Problem 46 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find \(\int \sec ^{3} x d x\) as follows: Use integration by parts with \(u=\sec x .\) The resulting new integral will contain \(\tan ^{2} x .\) Replace \(\tan ^{2} x\) by \(\sec ^{2} x-1\) and split the integral into the difference of two integrals, \(\int \sec ^{3} x d x-\int \sec x d x .\) Integrate the latter and solve algebraically for the former.

Step-by-Step Solution

Answer

\(\int \sec^{3} x dx = \sec x \tan x - \ln |\sec x + \tan x| + C\)

Step 1: Using Integration by Parts

The formula for integration by parts is \(\int u dv = u v - \int v du\). Let's choose \(u = \sec x\) and \(dv = \sec^{2} x dx\). We then need to find \(du\) and \(v\). Differentiating \(u\) gives \(du = \sec x \tan x dx\), and integrating \(dv\) gives \(v = \tan x\). Applying the formula, we get: \(\int \sec ^3 x dx = \sec x \tan x - \int \tan x \sec x dx\).

Step 2: Simplifying the New Integral

Notice that \(\tan{x}^2 = \sec{x}^2 - 1\). So we can write the integral as follows: \(\int \tan x \sec x dx = \int (\sec ^{2} x - 1) \sec x dx = \int \sec^{3} x dx - \int \sec x dx \). This is a simple linear equation of which the solution \(\int \sec^{3} x dx\) can be easily found.

Step 3: Solving for the Original Integral

To solve for the original integral, we first integrate \(\sec x dx\) to get \(\ln |\sec x + \tan x|\). We then solve the linear equation for \(\int \sec^{3} x dx\) as follows: \(\int \sec^{3} x dx = \sec x \tan x - \ln |\sec x + \tan x| + C\), where \(C\) is the constant of integration.

Key Concepts

Integration by Parts

Integration by parts is a technique used to find the integral of products of functions. It is particularly useful when direct integration is challenging. The integration by parts formula is given by: \[\int u \, dv = uv - \int v \, du\]where:

• \( u \) is a function chosen based on simplicity when differentiated
• \( dv \) is another function that is easy to integrate

For the problem at hand, we choose \( u = \sec x \) since differentiating the secant function is straightforward. We obtain \( du = \sec x \tan x \, dx \). The function \( dv \) is \( \sec^{2} x \, dx \), so its integral is \( \tan x \), leading to \( v = \tan x \).
By plugging these into the integration by parts formula, we get:\[\int \sec^{3} x \, dx = \sec x \tan x - \int \tan x \sec x \, dx\]This step initiates the breaking down of a complex integral into simpler components.

Trigonometric Integrals

Trigonometric integrals involve integrals of trigonometric functions and often require specific manipulations. A common approach is transforming or substituting one trigonometric identity into another. In our problem, we use the identity:\[\tan^{2} x = \sec^{2} x - 1\]This identity helps simplify the integral expression \( \int \tan x \sec x \, dx \) to:\[\int (\sec^{2} x - 1) \sec x \, dx = \int \sec^{3} x \, dx - \int \sec x \, dx\]This substitution not only reconfigures the original equation but also separates it into two more manageable integrals. It shows how transforming one type of function into another can aid in solving the integral effectively.
By separating the terms, we set up an equation that enables us to isolate the integral we are trying to solve for.

Techniques of Integration

A variety of techniques can be applied to integration problems, and understanding which technique to use is vital. In the given problem, multiple techniques are utilized:

• Integration by Parts: Serves as a primary strategy to simplify complex products into manageable parts.
• Trigonometric Identities: Employ identities to rework integrals into a solvable format.
• Logarithmic Integration: Used at the final stage of resolving the integral of \( \sec x \, dx \). For this, the result is \( \ln |\sec x + \tan x| \).

To solve the linear equation:\[\int \sec^{3} x \, dx = \sec x \tan x - \ln |\sec x + \tan x| + C\]This combination shows how the strategic application of various integration techniques can allow a seemingly challenging integral to be systematically broken down and resolved, emphasizing the power of technique variety in calculus.