Question

Use the double-angle formulas to evaluate the following integrals. \(\int \cos ^{2} 3 x d x\)

Short answer

The integral evaluates to \( \frac{1}{2}x + \frac{1}{12}\sin(6x) + C \).

Step-by-step solution

Identify the Double-Angle Formula for Cosine

The double-angle formula for cosine that will help simplify the expression is \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \). In this problem, the theta \( \theta \) is \( 3x \).

Apply the Double-Angle Formula

Substitute \( \cos^2(3x) \) in the integral using the double-angle identity: \( \cos^2(3x) = \frac{1 + \cos(6x)}{2} \). Thus, \( \int \cos^2(3x) \, dx \) becomes \( \int \frac{1 + \cos(6x)}{2} \, dx \).

Simplify and Split the Integral

Split the integral into two separate integrals: \( \int \frac{1}{2} \, dx + \int \frac{\cos(6x)}{2} \, dx \). This makes the integration process straightforward.

Integrate Both Terms

For the first integral, \( \int \frac{1}{2} \, dx \), the integral is \( \frac{1}{2}x + C_1 \). For the second integral, \( \int \frac{\cos(6x)}{2} \, dx \), use the substitution \( u = 6x \), which leads to \( du = 6 \, dx \) or \( dx = \frac{du}{6} \). The integral becomes \( \frac{1}{12} \sin(6x) + C_2 \).

Combine the Results from Integration

Combine the integrals from both parts: \( \frac{1}{2}x + \frac{1}{12}\sin(6x) + C \). Combining constants gives the general constant of integration \( C \).

Key concepts

calculus integration techniques

The art of integration in calculus involves various techniques, each suited to different forms of functions. One of the central methods is the use of substitution, where you replace a part of the integrand with a variable to simplify the integration process. In this exercise, we used substitution on the term involving cosine. To do so, we identified that making a substitution of the form \(u = 6x\) simplifies the integration of \( \cos(6x) \). By finding \(du = 6 \, dx\) or \(dx = \frac{du}{6}\), integration becomes manageable and straightforward. This technique often appears alongside other methods such as integration by parts or partial fraction decomposition, which are valuable when dealing with more complex integrals. Substitution is especially useful when a derivative of a function appears elsewhere in the integrand, making it easier to integrate by reversing the chain rule. It's crucial to practice these techniques to handle a variety of calculus problems efficiently.

integrals of trigonometric functions

Integrating trigonometric functions requires a solid understanding of how these functions behave. Trigonometric integrals, like \(\cos^2(3x)\) in our exercise, often involve the use of identities to simplify them. These functions can sometimes be systematically handled by recognizing patterns and applying specific formulas or substitutions.The fundamental goal in these scenarios is to transform the trigonometric expression into a form that readily integrates. For example, the expression \(\int \cos(2\theta)\, d\theta\) integrates to \(\frac{1}{2} \sin(2\theta)\), anchoring the importance of knowing integral forms of trig functions. Additionally, certain trigonometric integrals benefit from symmetry properties, which can simplify calculations by adjusting integration limits or recognizing even or odd functional properties.

trigonometric identities in calculus

Trigonometric identities are essential tools in calculus, especially when dealing with integrals of trigonometric functions. In this exercise, the double-angle identity \(\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}\) played a pivotal role. By applying this identity, we simplified the original expression and split the integral. Another key identity often used is the Pythagorean identity, \(\sin^2 \theta + \cos^2 \theta = 1\), which helps in transforming expressions involving both sine and cosine. Similarly, other identities like the angle-sum or difference identities facilitate the integration process by reducing complex trigonometric forms into manageable pieces.Becoming comfortable with these identities aids in both integration and differentiation problems in calculus, enhancing problem-solving efficiency and allowing for more guided and intelligent manipulations of equations.