Problem 36 - Calculus Single Variable

Question

Integrate by parts to evaluate the given definite integral. $$ \int_{\sqrt{2}}^{2} 2 x \operatorname{arcsec}(x) d x $$

Step-by-Step Solution

Answer

The integral evaluates to approximately \( \frac{4\pi}{3} - \frac{\pi\sqrt{2}}{4} \).

Step 1: Identify Parts for Integration by Parts

In integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). Here, we choose \( u = \operatorname{arcsec}(x) \) and \( dv = 2x \, dx \). This means \( du = \frac{1}{x \sqrt{x^2 - 1}} \, dx \) and integrating \( dv \), we get \( v = x^2 \).

Step 2: Apply Integration by Parts Formula

Substitute into the integration by parts formula: \[ \int 2x \operatorname{arcsec}(x) \, dx = x^2 \operatorname{arcsec}(x) - \int x^2 \frac{1}{x \sqrt{x^2 - 1}} \, dx. \] Simplify the remaining integral to \( \int \frac{x}{\sqrt{x^2 - 1}} \, dx \).

Step 3: Reevaluate Remaining Integral

Use substitution to solve \( \int \frac{x}{\sqrt{x^2 - 1}} \, dx \). Let \( w = x^2 - 1 \), then \( dw = 2x \, dx \). The integral becomes \( \int \frac{1}{2}\cdot\frac{1}{\sqrt{w}} \, dw \), which simplifies to \( \frac{1}{2} \int w^{-1/2} \, dw \).

Step 4: Compute Integral After Substitution

Integrate \( \frac{1}{2} \int w^{-1/2} \, dw \) to obtain \( \frac{1}{2} \cdot 2w^{1/2} = \sqrt{w} \). Substitute back, \( w = x^2 - 1 \), giving the result \( \sqrt{x^2 - 1} \).

Step 5: Apply Limits to Definite Integral

Evaluate \( x^2 \operatorname{arcsec}(x) - \sqrt{x^2 - 1} \) from \( \sqrt{2} \) to \( 2 \). Calculate each term separately: at \( x = 2 \), the expression becomes \( 4 \cdot \pi/3 - 1 \) and at \( x = \sqrt{2} \), it becomes \( 2 \cdot \pi/4 - 1 \).

Step 6: Subtract for Final Value

Subtract the lower limit expression from the upper limit: \( \left( 4 \frac{\pi}{3} - 1 \right) - \left( \sqrt{2} \frac{\pi}{4} - 1 \right) \), simplifying to obtain the final value.

Key Concepts

Definite Integral

A definite integral is a way to find the area under a curve within a specific interval. In the context of this exercise, the definite integral evaluates the area under the curve of \( f(x) = 2x \operatorname{arcsec}(x) \) from \( x = \sqrt{2} \) to \( x = 2 \).
This process involves two major steps:

• Finding the antiderivative of the function, which is the integral without limits.
• Subtracting the value of the antiderivative at the lower limit from its value at the upper limit.

This allows us to calculate the total accumulated value of the function over the given interval, giving us a single numerical result rather than a function.

Substitution Method

The substitution method is a technique used to simplify complex integrals, especially when parts of the function can be replaced with a single variable. For the integral in the exercise, after using integration by parts, we're left with a simpler integral:

• \( \int \frac{x}{\sqrt{x^2 - 1}} \, dx \)

To tackle this, we chose a substitution to simplify the expression. Letting \( w = x^2 - 1 \) meant the derivative \( dw = 2x \, dx \), which simplified the integral to a form of:

• \( \frac{1}{2} \int w^{-1/2} \, dw \)

This substitution transforms a complicated integral into a simpler one, leading to an easier calculation and thus, a straightforward path to the solution.

Arcsec Function

The arcsec function, denoted \( \operatorname{arcsec}(x) \), is the inverse of the secant trigonometric function. It's important in calculus because it allows us to handle integrals involving inverse trigonometric functions. Arcsec is defined for \( |x| \geq 1 \), because for the secant function, values occur for these domains.
When finding its integral, the arcsec function was chosen as \( u \) in the integration by parts method thus making it easier to differentiate. Knowing that the arcsec involves trigonometric identities helps in simplifying integrals that seem complex at first sight.

Derivative of Arcsec

The derivative of the arcsec function is key when differentiating \( u \) in our integration by parts process. For a function \( u = \operatorname{arcsec}(x) \), its derivative \( du \) is calculated as follows:

• \( du = \frac{1}{x \sqrt{x^2 - 1}} \, dx \)

This expression results from the fact that, for arcsec, the output angle's cosine is the reciprocal of \( x \).
Understanding this derivative allows us to actually manipulate the terms in the integral properly and are pivotal in executing the integration by parts correctly. This relationship reveals how essential it is to have a grasp on inverse trigonometric derivatives for evaluating integrals.