Problem 19 - Thomas Calculus

Question

Evaluate the integrals. \(\int_{2 / \sqrt{3}}^{2} t \sec ^{-1} t d t\)

Step-by-Step Solution

Answer

The integral evaluates to \(2\sec^{-1}(2) - \frac{2}{3}\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) - \sqrt{3} + \frac{2\sqrt{3}}{3}\).

Step 1: Identify the type of integration

This problem requires us to evaluate the definite integral of the function \(t \sec^{-1}(t)\) with respect to \(t\). The function involves an inverse trigonometric function which hints at integration techniques involving integration by parts.

Step 2: Set up integration by parts

Use integration by parts, where \(\int u \, dv = uv - \int v \, du\). Choose \(u = \sec^{-1}(t)\) and thus \(du = \frac{1}{t \sqrt{t^2 - 1}} \, dt\). Let \(dv = t \, dt\) so that \(v = \frac{t^2}{2}\).

Step 3: Apply integration by parts formula

Apply the formula: \(uv - \int v \frac{1}{t \sqrt{t^2 - 1}} \, dt\). Substitute \(u\) and \(v\) to get: \[\frac{t^2}{2} \sec^{-1}(t) - \int \frac{t^2}{2} \cdot \frac{1}{t \sqrt{t^2 - 1}} \, dt\].

Step 4: Simplify the integral

The integral \(\int \frac{t}{2 \sqrt{t^2 - 1}} \, dt\) remains. This simplifies to \(\frac{1}{2}\int \frac{t}{\sqrt{t^2 - 1}} \, dt\), solvable by substitution: \(w = t^2 - 1\), \(dw = 2t \, dt\).

Step 5: Substitute and solve

Substitute \(w = t^2 - 1\) so \(dt = \frac{dw}{2t}\). The integral becomes \(\frac{1}{4} \int \frac{dw}{\sqrt{w}}\), solving as \(\frac{1}{4}(2\sqrt{w}) = \frac{1}{2}\sqrt{t^2 - 1}\).

Step 6: Evaluate the expression at bounds

Re-write the original expression: \[\frac{t^2}{2} \sec^{-1}(t) - \frac{1}{2}\sqrt{t^2 - 1}\]. Evaluate this from \(t = \frac{2}{\sqrt{3}}\) to \(t = 2\).

Step 7: Calculate the definite integral

Plug in the upper and lower limits: For \(t=2\), calculate \(2 \cdot \sec^{-1}(2) - \sqrt{3}\). For \(t=\frac{2}{\sqrt{3}}\), evaluate \(\frac{2}{3} \cdot \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) - \sqrt{\left(\frac{2}{\sqrt{3}}\right)^2 - 1}\).

Step 8: Final calculation and result

Use the antiderivative results: \[ \left[2 \cdot \sec^{-1}(2) - \sqrt{3} \right] - \left[ \frac{2}{3} \cdot \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) - \frac{2\sqrt{3}}{3} \right] \]. Calculate these terms for the final result.

Key Concepts

Definite Integral

The concept of a definite integral is fundamental in calculus as it allows us to calculate the area under a curve between two specific points. In this exercise, we're dealing with the definite integral of the function \( t \sec^{-1}(t) \) from \( \frac{2}{\sqrt{3}} \) to \( 2 \). This means we need to find the area under the curve of this function within these bounds.

Definite integrals have specific boundaries, known as the upper and lower limits. These boundaries set where we start and stop calculating the area. Unlike an indefinite integral, which represents a family of functions, a definite integral computes a specific numerical value.

• The solution involves evaluating the antiderivative at both limits.
• The value of the definite integral is the difference between the antiderivative at the upper limit and at the lower limit.

This concept is key in applications ranging from physics to engineering, where understanding areas or accumulations is vital.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find angles based on trigonometric ratios. In this exercise, we deal with \( \sec^{-1}(t) \), the inverse of the secant function, which is crucial in trigonometry.

When you encounter \( \sec^{-1}(t) \), it tells you the angle whose secant is \( t \). In integration, inverse trigonometric functions often require special techniques such as integration by parts, which we use here to handle the complexity they introduce.

• Inverse trigonometric functions can introduce limits that relate directly to geometric interpretations, like angles and lengths.
• Due to their unique properties, these functions have specific derivatives and integrals, which are useful for solving problems involving angles.

Understanding how to manipulate and integrate these functions is key to solving integrals that involve them.

Substitution Method

The substitution method is a powerful technique in integration that simplifies a complex integral into a more manageable form. It is akin to reverse chain rule or u-substitution, where we change variable names to untangle a complex function.

For example, in the integral \( \int \frac{t}{2 \sqrt{t^2 - 1}} \, dt \), we used the substitution \( w = t^2 - 1 \). This simplifies the integral process and makes it solvable. The steps include:

• Identifying a part of the integral to replace with a single variable \( w \).
• Rewriting the differential \( dt \) in terms of \( dw \).
• Solving the integral in terms of \( w \) and then substituting back to get the original variable.

This technique is especially useful for integrals that contain a nested or composite function, making the problem straightforward and manageable. It helps us see through the complex expressions by shifting perspective temporarily.