Problem 77

Question

Find the indefinite integrals. $$\int(10+8 \sin (2 x)) d x$$

Step-by-Step Solution

Verified
Answer
\(10x - 4\cos(2x) + C\)
1Step 1: Identify the components of the integral
We start with the integral \( \int (10 + 8 \sin(2x)) \, dx \). This means we need to integrate each term separately. The components are 10 and \( 8 \sin(2x) \).
2Step 2: Integrate the constant term
The integral of a constant \( c \) with respect to \( x \) is \( cx + C_1 \), where \( C_1 \) is the integration constant. Thus, \( \int 10 \, dx = 10x + C_1 \).
3Step 3: Integrate the sine function
The integral of \( \sin(ax) \) is \( -\frac{1}{a}\cos(ax) + C_2 \). Here, we integrate \( 8 \sin(2x) \), so \( \int 8 \sin(2x) \, dx = -\frac{8}{2} \cos(2x) + C_2 = -4\cos(2x) + C_2 \).
4Step 4: Combine the integrated parts
After integrating both terms separately, add the results: \( 10x - 4\cos(2x) + C \), where \( C = C_1 + C_2 \) is the overall integration constant.

Key Concepts

Integration of Trigonometric FunctionsIntegration of Constant TermsCalculus Steps for Integration
Integration of Trigonometric Functions
When faced with the integration of trigonometric functions, it's important to understand the standard forms and transformations that one can use. Trigonometric functions like sine and cosine appear frequently in calculus and have well-defined integration rules. One basic rule is:
  • The integral of \( \sin(ax) \) is \( -\frac{1}{a} \cos(ax) + C \).
  • Similarly, the integral of \( \cos(ax) \) is \( \frac{1}{a} \sin(ax) + C \).
These rules arise from differentiation rules and applying the reverse process. In our specific case, the expression \( 8 \sin(2x) \) needs integration. We can rewrite the coefficient inside the integral to factor in the 2 from \( \sin(2x) \), this involves dividing 8 by 2. Hence, its integral becomes \( -4 \cos(2x) + C_2 \). Splitting the following multiplication and adjusting the trigonometric function using the rule helps achieve the final result. This process makes dealing with complex trigonometric forms much more manageable.
Integration of Constant Terms
The integration of constant terms is one of the simplest processes within calculus. When you have a constant \( c \) within an integral, you can think of it as merely being multiplied by the variable of integration. The rule here is straightforward:
  • The integral of a constant \( c \) is \( cx + C \).
A constant term like 10, therefore, integrates to \( 10x + C_1 \). This rule is derived from the fact that the derivative of \( cx \) with respect to \( x \) is simply \( c \), so integration being the reverse operation yields this result. Any constant term, when integrated, will always add such a linear term to the indefinite integral. This allows for routine handling of constants without reevaluating each time. Simple and direct, it's a foundational calculus concept utilized in numerous problems.
Calculus Steps for Integration
The process of integrating requires methodical application of different rules, and it usually follows several specific steps. For indefinite integrals, here’s a simplified approach:
  • **Identify each term**: Break down the expression into individual parts that are easier to handle.
  • **Apply rules**: Use the relevant formulas for each component, such as constants or trigonometric forms.
  • **Factor and simplify**: Sometimes, factoring expressions make them easier to integrate, particularly for trigonometric functions.
  • **Add a constant of integration**: Remember the constant \( C \) which is crucial in indefinite integrals.
  • **Combine the results**: After integrating, sum up the parts for the complete solution.
In this example, integrating \( \int (10 + 8 \sin(2x)) \, dx \), we followed these steps to arrive at the solution: first integrating \( 10 \) to \( 10x \), and \( 8 \sin(2x) \) to \( -4\cos(2x) \). The last step was assembling these components along with the constant \( C = C_1 + C_2 \) to form the comprehensive indefinite integral result: \( 10x - 4\cos(2x) + C \). Each problem-solving step ensures accuracy and completeness in tackling calculus exercises.