Question

Perform the indicated integrations. $$ \int \sin ^{4} 6 x d x $$

Short answer

The integral is \( \frac{3}{8}x - \frac{1}{24} \sin(12x) + \frac{1}{192} \sin(24x) + C \).

Step-by-step solution

Use a Trigonometric Identity

To simplify the integration of \( \sin^4(6x) \), we use the identity \( \sin^2(\theta) = \frac{1}{2} (1 - \cos(2\theta)) \) to rewrite the expression. Therefore, \( \sin^4(6x) = (\sin^2(6x))^2 = \left( \frac{1}{2} (1 - \cos(12x)) \right)^2 \).

Expand and Simplify the Expression

Simplify \( \left( \frac{1}{2} (1 - \cos(12x)) \right)^2 \) by squaring the terms: \( \frac{1}{4} (1 - 2 \cos(12x) + \cos^2(12x)) \). The expression now becomes a combination of constants and simpler trigonometric terms.

Use the Cosine-Squared Identity

We can further reduce \( \cos^2(12x) \) using the identity \( \cos^2(\theta) = \frac{1}{2} (1 + \cos(2\theta)) \). Thus, \( \cos^2(12x) = \frac{1}{2} (1 + \cos(24x)) \). Substitute this back into the expression: \( \frac{1}{4} (1 - 2 \cos(12x) + \frac{1}{2} (1 + \cos(24x))) \).

Simplify the Expression

Combine all terms: \( \frac{1}{4} \left[ 1 - 2 \cos(12x) + \frac{1}{2} + \frac{1}{2} \cos(24x) \right] = \frac{3}{8} - \frac{1}{2} \cos(12x) + \frac{1}{8} \cos(24x) \).

Integrate Each Term Separately

Integrate each term: \( \int \left( \frac{3}{8} - \frac{1}{2} \cos(12x) + \frac{1}{8} \cos(24x) \right) dx \). This results in: \( \frac{3}{8}x - \frac{1}{2} \frac{1}{12} \sin(12x) + \frac{1}{8} \frac{1}{24} \sin(24x) + C \).

Simplify the Integral

Combine the constants with the integral to obtain: \( \frac{3}{8}x - \frac{1}{24} \sin(12x) + \frac{1}{192} \sin(24x) + C \), where \( C \) is the integration constant. This is the final simplified form of the integrated expression.

Key concepts

Trigonometric Identities

Trigonometric identities are important tools in calculus, especially in trigonometric integration. They help simplify complex trigonometric expressions into forms that are easier to integrate. For instance, the identity \( \sin^2(\theta) = \frac{1}{2} (1 - \cos(2\theta)) \) is a powerful tool for reducing powers of sine and cosine into more manageable terms.
When facing an integral such as \( \int \sin^4(6x) \, dx \), breaking down the higher power of sine becomes essential. Here, the identity transforms \( \sin^4(6x) = (\sin^2(6x))^2 \) into \( \left( \frac{1}{2} (1 - \cos(12x)) \right)^2 \). This expression is far easier to work with since it has been converted from a single trigonometric function raised to a high power into a sum of functions with lower powers.
Using trigonometric identities in integration helps:

• Transform functions into simpler equivalent forms
• Reduce the complexity of operations and calculations
• Introduce new functions that might be easier to integrate

Understanding these identities and their applications is crucial for mastering integral calculus, as they often provide the first step in solving complex integrals.

Integral Calculus

Integral calculus focuses on finding the antiderivative or the area under a curve. It involves techniques and formulas to integrate various types of functions, translating real-world problems into calculable forms. Key to solving these integrals is not just recognizing when to apply basic rules, but also using identities and transformations, as needed, to simplify functions into forms that can be easily integrated.
In the solution of \( \int \sin^4(6x) \, dx \), after applying trigonometric identities, the expression can be expanded and simplified into a sum of terms: \( \frac{3}{8} - \frac{1}{2} \cos(12x) + \frac{1}{8} \cos(24x) \). Each term here is simpler and can be integrated separately. For example:

• \( \frac{3}{8} \, x \) is straightforward.
• \( \int \cos(nx) \, dx = \frac{1}{n} \sin(nx) \) transforms cosine terms into sine.

Thus, integral calculus enables us to dissect complex integrals into manageable pieces using rules, identities, and simplifications. By employing these techniques, the evaluation of integrals becomes structured and systematic.

Definite and Indefinite Integrals

In integral calculus, there are two main types of integrals: definite and indefinite. Each serves a different purpose but is connected through their calculation methods.

• Indefinite Integrals: These involve finding the antiderivative of a function and include a constant of integration \( C \). For example, when we solve \( \int \sin^4(6x) \, dx \), the final expression, \( \frac{3}{8}x - \frac{1}{24} \sin(12x) + \frac{1}{192} \sin(24x) + C \), is an indefinite integral. It describes a family of functions differing only by a constant value.
• Definite Integrals: These calculate the area under a curve between two specific limits \( a \) and \( b \). The result is a numerical value, representing the accumulated quantity between the limits, often applied in real-world problems involving measurements of total quantities.

Understanding the distinction between these integrals is crucial. With indefinite integrals, we focus on deriving function families, while definite integrals provide concrete values through limits, often requiring evaluation of the definite antiderivatives at specified bounds. Mastery of both is imperative for a comprehensive understanding of integral calculus.