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 \(\int \sec x \, d x=\frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\)

Short answer

Question: Verify the identity sec(x) = (cos x)/(1-sin^2 x) and then evaluate the integral of sec(x) with respect to x. Answer: The identity is verified as sec(x) = (cos x)/(1-sin^2 x). The integral of sec(x) with respect to x is (1/2)ln|(1+sin x)/(1-sin x)| + C.

Step-by-step solution

Verification of the Identity:

First, we need to verify the given identity by showing that sec\((x) = \frac{\cos x}{1-\sin^2 x}\). We know that sec\((x) = \frac{1}{\cos x}\). Now, let's rewrite the expression for sec\((x)\) in terms of \(\sin x\) and \(\cos x\): $$\sec x = \frac{1}{\cos x} = \frac{\cos x}{\cos^2 x}.$$ Recalling the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), we can write: $$\cos^2 x = 1 - \sin^2 x.$$ Now replacing the denominator in our previous equation: $$\sec x = \frac{\cos x}{1 - \sin^2 x}.$$ Hence, the provided identity has been verified.

Integration Using the Verified Identity:

Now, we have to evaluate the integral \(\int \sec x\, dx\) using the identity sec\((x) = \frac{\cos x}{1-\sin^2 x}\). We write the integral as follows: $$\int \sec x\, dx = \int \frac{\cos x}{1-\sin^2 x}\, dx.$$ To continue, we will use substitution method. Let's make the following substitution: $$u = \sin x \implies du = \cos x \, dx.$$ Our integral now becomes: $$\int \frac{\cos x}{1-\sin^2 x}\, dx = \int \frac{1}{1-u^2}\, du.$$ This integral can be rewritten and solved using partial fraction decomposition. We begin by rewriting the integral as follows: $$\int \frac{1}{1-u^2}\,du = \frac{1}{2}\left[\int \frac{1}{1-u}\,du - \int \frac{1}{1+u}\,du\right].$$ Now we can integrate the two separate integrals: $$\frac{1}{2}\left[\int \frac{1}{1-u}\,du - \int \frac{1}{1+u}\,du\right] = \frac{1}{2}\left[\ln|1-u| - \ln|1+u|\right] + C.$$ We then use the logarithmic property of difference to combine the two logarithms: $$\frac{1}{2}\left[\ln|1-u| - \ln|1+u|\right] + C = \frac{1}{2}\ln\left|\frac{1-u}{1+u}\right| + C.$$ As the last step, we substitute back \(u = \sin x\): $$\frac{1}{2}\ln\left|\frac{1-\sin x}{1+\sin x}\right| + C.$$ The final solution for the integral is: $$\int \sec x\, dx = \frac{1}{2}\ln\left|\frac{1+\sin x}{1-\sin x}\right| + C.$$

Key concepts

Pythagorean Identity

In calculus and trigonometry, the Pythagorean identity is an essential concept. It states that \( \sin^2 x + \cos^2 x = 1 \). This identity is fundamental because it establishes a relationship between sine and cosine functions. When dealing with problems involving trigonometric functions, such as integration, the Pythagorean identity can simplify expressions significantly.

Here’s how it works: if you know one of the two trigonometric functions (sine or cosine), you can easily find the other. For example, you can express \( \cos^2 x \) as \( 1 - \sin^2 x \) using this identity. This transformation is vital when you need to simplify or manipulate trigonometric integrals, as seen in the given exercise where \( \sec x = \frac{1}{\cos x} \) is rewritten to involve \( \sin x \).

Using such identities can make complex integrals more manageable, allowing for easier integration methods, such as substitution and partial fraction decomposition.

Substitution Method

The substitution method is a powerful integration technique that simplifies complex integrals by changing variables. Think of it as a way to transform a difficult integral into a simpler form, much like using a map to navigate a complicated city.

In the exercise we examined, substitution was used to simplify the integral of \( \sec x \). By setting \( u = \sin x \), the integral \( \int \frac{\cos x}{1 - \sin^2 x}\, dx \) was converted into \( \int \frac{1}{1-u^2}\, du \). This step crucially transforms a trigonometric function into a form suitable for further manipulation, specifically with partial fraction decomposition.

Using substitution not only breaks down complex integrals but also allows us to use standard integration formulas more effectively. The key is to choose a substitution that simplifies the integration process, focusing on simplifying the expression or its derivative. After computing the integral in terms of the new variable, the final step is to convert back to the original variable to find the solution to the integral.

Partial Fraction Decomposition

Partial fraction decomposition is another helpful technique used in finding integrals, specifically those involving rational functions. It involves breaking down a complex fraction into simpler fractions, which are easier to integrate. This method is particularly useful when dealing with integrals where the denominator can be factored into simpler polynomials.

In our exercise, after using the substitution \( u = \sin x \), we obtained the integral \( \int \frac{1}{1-u^2}\, du \). Recognizing that \( 1-u^2 \) can be factored into \((1-u)(1+u)\), the integral was rewritten as a sum of two simpler fractions: \( \frac{1}{2}\left[\int \frac{1}{1-u}\,du - \int \frac{1}{1+u}\,du\right] \).

Each part can then be integrated individually, resulting in simpler solutions that can be recombined. This method provides a systematic way to tackle integrals that initially seem daunting. The principle behind partial fraction decomposition is balancing between simplifying integrals and ensuring accuracy, leading to results that are easier to interpret and apply.