Question

What is the indefinite integral of 6sin(3t) dt?

Answer 1

`\displaystyle \int {6sin(3t)} \, dt = -2cos(3t) + C`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:
`\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals
• [Indefinite Integrals] Integration Constant C

Integration Property [Multiplied Constant]:
`\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx`

U-Substitution

Explanation:

Step 1: Define

Identify

`\displaystyle \int {6sin(3t)} \, dt`

Step 2: Integrate Pt. 1

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {6sin(3t)} \, dt = 6\int {sin(3t)} \, dt`

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

1. Set u:
   `\displaystyle u = 3t`
2. [u] Differentiate [Basic Power Rule, Multiplied Constant]:
   `\displaystyle du = 3 \ dt`

Step 4: integrate Pt. 3

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {6sin(3t)} \, dt = 2\int {3sin(3t)} \, dt`
2. [Integral] U-Substitution:
   `\displaystyle \int {6sin(3t)} \, dt = 2\int {sin(u)} \, du`
3. [Integral] Trigonometric Integration:
   `\displaystyle \int {6sin(3t)} \, dt = -2cos(u) + C`
4. Back-Substitute:
   `\displaystyle \int {6sin(3t)} \, dt = -2cos(3t) + C`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration