Mathematics, 30.07.2019 02:40 KittyLitty

Find the indefinite integral. (use c for the constant of integration.) sin xcos5 x dx

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Answer from: DWASS

$\displaystyle \int {sin(x)cos(5x)} \, dx = \frac{cos(4x)}{8} - \frac{cos(6x)}{12} + C$

General Formulas and Concepts:

Algebra I

Terms/CoefficientsExpanding/Factoring

Pre-Calculus

Trigonometric Identities

Product-to-Sum Formula: $\displaystyle sin(x)cos(y) = \frac{sin(y + x) - sin(y - x)}{2}$

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

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int {sin(x)cos(5x)} \, dx$

Step 2: Integrate Pt. 1

[Integrand] Rewrite [Product-to-Sum Formula]: $\displaystyle \int {sin(x)cos(5x)} \, dx = \int {\frac{sin(6x) - sin(4x)}{6}} \, dx$ [Integrand] Rewrite: $\displaystyle \int {sin(x)cos(5x)} \, dx = \int {\Big( \frac{sin(6x)}{2} - \frac{sin(4x)}{2} \Big)} \, dx$ [Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int {sin(x)cos(5x)} \, dx = \int {\frac{sin(6x)}{2}} \, dx - \int {\frac{sin(4x)}{2}} \, dx$ [Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {sin(x)cos(5x)} \, dx = \frac{1}{2}\int {sin(6x)} \, dx - \frac{1}{2}\int {sin(4x)} \, dx$ Factor: $\displaystyle \int {sin(x)cos(5x)} \, dx = \frac{1}{2} \bigg[ \int {sin(6x)} \, dx - \int {sin(4x)} \, dx \bigg]$

Step 3: integrate Pt. 2

Identify variables for u-substitution.

Integral 1:

Set u: $\displaystyle u = 6x$ [u] Differentiate [Basic Power Rule, Multiplied Constant]: $\displaystyle du = 6 \ dx$

Integral 2:

Set z: $\displaystyle z = 4x$ [z] Differentiate [Basic Power Rule, Multiplied Constant]: $\displaystyle dz = 4 \ dx$

Step 4: Integrate Pt. 3

[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {sin(x)cos(5x)} \, dx = \frac{1}{2} \bigg[ \frac{1}{6}\int {6sin(6x)} \, dx - \frac{1}{4}\int {4sin(4x)} \, dx \bigg]$ [Integrals] U-Substitution: $\displaystyle \int {sin(x)cos(5x)} \, dx = \frac{1}{2} \bigg[ \frac{1}{6}\int {sin(u)} \, du - \frac{1}{4}\int {sin(z)} \, dz \bigg]$ [Integrals] Trigonometric Integration: $\displaystyle \int {sin(x)cos(5x)} \, dx = \frac{1}{2} \bigg[ \frac{1}{6}[-cos(u)] - \frac{1}{4}[-cos(z)] \bigg] + C$ Simplify: $\displaystyle \int {sin(x)cos(5x)} \, dx = \frac{1}{2} \bigg[ \frac{cos(z)}{4} - \frac{cos(u)}{6} \bigg] + C$ Expand: $\displaystyle \int {sin(x)cos(5x)} \, dx = \frac{cos(z)}{8} - \frac{cos(u)}{12} + C$ Back-Substitute: $\displaystyle \int {sin(x)cos(5x)} \, dx = \frac{cos(4x)}{8} - \frac{cos(6x)}{12} + C$

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

Unit: Integration

Answer from: Quest

4should be the last one n=4

Answer from: Quest

answer: it's upside down and i can't read it

step-by-step explanation: