Question

a. Verify the identity \(\sec x=\frac{\cos x}{1-\sin ^{2} x}\) b. Use the identity in part (a) to verify that \(\int \sec x d x=\frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\) (Source: The College Mathematics Joumal \(32,\) No. 5 (November 2001))

Short answer

Question: Verify the trigonometric identity \(\sec x=\frac{\cos x}{1-\sin^2 x}\), and use it to show that the following integral formula is true: \(\int \sec x dx = \frac{1}{2}\ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\). Answer: We verified the given trigonometric identity using fundamental trigonometric identities and the definition of secant. Then, we used the verified identity to substitute into the given integral and performed integration using substitution and partial fraction decomposition methods. As a result, we showed that the given integral formula is indeed true.

Step-by-step solution

Part a: Verifying the trigonometric identity

1. Start by writing the given identity and the fundamental trigonometric identity \(\sin^2x + \cos^2x = 1\): \(\sec x = \frac{\cos x}{1-\sin^2 x}\) and \(\sin^2x + \cos^2x = 1\). 2. Express \(\cos^2x\) in terms of \(\sin^2x\) using the fundamental identity: \(\cos^2x = 1 - \sin^2x\). 3. Substitute the expression of \(\cos^2x\) into the identity we want to prove: \(\sec x = \frac{\cos x}{\cos^2x} = \frac{\cos x}{1-\sin^2 x}\). 4. Recall the definition of \(\sec x = \frac{1}{\cos x}\) and substitute it into the expression: \(\frac{1}{\cos x} = \frac{\cos x}{1-\sin^2 x}\). By simplifying, we have shown that the given identity is true.

Part b: Verifying the integral

1. Start by writing down the given integral and the identity from part (a): \(\int \sec x dx\) and \(\sec x = \frac{\cos x}{1 - \sin^2 x}\). 2. Substitute the identity into the integral: \(\int \frac{\cos x}{1 - \sin^2 x} dx\). 3. Make a substitution in the integral: let \(u = \sin x\), then \(du = \cos x dx\). This gives the new integral: \(\int \frac{1}{1 - u^2} du\). 4. Integrate the expression with respect to \(u\) using partial fraction decomposition: \(\frac{1}{1 - u^2} = \frac{1}{(1 + u)(1 - u)} = \frac{A}{1 + u} + \frac{B}{1 - u}\). 5. Solve for \(A\) and \(B\): \(A(1 - u) + B(1 + u) = 1\). By setting \(u = 1\), we get \(A = 0\) and by setting \(u = -1\), we get \(B = \frac{1}{2}\) and \(A = \frac{1}{2}\). 6. Substitute \(A\) and \(B\) into the integral: \(\int \frac{1}{1 - u^2} du = \frac{1}{2} \int (\frac{1}{1 + u} + \frac{1}{1 - u}) du\). 7. Integrate with respect to \(u\): \(\frac{1}{2}\int \frac{1}{1 + u} du + \frac{1}{2}\int \frac{1}{1 - u} du = \frac{1}{2} \ln(|1 + u|) + \frac{1}{2} \ln(|1 - u|) + C\). 8. Reverse the substitution of \(u = \sin x\): \(\frac{1}{2} \ln(|1 + \sin x|) + \frac{1}{2} \ln(|1 - \sin x|) + C = \frac{1}{2}\ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\). By following these steps, we have verified that the given integral formula is true.

Key concepts

Fundamental Trigonometric Identities

Understanding trigonometric identities is crucial for solving many mathematical problems, especially those involving trigonometric functions. One of the core identities is

• the Pythagorean identity: \[ \sin^2x + \cos^2x = 1 \] It expresses a fundamental relation between sine and cosine.
• Using this identity, we can find expressions for other trigonometric functions, like secant.

In the exercise given, we needed to verify the identity \[ \sec x = \frac{\cos x}{1-\sin^2 x} \]By expressing \( \cos^2x \) in terms of \( \sin^2x \), as \( 1 - \sin^2x \), we leverage the Pythagorean identity to simplify the expression for \( \sec x \). Since \( \sec x \) is defined as \( \frac{1}{\cos x} \), setting it equal to \( \frac{\cos x}{\cos^2x} \) confirms the given trigonometric identity. Importantly, this demonstrates how these fundamental identities help in simplifying and proving more complex expressions.

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the total accumulation of quantities, commonly represented as the area under a curve. One important skill is integrating trigonometric functions, often requiring substitutions to simplify the expression.

• In part b of the exercise, we were tasked with integrating \( \sec x \), known to be a challenging integral.
• The substitution \( u = \sin x \) was key, transforming the integral into a more recognizable form.

The effect of this substitution is to change the perspective of the problem by turning it into an integral involving rational functions, which can often be more easily solved. The ultimate goal is to determine the antiderivative, which in this case, results in a logarithmic function:\[ \int \sec x dx = \frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right| + C \]This showcases the power of integral calculus in resolving complex trigonometric integrations by converting them into simpler forms through substitution.

Partial Fraction Decomposition

Partial fraction decomposition is a method used in algebra and calculus to break down complex rational expressions into a sum of simpler fractions. This is especially helpful in integrating expressions that otherwise appear intractable.

• In the exercise, once we substituted \( u = \sin x \), we transformed the integral into: \[ \int \frac{1}{1-u^2} du \]
• Recognizing the expression \( \frac{1}{1-u^2} \) as a difference of squares enables us to apply partial fraction decomposition.

The decomposition results in: \[ \frac{1}{1-u^2} = \frac{1}{(1+u)(1-u)} \]Allowing us to express it as: \[ \frac{1}{2} \left( \frac{1}{1+u} + \frac{1}{1-u} \right) \]This transforms the problem into two simpler integrals, each of which is a standard logarithmic form. Consequently, partial fraction decomposition allows us to verify the given integral formula, showcasing its utility in breaking down and solving complex integrals.