Problem 75 - Calculus Early Transcendentals

Question

Another form of \(\int \sec x d x\) a. Verify the identity \(\sec x=\frac{\cos x}{1-\sin ^{2} x}\) b. Use the identity in part (a) to verify that \(\int \sec x d x=\frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\)

Step-by-Step Solution

Answer

Q: Verify the identity: \(\sec x = \frac{\cos x}{1-\sin^2 x}\). A: We can verify the identity by rewriting the left-hand side of the equation in terms of cosine and using the Pythagorean identity. Start by expressing the secant function in terms of cosine: \(\sec x = \frac{1}{\cos x}\). Now, to make this expression match the right-hand side, multiply by \(\frac{1-\sin^2 x}{1-\sin^2 x}\): \(\frac{1}{\cos x} \cdot \frac{1-\sin^2 x}{1-\sin^2 x} = \frac{1-\sin^2 x}{\cos x(1-\sin^2 x)}\). Recall the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), which can be rewritten as \(1-\sin^2 x = \cos^2 x\). Substitute this into the expression: \(\frac{1-\sin^2 x}{\cos x(1-\sin^2 x)} = \frac{\cos^2 x}{\cos x\cdot \cos^2 x} = \frac{\cos x}{1-\sin^2 x}\). Thus, we have verified that \(\sec x = \frac{\cos x}{1-\sin^2 x}\). Q: Evaluate the integral: \(\int \sec x\ dx\). A: Use the verified identity to evaluate the integral: \(\int \sec x dx = \int \frac{\cos x}{1 - \sin^2 x} dx\) Now make a substitution: let \(u = \sin x\) and \(du = \cos x dx\), giving us: \(\int \frac{\cos x}{1 - \sin^2 x} dx = \int \frac{du}{1-u^2}\). Using partial fraction decomposition, rewrite the integral: \(\int \frac{du}{1-u^2} = \int \left(\frac{1}{2}\frac{1}{1-u}+\frac{1}{2}\frac{1}{1+u}\right)du\). Then integrate: \(\int \frac{du}{1-u^2} = \frac{1}{2}\ln|1-u|+\frac{1}{2}\ln|1+u|+C\). Finally, substitute back the original variable \(u = \sin x\): \(\int \sec x dx = \frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\).

Step 1: Verify the identity \(\sec x = \frac{\cos x}{1-\sin^2 x}\)

To verify this identity, we'll first rewrite \(\sec x\) in terms of \(\cos x\). Recall that \(\sec x=\frac{1}{\cos x}\). Now, let's verify the given identity: $$ \frac{1}{\cos x}=?\frac{\cos x}{1-\sin^2 x} $$ To make the left side similar to the right side, we need to have the same denominator. We can do that by multiplying the left side by \(\frac{1-\sin^2 x}{1-\sin^2 x}\): $$ \frac{1}{\cos x} \cdot \frac{1-\sin^2 x}{1-\sin^2 x} = \frac{1-\sin^2 x}{\cos x(1-\sin^2 x)} $$ Recall the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). We can rewrite it as \(1-\sin^2 x = \cos^2 x\). So, $$ \frac{1-\sin^2 x}{\cos x(1-\sin^2 x)}=\frac{\cos^2 x}{\cos x\cdot \cos^2 x}=\frac{\cos x}{1-\sin^2 x} $$ Therefore, we have verified the identity \(\sec x = \frac{\cos x}{1-\sin^2 x}\).

Step 2: Use the identity to find \(\int \sec x\ dx\)

Now that we have verified the identity, let's use it to find the integral of \(\sec x\). We have $$ \int \sec x dx = \int \frac{\cos x}{1 - \sin^2 x} dx $$ Make a substitution: \(u = \sin x, du = \cos x dx\). Now we have $$ \int \frac{\cos x}{1 - \sin^2 x} dx = \int \frac{du}{1-u^2} $$ This is a standard integral, which we can solve using partial fraction decomposition. Write the integral in the form of partial fractions: $$ \frac{1}{1 - u^2} = \frac{A}{1 + u} + \frac{B}{1 - u} $$ Solve for A and B: $$ A(1 - u) + B(1 + u) = 1 $$ Let \(u = -1\): \(2A = 1 \implies A = \frac{1}{2}\). Let \(u = 1\): \(2B = 1 \implies B = \frac{1}{2}\). Now rewrite the integral: $$ \int \frac{du}{1-u^2}=\int \left(\frac{1}{2}\frac{1}{1-u}+\frac{1}{2}\frac{1}{1+u}\right)du $$ Integrate: $$ \int \frac{du}{1-u^2} = \frac{1}{2}\ln|1-u|+\frac{1}{2}\ln|1+u|+C $$ Now we need to go back to the original variable, x. Recall that \(u=\sin x\): $$ \frac{1}{2}\ln|1-u|+\frac{1}{2}\ln|1+u|+C = \frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C $$ Thus, we have found that $$ \int \sec x dx = \frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C $$

Key Concepts

Trigonometric Identities

Trigonometric identities are essential mathematical tools that simplify complex expressions and facilitate integration. They relate the six fundamental trigonometric functions: sine, cosine, tangent, cotangent, secant, and cosecant, often transforming one function into another. In our exercise, we dealt with

• The Secant function, defined as: \( \sec x = \frac{1}{\cos x} \)
• And the Pythagorean identity: \( \sin^2 x + \cos^2 x = 1 \)

This identity can be reorganized to express either the sine or cosine in terms of each other. Specifically, \( 1 - \sin^2 x = \cos^2 x \) is crucial for transforming the secant identity into another form, as proved by verifying \( \sec x = \frac{\cos x}{1-\sin^2 x} \). By mastering these trigonometric identities, it becomes easier to manipulate and solve integrals involving trigonometric functions.

Integration by Substitution

Integration by Substitution is a technique useful in transforming a difficult integral into a simpler one. This procedure involves a change of variable, typically denoted as \( u \), helping to simplify the integrand.

Consider our integral problem: \( \int \sec x \, dx = \int \frac{\cos x}{1 - \sin^2 x} \, dx \). We employed substitution by setting:

• \( u = \sin x \)
• Thus, \( du = \cos x \, dx \)

This substitution helped facilitate the conversion of the complex integral into the simpler form: \( \int \frac{du}{1-u^2} \).

Integration by substitution is akin to reverse chain rule. Replacing expressions in terms of \( u \) can make the function easier to integrate, especially when dealing with composite functions or polynomials inside basic integrals. This method is a cornerstone in calculus, broadening the scope of functions we can integrate easily.

Partial Fraction Decomposition

Partial Fraction Decomposition is a method used to break down complex rational expressions into simpler fractions, making them easier to integrate. It's particularly useful when dealing with expressions that involve fractions with polynomials in the denominator.

In our exercise, we encountered the integral: \( \int \frac{du}{1-u^2} \). To solve this, we use partial fraction decomposition, dividing the expression: \( \frac{1}{1-u^2} \) into

• \( \frac{A}{1+u} \)
• \( \frac{B}{1-u} \)

For constants \( A \) and \( B \). By solving the system of equations derived from matching coefficients, we find \( A = \frac{1}{2} \) and \( B = \frac{1}{2} \). The integral is then taken in parts: \( \int \left( \frac{1}{2} \cdot \frac{1}{1-u} + \frac{1}{2} \cdot \frac{1}{1+u} \right)du \), yieldingthe result: \( \frac{1}{2} \ln|1-u| + \frac{1}{2} \ln|1+u| + C \).

Utilizing partial fraction decomposition makes seemingly complex, non-factorable polynomials manageable and integrable.