Question

More than one way Occasionally, two different substitutions do the job. Use each substitution to evaluate the following integrals. $$\int \sec ^{3} \theta \tan \theta d \theta(u=\cos \theta \text { and } u=\sec \theta)$$

Short answer

In this exercise, we evaluated the integral \(\int \sec^3 \theta \tan \theta d \theta\) using two different substitution methods: \(u=\cos \theta\) and \(u=\sec \theta\). By using the first substitution method, we obtained the result \(-\frac{1}{2}\sec^2 \theta + C_1\). With the second substitution method, we got the result \(\frac{1}{3}\sec^3 \theta + C_2\). Although these solutions are different, they both satisfy the original integrand, which confirms the validity of both substitution methods. The difference between the solutions is due to an arbitrary constant of integration that can be adjusted to make the two results match.

Step-by-step solution

Substitution 1: \(u = \cos \theta\)

Let \(u = \cos \theta\). Then, \(du = -\sin \theta d\theta\). Additionally, we have the trigonometric identity \(\sec^2\theta = 1 + \tan^2\theta\). As for the given integral, we can rewrite it in terms of sine and cosine: $$\int \sec^3 \theta \tan \theta d \theta = \int \frac{\sin \theta}{\cos^3 \theta} d\theta$$ Now, substitute \(u\) and \(du\) in the integral: $$-\int \frac{1}{u^3} du$$ Integrate with respect to \(u\): $$-\int u^{-3} du = -\frac{1}{2}u^{-2} + C$$ Now, substitute back \(\cos \theta\) for \(u\): $$-\frac{1}{2}(\cos \theta)^{-2} + C = -\frac{1}{2}\sec^2 \theta + C_1$$

Substitution 2: \(u = \sec \theta\)

Let \(u = \sec \theta\). Then, \(du = \sec \theta \tan \theta d\theta\). We can rewrite the given integral as: $$\int \sec^3 \theta \tan \theta d \theta = \int u^2 \sec \theta \tan \theta d\theta$$ Now, substitute \(u\) and \(du\) in the integral: $$\int u^2 du$$ Integrate with respect to \(u\): $$\int u^2 du = \frac{1}{3}u^3 + C$$ Now, substitute back \(\sec \theta\) for \(u\): $$\frac{1}{3}(\sec \theta)^3 + C = \frac{1}{3}\sec^3 \theta + C_2$$

Comparing results

Now, we have two different results from the two substitution methods: \(-\frac{1}{2}\sec^2 \theta + C_1\) and \(\frac{1}{3}\sec^3 \theta + C_2\). However, if we take the derivative of both results with respect to \(\theta\), we will find that both results satisfy the original integrand \(\sec^3 \theta \tan \theta\). Therefore, both results are correct up to a constant, and the two methods indeed provided valid solutions to the integral. Final results (up to a constant): $$-\frac{1}{2}\sec^2 \theta + C_1 \quad \text{and} \quad \frac{1}{3}\sec^3 \theta + C_2$$

Key concepts

Trigonometric Substitution

Trigonometric substitution is an integration technique where we substitute trigonometric functions for other expressions in an integral. This is particularly useful when the integrand involves radicals, or in cases where the direct integration is not straightforward. The goal is to simplify the integral into a form that is easier to work with.

In the example integral \(\int \sec^3\theta \tan\theta d\theta\), two different trigonometric substitutions are used to find the solution: \(u = \cos\theta\) and \(u = \sec\theta\). Each substitution transforms the integral into a simpler form—a power function of \(u\)—which can then be integrated using basic integration rules. The key in trigonometric substitution is to also find the differential \(du\) corresponding to our substitution, which allows us to completely re-express the original integral in terms of \(u\).

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, represent the general form of the integral of a function, without specific bounds. The notation for an indefinite integral is \(\int f(x)dx\), and the result is a function plus an arbitrary constant, \(C\), since the derivative of \(C\) is zero.

In our example, after performing the substitutions, we find the indefinite integrals \(\int u^{-3} du\) and \(\int u^{2} du\). When we integrate these expressions, we get a function of \(u\) plus a constant. Since \(u\) is a trigonometric substitution for \(\theta\), the final step is to revert back to \(\theta\) to express the results in terms of the original variable, incorporating the constant of integration.

Integration Techniques

There are various integration techniques used to solve integrals that are not readily integrable. Each technique has its specific applications and is used to simplify different kinds of integrals. Some common techniques include substitution, integration by parts, partial fractions, and trigonometric substitution.

For the integral \(\int \sec^3\theta \tan\theta d\theta\), trigonometric substitution is applied with two different approaches. These approaches illustrate the versatility and creativity required in choosing the right technique for integration. It is a valuable skill to recognize which technique to apply to a given problem and how to implement it effectively. Mastery of integration techniques is not only about knowing steps but also understanding how they transform the integral into a form that is manageable.

Trigonometric Identities

Trigonometric identities are equalities involving trigonometric functions. These identities are crucial in simplifying the expressions involving trigonometric functions when we are working on integrals or other problems in calculus. Common identities include Pythagorean identities, angle identities, and double-angle identities.

In solving the integral \(\int \sec^3\theta \tan\theta d\theta\), we encounter the identity \(\sec^2\theta = 1 + \tan^2\theta\), which is derived from the Pythagorean identity. Understanding and applying trigonometric identities enable us to change the appearance of the integrand, as we did in our example, where \(\tan\theta\) and \(\sec\theta\) were manipulated for the substitutions. This underscores the importance of memorizing these identities as they are fundamental tools in solving calculus problems.