Problem 15 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int x \sec ^{2} x d x $$

Step-by-Step Solution

Answer

The integral \( \int x \sec^2 x \, dx \) evaluates to \( x \tan x + \ln |\cos x| + C \).

Step 1: Identify Functions for Integration by Parts

Integration by parts is based on the formula \( \int u \, dv = uv - \int v \, du \). To apply it, choose \( u \) and \( dv \). Let \( u = x \), resulting in \( du = dx \). Let \( dv = \sec^2 x \, dx \), leading to \( v = \tan x \) since the integral of \( \sec^2 x \) is \( \tan x \).

Step 2: Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx \]. We now need to evaluate \( \int \tan x \, dx \).

Step 3: Integrate \( \int \tan x \, dx \)

Recall that \( \tan x = \frac{\sin x}{\cos x} \). To integrate \( \int \tan x \, dx \), note that \( \int \tan x \, dx = -\ln |\cos x| + C \), using a substitution \( u=\cos x \) where \( du = -\sin x \, dx \).

Step 4: Substitute Back and Simplify

Substituting back gives \( x \tan x - (-\ln |\cos x|) \), which simplifies to \[ x \tan x + \ln |\cos x| + C \]. The integration constant \( C \) is added to indicate the family of antiderivatives.

Key Concepts

Integrals

Integrals play a critical role in calculus by helping us find the area under curves, among other applications. They are essentially the inverse operation of derivatives.
In this exercise, we are using integration by parts to solve an integral involving the function \( x\sec^2 x \). The integration by parts is derived from the product rule in differentiation. When two functions are multiplied, like in our exercise, the formula for integration by parts comes into play:
\[\int u \, dv = uv - \int v \, du\]

Integration by parts requires choosing a part of the function to differentiate \((u)\) and part to integrate \((dv)\). The goal is to simplify the integral so that it becomes easier to solve. Choosing the right \(u\) and \(dv\) is key to making the process efficient. Once we apply integration by parts, as shown in the solution, it reduces the complexity of the provided integral.

Trigonometric Functions

Trigonometric functions like \(\sec^2 x\) and \(\tan x\) make calculus problems more challenging due to their complex integrals. Understanding how these functions work is essential to solving integrals involving them.
One of the basic rules is that the derivative of \(\tan x\) is \(\sec^2 x\), which is useful in solving our given problem.
We set \(dv = \sec^2 x\, dx\) knowing that \(v\) then becomes \(\tan x\). Upon applying the integration by parts formula, we arrive at a simpler integral involving \(\tan x\).

Integrating \(\tan x\) involves another trick: rewriting it as \(\frac{\sin x}{\cos x}\) and using a substitution method. This transformation allows the integral to be solved using the natural logarithm function, resulting in \(-\ln |\cos x|\). Trigonometric identities and understanding how to manipulate these functions are crucial in calculus.

Calculus Problems

Calculus problems often test your ability to apply multiple concepts to find solutions. The problem \(\int x \sec^2 x \, dx\) is a great example of integrating trigonometric functions using integration by parts.
It requires several steps and decisions:

• Choosing the functions \(u\) and \(dv\) correctly
• Applying the integration by parts formula
• Performing substitutions
• Recognizing simplifications to solve the integral

Each step builds on the previous, demanding careful attention to detail. Understanding the mechanics of integration by parts and the properties of trigonometric functions is essential. This ensures not only solving particular integrals but also building a foundation for tackling even more complex calculus challenges. The process involves both analytical thinking and creative problem-solving, typical of many calculus problems.