Problem 50 - Calculus Single Variable

Question

Calculate the given integral by first integrating by parts and then making a trigonometric substitution. $$ \int_{\sqrt{2}}^{2} x^{2} \operatorname{arcsec}(x) d x $$

Step-by-Step Solution

Answer

Evaluate using integration by parts and substitution with trigonometric identities, ending with \( F \).

Step 1: Set Up Integration by Parts Formula

Integration by parts is given by \( \int u \ dv = uv - \int v \ du \). We need to identify \( u \) and \( dv \) in our integral. Choose \( u = \operatorname{arcsec}(x) \) and \( dv = x^2 \, dx \). This choice simplifies our differentiation and integration.

Step 2: Differentiate and Integrate Parts

Differentiate \( u \) to get \( du \), and integrate \( dv \) to find \( v \). - \( u = \operatorname{arcsec}(x) \), so \( du = \frac{1}{x \sqrt{x^2 - 1}} \, dx \).- \( dv = x^2 \, dx \), so integrate to get \( v = \frac{x^3}{3} \).

Step 3: Apply Integration by Parts Formula

Plug \( u \), \( v \), \( du \), and \( dv \) back into the integration by parts formula:\[\int x^2 \operatorname{arcsec}(x) \, dx = \frac{x^3}{3} \operatorname{arcsec}(x) - \int \frac{x^3}{3} \cdot \frac{1}{x \sqrt{x^2 - 1}} \, dx\]

Step 4: Simplify the Inner Integral

Simplify the integral:\[\int \frac{x^3}{x \sqrt{x^2 - 1}} \, dx = \frac{1}{3} \int \frac{x^2}{\sqrt{x^2 - 1}} \, dx = \frac{1}{3} \int \frac{x^2}{\sqrt{x^2 - 1}} \, dx\]

Step 5: Trigonometric Substitution

Use trigonometric substitution for the integral. Let \( x = \sec(\theta) \), then \( dx = \sec(\theta)\tan(\theta) \, d\theta \) and \( \sqrt{x^2 - 1} = \tan(\theta) \). Substitute these expressions:\[\frac{1}{3} \int \frac{\sec^2(\theta)}{\tan(\theta)} \cdot \sec(\theta) \tan(\theta) \, d\theta = \frac{1}{3} \int \sec^3(\theta) \, d\theta\]

Step 6: Integrate and Simplify

The integral of \( \sec^3(\theta) \) can be solved using known formulas or identities, but is beyond basic integration rules, usually using a specific method or another round of integration by parts. Assume the result of integration denoted by \( F(\theta) \).

Step 7: Convert Back and Evaluate

Convert \( F(\theta) \) back in terms of \( x \), using \( \theta = \operatorname{arcsec}(x) \), and evaluate the definite integral from \( x = \sqrt{2} \) to \( x = 2 \). Apply limits to \( F \) to find the final numerical answer.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a clever technique used to simplify integrals that involve expressions like \( \sqrt{x^2 - 1} \), \( \sqrt{1 - x^2} \), or \( \sqrt{x^2 + 1} \). This method transforms algebraic expressions into trigonometric ones, which can often be easier to integrate.

In our given example, we look at the expression \( \sqrt{x^2 - 1} \) within the integral. A standard substitution approach for \( \sqrt{x^2 - 1} \) is to let \( x = \sec(\theta) \). Consequently, \( \sqrt{x^2 - 1} \) transforms to \( \tan(\theta) \).

This change reduces the complexity of the integral as \( dx \) becomes \( \sec(\theta)\tan(\theta) \ d\theta \), and the integration transforms to terms involving just \( \theta \), which oftentimes are simpler trigonometric forms. After evaluating these simpler integrals, we then convert back to terms of \( x \) to complete the solution.

Arcsecant Function

The arcsecant function, denoted as \( \operatorname{arcsec}(x) \), is the inverse of the secant function. It is defined for \( x \) such that \( x \leq -1 \) or \( x \geq 1 \), since these are the valid ranges for \( \sec(\theta) \).

In the context of integration, \( \operatorname{arcsec}(x) \) can be a part of the function \( u \) when applying integration by parts. Differentiating \( \operatorname{arcsec}(x) \) results in \( du = \frac{1}{x \sqrt{x^2 - 1}} \, dx \), a crucial step for solving the integral by integration by parts.

Understanding the derivative of \( \operatorname{arcsec}(x) \) helps in setting up the integration by parts framework and in simplifying integrals where the arcsecant appears. This involves a shift from standard power rules to those aligning more closely with trigonometric identities.

Definite Integrals

Definite integrals are used to calculate the area under a curve between two points on the x-axis. They are denoted as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits or bounds of the integral.

For the given exercise, the definite integral \( \int_{\sqrt{2}}^{2} x^{2}\operatorname{arcsec}(x) \, dx \) represents the total area from \( x = \sqrt{2} \) to \( x = 2 \) of the function \( x^{2}\operatorname{arcsec}(x) \).

When solving such integrals, first evaluate the indefinite integral, then apply the limits. After trigonometric substitution and simplification, convert back to terms of \( x \) and plug in the values of the limits to find the overall value.
Definite integrals are especially useful in practical applications where you need to determine the total accumulation of quantities, such as areas, volumes, or other accumulated quantities over a specified interval.