiate \( u \) to find \( du \): \( du = dx \). Integrate \( dv \) to find \( v \): \( v = -\cot(x) \), because \( \int \csc^2(x) \, dx = -\cot(x) \).

Step 3: Apply Integration by Parts Formula

Substitute \( u = x \), \( dv = \csc^2(x) \, dx \), \( v = -\cot(x) \), and \( du = dx \) into the integration by parts formula: \( \int x \csc^2(x) \, dx = -x \cot(x) - \int (-\cot(x)) \, dx \).

Step 4: Simplify the Expression

The expression becomes \( -x \cot(x) + \int \cot(x) \, dx \). The integral of \( \cot(x) \) is \( \ln |\sin(x)| \), so you have: \( -x \cot(x) + \ln |\sin(x)| + C \).

Step 5: Evaluate the Definite Integral

Evaluate the expression \( -x \cot(x) + \ln |\sin(x)| \) from \( \pi/6 \) to \( \pi/2 \). Substitute \( x = \pi/2 \) and \( x = \pi/6 \) into the antiderivative and find the difference.At \( x = \pi/2 \), \(-\left(\frac{\pi}{2}\right)\cot\left(\frac{\pi}{2}\right) + \ln |\sin\left(\frac{\pi}{2}\right)| = 0 + \ln(1) = 0 \).At \( x = \pi/6 \), \( -\left(\frac{\pi}{6}\right)\cot\left(\frac{\pi}{6}\right) + \ln |\sin\left(\frac{\pi}{6}\right)| = -\left(\frac{\pi}{6}\right)\left(\sqrt{3}\right) + \ln\left(\frac{1}{2}\right) = -\frac{\pi\sqrt{3}}{6} + \ln\left(\frac{1}{2}\right) \).

Step 6: Calculate the Result

Combine the results from Step 5: \( 0 - \left(-\frac{\pi\sqrt{3}}{6} - \ln 2\right) = \frac{\pi\sqrt{3}}{6} + \ln 2 \).

Key Concepts

Definite Integral

A definite integral represents the area under the curve of a function between two specific points. It is computed using limits and can have several applications in physics, engineering, and mathematics. In this specific problem, we evaluate the integral from \( \pi/6 \) to \( \pi/2 \). The limits tell us where we start and stop calculating.

The evaluation involves substituting these limits into the antiderivative function that results from integrating. You calculate the antiderivative at the upper limit and subtract the antiderivative at the lower limit. This gives the net area under the curve of the function between those two points.

Evaluating a definite integral effectively turns the indefinite integral, which includes the constant of integration \( C \), into a specific numerical answer without needing \( C \).

Trigonometric Integrals

Trigonometric integrals involve the integration of trigonometric functions. They frequently appear in calculus and require specific techniques for simplification.

In this exercise, the trigonometric function \( \csc^2(x) \) is involved, making it necessary to recall that its integral is \( -\cot(x) \). Knowing these basic integrals can significantly simplify the process.

Trigonometric identities can also play a crucial role. For example, rearranging these identities can sometimes help to simplify the integrand or make integration by parts more manageable. These integrals often test your familiarity with both fundamental trigonometric functions and their relationships.

Integration Techniques

Integration techniques, such as integration by parts, are strategies employed to solve more complex integrals that cannot be integrated directly. The technique relies on the formula \( \int u \, dv = uv - \int v \, du \). The choice of \( u \) and \( dv \) can significantly affect the simplicity of solving the integral.

In this step-by-step solution, we set \( u = x \) and \( dv = \csc^2(x) \, dx \). By differentiating \( u \) and integrating \( dv \), we arrived at \( du = dx \) and \( v = -\cot(x) \).

Substituting these into the integration by parts formula allowed us to break the original complex integral into a simpler expression that was easier to solve. The choice of parts was crucial, as picking differently might have led to more complicated integrals later in the process.