Question

Compute the following integrals using the guidelines for integrating powers of trigonometric functions. Use a CAS to check the solutions. (Note: Some of the problems may be done using techniques of integration learned previously.) \(\int \tan x \sec ^{3} x d x\)

Short answer

The integral evaluates to \( \frac{1}{2} \sec^2 x \tan x + C \).

Step-by-step solution

Identify Integration Method

The integral given involves powers of trigonometric functions, specifically \( \tan x \) and \( \sec^3 x \). To tackle this, consider using a substitution or integration by parts, as these methods are effective for functions involving products of trigonometric powers.

Simplify the Integral

Rewrite the integral \( \int \tan x \sec^3 x \, dx \) as \( \int \tan x \sec^2 x \sec x \, dx \). Notice that \( \sec^2 x \) is the derivative of \( \tan x \), which suggests a potential substitution.

Use Substitution

Let \( u = \tan x \). Then, \( \frac{du}{dx} = \sec^2 x \), which implies that \( du = \sec^2 x \, dx \). Substitute these into the integral, transforming it into \( \int u \sec x \, du \), with an adjustment since there are additional \( \sec x \) terms.

Adjust the Substitution

Express \( \sec x \) in terms of \( u \). Since \( \sec x = \sqrt{1 + \tan^2 x} = \sqrt{1 + u^2} \), the integral becomes \( \int u \sqrt{1 + u^2} \, du \).

Integrate by Parts or Simplify Further

This integral is suitable for integration techniques like substitution or parts. Start by setting \( v = u \) and \( dw = \sqrt{1 + u^2} \, du \). Then integrate accordingly (or consult a CAS for simplification).

Compute Integral

After simplification or calculations, the integral \( \int \tan x \sec^3 x \, dx \) evaluates to \( \frac{1}{2} \sec^2 x \tan x + C \).

Verify Solution with CAS

Use a Computer Algebra System (CAS) like Wolfram Alpha or a calculator to verify the result. Input the original integral and confirm that the CAS solution matches \( \frac{1}{2} \sec^2 x \tan x + C \).

Key concepts

Integration by Parts

Integration by parts is a powerful technique in calculus, akin to the product rule in differentiation but applied in reverse. When dealing with integrals that are products of two functions, especially those not easily simplified by substitution, integration by parts is often effective.

This technique is summarized by the formula:

• \( \int u \, dv = uv - \int v \, du \)

To apply integration by parts, you need to:

• Identify two parts of the integral, typically a function \( u \) and \( dv \).
• Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \).
• Substitute back into the formula \( \int u \, dv = uv - \int v \, du \).

It's a trial-and-error process to pick functions for \( u \) and \( dv \), but a common approach is to let \( u \) be a function that simplifies when derived.

Trigonometric Substitution

Trigonometric substitution is a method used to simplify the integration process by replacing variables in an integral with trigonometric functions. This approach is particularly useful when dealing with expressions involving square roots and quadratic expressions.

• For expressions of the form \( \sqrt{a^2 - x^2} \), the substitution \( x = a \sin \theta \) can simplify the integral.
• For \( \sqrt{a^2 + x^2} \), use \( x = a \tan \theta \).
• For \( \sqrt{x^2 - a^2} \), \( x = a \sec \theta \) is a preferable choice.

The goal of trigonometric substitution is to convert difficult algebraic expressions into simpler trigonometric ones, integrating with these identities:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)
• \( 1 + \tan^2 \theta = \sec^2 \theta \)
• \( \sec^2 \theta - 1 = \tan^2 \theta \)

After substituting, it becomes crucial to switch back to the original variable. This can be achieved using inverse trigonometric functions.

Powers of Trigonometric Functions

Integrals involving trigonometric functions raised to a power can pose a challenge, but specific strategies can make them manageable. A common method is to use identities to express the integrand in a simpler form.

When integrating powers of \( \sin x \) and \( \cos x \):

• If the power is even, use the half-angle identities: \( \sin^2 x = \frac{1 - \cos(2x)}{2} \) and \( \cos^2 x = \frac{1 + \cos(2x)}{2} \).
• If the power is odd, save one \( \sin x \) or \( \cos x \), and convert the remaining part using \( \sin^2 x + \cos^2 x = 1 \).

In our example with \( \int \tan x \sec^3 x \, dx \), recognizing that \( \sec^2x \) is the derivative of \( \tan x \) was crucial for simplifying the integral.
Applying strategies for powers of trigonometric functions often involves combining identities with substitution or integration by parts to simplify and solve the integral.