Problem 46 - Calculus Single Variable

Question

Evaluate the given integral by converting the integrand to an expression in sines and cosines. $$ \int 3 \tan (x) \sec ^{3}(x) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \sec^3(x) + C \).

Step 1: Rewrite Tangent and Secant in Terms of Sine and Cosine

The functions tangent and secant can be expressed in terms of sine and cosine. We have \( \tan(x) = \frac{\sin(x)}{\cos(x)} \) and \( \sec(x) = \frac{1}{\cos(x)} \). Substitute these into the integral: \[ \int 3 \tan(x) \sec^3(x) \, dx = \int 3 \frac{\sin(x)}{\cos(x)} \left(\frac{1}{\cos(x)}\right)^3 \, dx = \int 3 \frac{\sin(x)}{\cos^4(x)} \, dx.\]

Step 2: Simplify the Integrand

Now that we have everything in terms of sine and cosine, combine the terms:\[ \int 3 \frac{\sin(x)}{\cos^4(x)} \, dx = \int 3 \sin(x) \cos^{-4}(x) \, dx. \]

Step 3: Substitution

Notice that \( \cos(x) \) and \( \sin(x) \) are in the integrand. Use substitution: let \( u = \cos(x) \) thus \( du = -\sin(x) \, dx \). Therefore, this substitution gives:\[ dx = -\frac{du}{\sin(x)} \]So the integral becomes:\[ \int 3 \frac{\sin(x)}{u^4} \left(-\frac{du}{\sin(x)}\right) = -3 \int u^{-4} \, du. \]

Step 4: Integrate with Respect to u

Now integrate the expression with respect to \( u \): \[ -3 \int u^{-4} \, du = -3 \left( -\frac{1}{3}u^{-3} \right) + C = u^{-3} + C. \]This simplifies to \[ \frac{1}{u^3} + C. \]

Step 5: Substitute Back to Terms of x

Since \( u = \cos(x) \), replace \( u \, \text{with} \, \cos(x) \) to return the expression in terms of \( x \):\[ \frac{1}{\cos^3(x)} + C. \]

Step 6: Express in Terms of Secant

Recognize that \( \frac{1}{\cos^3(x)} \) is \( \sec^3(x) \), so the integral simplifies to:\[ \sec^3(x) + C. \]

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in integration to simplify integrals. It is especially helpful when dealing with integrands that involve expressions like \(a^2 - x^2\), \(a^2 + x^2\), or \(x^2 - a^2\). These expressions can be transformed using the relationships between trigonometric functions and the Pythagorean identity.To apply trigonometric substitution:

• Identify the form of the integral that matches a trigonometric identity.
• Choose the appropriate trigonometric substitution, such as \(x = a \sin(\theta)\), \(x = a \tan(\theta)\), or \(x = a \sec(\theta)\).
• Replace the variable and differentials in the integral using the substitution.
• Simplify the integral using trigonometric identities if necessary.

In the exercise, we didn't use a direct trigonometric substitution for forms like \(a^2 - x^2\) but the idea of substitution showed up when replacing components of the trigonometric functions with sine and cosine, paving the way to simplify and solve the integral.

Trigonometric Identities

Understanding trigonometric identities is crucial when working through integrals that involve trigonometric functions. These identities allow us to rewrite expressions and simplify integrals, making them easier to evaluate. Some frequently used trigonometric identities include:

• \(\sin^2(x) + \cos^2(x) = 1\)
• \(\sec(x) = \frac{1}{\cos(x)}\)
• \(\tan(x) = \frac{\sin(x)}{\cos(x)}\)

The given integral in the original exercise required rewriting the tangent and secant functions in terms of sine and cosine. By expressing \(\tan(x)\) as \(\frac{\sin(x)}{\cos(x)}\) and \(\sec(x)\) as \(\frac{1}{\cos(x)}\), we were able to transform the integral into a more manageable form. Recognizing and leveraging these identities can simplify an integral, and practice with these will deepen understanding and improve problem-solving skills.

Definite and Indefinite Integrals

When dealing with calculus, especially integrals, it's important to distinguish between definite and indefinite integrals. They serve different purposes and have distinct properties.- **Indefinite Integrals** are the most basic form, representing a family of functions. An indefinite integral, \(\int f(x) \, dx\), gives us the antiderivative of \(f(x)\) plus an arbitrary constant \(C\), because integration is essentially finding the reverse of differentiation.- **Definite Integrals** are calculated over a specific interval \([a, b]\). They evaluate the net area under the curve defined by a function between these two points. The result is a number, not a function. Definite integrals are computed using the Fundamental Theorem of Calculus by calculating \(F(b) - F(a)\), where \(F(x)\) is the antiderivative of \(f(x)\).In the exercise provided, we focused on finding an indefinite integral. The arbitrary constant \(C\) is added until boundary conditions or specific limits convert it into a definite integral. Understanding these differences will help students determine the approach and methods needed for each type they encounter.