Problem 91 - Calculus Early Transcendentals

Question

a. Multiply the numerator and denominator of sec \(x\) by \(\sec x+\tan x ;\) then use a change of variables to show that $$\int \sec x d x=\ln |\sec x+\tan x|+C$$ b. Use a change of variables to show that $$\int \csc x d x=-\ln |\csc x+\cot x|+C$$

Step-by-Step Solution

Answer

Answer: After applying the given techniques, the integrals are found as follows: a. The integral of \(\sec(x)\) is: $$\int \sec(x)\, dx=\ln|\sec(x)+\tan(x)|+C$$ b. The integral of \(\csc(x)\) is: $$\int \csc(x)\, dx=-\ln|\csc(x) - \cot(x)|+C$$

Step 1: Part (a)

Step 1: Multiply the numerator and denominator of \(\sec(x)\)

Step 1: Multiply the numerator and denominator of \(\sec(x)\)

We start by multiplying both the numerator and denominator of \(\sec(x)\) by \((\sec(x)+\tan(x))\): $$\frac{\sec(x)}{1} \cdot \frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)} = \frac{\sec^2(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}$$ Now we have to find the integral of this expression: $$\int \frac{\sec^2(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}\, dx$$ Step 2: Change of Variables

Step 2: Change of Variables

Let us substitute \(u = \sec(x) + \tan(x)\). Then, we can differentiate \(u\) with respect to \(x\) to find \(du\): $$\frac{d}{dx}u = \frac{d}{dx}(\sec(x)+\tan(x))$$ $$du = (\sec^2(x)+\sec(x)\tan(x))\, dx$$ Since our integrand has the term \(\sec^2(x) + \sec(x)\tan(x)\) in the numerator, which is equal to \(du\), the integral will be in terms of \(u\) as follows: $$\int \frac{du}{u}$$ Step 3: Calculation of Integral

Step 3: Calculation of Integral

The integral of \(\frac{1}{u}\) is a known basic integral and it evaluates to: $$\int \frac{du}{u}=\ln|u|+C$$ Now we substitute back \(u = \sec(x) + \tan(x)\): $$\int \sec(x)\, dx=\ln|\sec(x)+\tan(x)|+C$$

Step 5: Part (b)

Step 1: Change of Variables

Step 1: Change of Variables

To find the integral of \(\csc(x)\), we can directly perform a change of variables. Let's substitute \(v = \csc(x) - \cot(x)\). Then, we can differentiate \(v\) with respect to \(x\) to find \(dv\): $$\frac{d}{dx}v = \frac{d}{dx}(\csc(x)-\cot(x))$$ $$dv = (-\csc(x) \cot(x) - \csc^2(x))\, dx$$ Notice that our desired integral involves \(\csc(x)\) and its derivative can be factored from \(dv\) as: $$-\csc(x)(\cot(x) + \csc(x))\, dx=dv$$ We can rewrite the integral as: $$\int \csc(x)\, dx =\int -\frac{dv}{v}$$ Step 2: Calculation of Integral

Step 2: Calculation of Integral

The integral of \(\frac{1}{v}\) is a known basic integral and it evaluates to: $$\int -\frac{dv}{v}=-\ln|v|+C$$ Now we substitute back \(v = \csc(x) - \cot(x)\): $$\int \csc(x)\, dx=-\ln|\csc(x)+(-\cot(x))|+C=-\ln|\csc(x) - \cot(x)|+C$$

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique to solve integrals, especially when the integrand becomes simpler after a substitution. Think of it as changing the variable to make the integration process easier.

Here's how it generally works: You choose a substitution, say let \( u = g(x) \), where \( g(x) \) is a function of \( x \). Once you have \( u \), you then compute \( du \), which represents the differential change in \( u \) with respect to \( x \). The equation becomes, \( du = g'(x) dx \).

This substitution allows us to rewrite the integral in terms of \( u \), often leading to a simpler form that can be easily integrated. After finding the integral in terms of \( u \), one substitutes back the original variable \( x \). This method is particularly useful when dealing with products of functions or composite functions. In the exercise provided, the substitution \( u = \sec{x} + \tan{x} \) simplifies the integration of \( \sec{x} \). It effectively reduces the integral to the basic form \( \int \frac{du}{u} \), which equals \( \ln|u| + C \) after integration.

Trigonometric Integrals

Trigonometric integrals involve functions such as sine, cosine, tangent, secant, cosecant, and cotangent. These integrals often require specific techniques or identities to simplify and solve them.

When integrating trigonometric functions, a common strategy is to manipulate the given expression using trigonometric identities. For example, in the case of integrating \( \sec{x} \), we use an identity by multiplying the expression by \( \sec{x} + \tan{x} \). This manipulation helps in forming a derivative inside the integral, making it easier to solve using substitution.

Different scenarios might require the use of different trigonometric identities or substitutions. Knowing your trigonometric identities, such as \( \sin^2{x} + \cos^2{x} = 1 \), is crucial. When integrating \( \csc{x} \), the substitution \( v = \csc{x} - \cot{x} \) helps transform the integral into a basic logarithmic form, similar to the integral of \( \sec{x} \). These methods reveal how trigonometric and substitution techniques work hand-in-hand for solving integrals.

Logarithmic Functions

Logarithmic functions often appear in calculus, particularly as solutions to certain integrals. A basic integral involving the natural logarithm is \( \int \frac{1}{x} \, dx = \ln|x| + C \).

In the context of integrating trigonometric functions, logarithmic expressions commonly emerge after applying substitution methods. In the exercise, you use substitution to convert the trigonometric expressions into a form involving the natural logarithm. For \( \int \sec{x} \, dx \), you reach \( \int \frac{du}{u} \) after substitution, giving \( \ln|u| + C \) as the result, where \( u = \sec{x} + \tan{x} \).

Similarly, to evaluate \( \int \csc{x} \, dx \), substitution yields an integral involving the negative of a logarithmic function, \( -\ln|v| + C \), with \( v = \csc{x} - \cot{x} \). Logarithmic solutions like these are frequently the result of integrals with specific forms or set-ups, especially after simplifying expressions containing derivatives of inverse functions or rational functions.