Problem 20 - Essential Calculus

Question

Find the integral involving secant and tangent. $$ \int \sec ^{6} 3 x d x $$

Step-by-Step Solution

Answer

The integral of \( \sec ^{6} 3 x dx \) is \( \frac{1}{3} [\tan 3x + \frac{2}{3}\tan^3 3x + \frac{1}{5}\tan^5 3x] + C \)

Step 1: Find a suitable substitution

A common strategy for integrals involving powers of secant or tangent is to rewrite everything in terms of a lower power and then make a substitution. We observe that \(\sec ^{6} x = (\sec ^{4} x)(\sec ^{2} x) = (1 + \tan ^{2} x)^2 (\sec ^{2} x)\). This suggests a substitution of \(u = \tan 3x\). Substituting, we find that the integral becomes \(\int (1 + u^2)^2 du\)

Step 2: Expand and simplify

Expanding and simplifying the term gets \(\int (1 + 2u^2 + u^4) du\) = \(u + \frac{2}{3}u^3 + \frac{1}{5}u^5 + C\).

Step 3: Substitute back

Substitute \(u\) back into the equation. Resulting in, \(= \tan 3x + \frac{2}{3} (\tan^3 3x) + \frac{1}{5} (\tan^5 3x) + C\).

Step 4: Final adjustment

Since we made the substitution \(u = \tan 3x\) from \(dx = du/3sec^2(3x)\), we need to adjust for the 3 in the tan. Therefore we will divide the entire integral by 3. The final integration take the form \(\frac{1}{3}(\tan 3x + \frac{2}{3}\tan^3 3x + \frac{1}{5}\tan^5 3x + C)\).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is an efficient technique for evaluating integrals involving squares, particularly when you see expressions like \( \sec^2 \) or \( \tan^2 \). The key idea is to make the substitution based on trigonometric identities you can use to simplify the integrand. For instance, in the integral \( \int \sec^6 3x \, dx \), trigonometric identity \( \sec^2 x = 1 + \tan^2 x \) is utilized. This allows rewriting \( \sec^6 3x \) as \( (1 + \tan^2 3x)^2 \cdot \sec^2 3x \). The substitution \( u = \tan 3x \) is then performed to transform the integral into an expression solely in \( u \), simplifying our computations.

When performing trigonometric substitution:

• Identify part of the integrand, leveraging trigonometric identities.
• Make an appropriate substitution to simplify the expression.
• Solve the resulting integral in simpler terms.

This method shines in calculus problems requiring simplification of complex rational functions, particularly when directly integrating seems impossible.

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is vital in calculus. An indefinite integral, represented without specific upper and lower limits, captures a family of functions whose derivatives result in the original function. In our example \( \int \sec^6 3x \, dx \), solving this involves simplifying to \( \int (1 + 2u^2 + u^4) \, du \), which is an indefinite integral.

Key concepts about integrals include:

• Indefinite Integrals: These integrals do not have boundary values and include a constant of integration \( C \).
• Definite Integrals: These involve specific limits, providing the net area under a curve between those points.
• General Rule: Integrating a function allows us to find functions whose derivatives give the integrand.

Indefinite integrals, like in our problem, lead you to express solutions in a generalized form, ready for specific applications or evaluations.

Integration by Parts

Integration by parts is a useful technique applicable when an integral is the product of two functions. This technique is often remembered through the formula \( \int u \, dv = uv - \int v \, du \). In problems involving higher mathematics like \( \int \sec^6 3x \, dx \), while not directly used here, integration by parts serves well when variables need alternating differentiation and integration steps.

Steps to use Integration by Parts effectively involve:

• Identifying \( u \) and \( dv \): Choose \( u \) to be a function that simplifies upon differentiation, and \( dv \) that remains integrable.
• Calculating \( du \) and \( v \): Differentiate \( u \) to get \( du \) and integrate \( dv \) to get \( v \).
• Apply the formula: Substitute in \( uv - \int v \, du \) and solve.

Although not essential for the original problem of integrating \( \sec^6 3x \), becoming fluent with methods like integration by parts builds a stronger foundation for handling complex integrals in broader calculus challenges.