Mathematics, 21.08.2019 00:40 BrainGeek4848

How do i evaluate the indefinite integral
∫sin(3x)⋅sin(6x)dxintsin(3x)*sin(6x )dx ?

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Answer from: newjersey4015

$\displaystyle \int {sin(3x)sin(6x)} \, dx = \frac{2sin^3(3x)}{9} + C$

General Formulas and Concepts:

Pre-Calculus

Trigonometric Identities

Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals[Indefinite Integrals] Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int {sin(3x)sin(6x)} \, dx$

Step 2: Integrate Pt. 1

Set variables for u-substitution.

Set u: $\displaystyle u = 3x$ [u] Differentiate [Basic Power Rule, Multiplied Constant]: $\displaystyle du = 3 \ dx$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {sin(3x)sin(6x)} \, dx = \frac{1}{3} \int {3sin(3x)sin(6x)} \, dx$ [Integral] U-Substitution: $\displaystyle \int {sin(3x)sin(6x)} \, dx = \frac{1}{3} \int {sin(u)sin(2u)} \, du$ [Integrand] Rewrite [Trigonometric Identities]: $\displaystyle \int {sin(3x)sin(6x)} \, dx = \frac{1}{3} \int {sin(u)[2cos(u)sin(u)]} \, du$ [Integral] Simplify: $\displaystyle \int {sin(3x)sin(6x)} \, dx = \frac{1}{3} \int {2sin^2(u)cos(u)} \, du$ [Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {sin(3x)sin(6x)} \, dx = \frac{2}{3} \int {sin^2(u)cos(u)} \, du$

Step 4: Integrate Pt. 3

Set variables for u-substitution #2.

Set z: $\displaystyle z = sin(u)$ [z] Differentiate [Trigonometric Differentiation]: $\displaystyle dz = -cos(u) \ du$

Step 5: Integrate Pt. 4

[Integral] U-Substitution: $\displaystyle \int {sin(3x)sin(6x)} \, dx = \frac{2}{3} \int {z^2} \, dz$ [Integral] Reverse Power Rule: $\displaystyle \int {sin(3x)sin(6x)} \, dx = \frac{2}{3} \Big( \frac{z^3}{3} \Big) + C$ Simplify: $\displaystyle \int {sin(3x)sin(6x)} \, dx = \frac{2z^3}{9} + C$ [z] Back-Substitute: $\displaystyle \int {sin(3x)sin(6x)} \, dx = \frac{2sin^3(u)}{9} + C$ [u] Back-Substitute: $\displaystyle \int {sin(3x)sin(6x)} \, dx = \frac{2sin^3(3x)}{9} + C$

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

Unit: Integration

Answer from: Quest

D. i think it is the savings

Answer from: Quest

a: 1/6, j: 1/4

step-by-step explanation: you are welcome

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