Respuesta :
Answer:
See Explanation.
General Formulas and Concepts:
Pre-Algebra
- Distributive Property
- Equality Properties
Algebra I
- Combining Like Terms
- Factoring
Calculus
- Derivative 1: [tex]\frac{d}{dx} [e^u]=u'e^u[/tex]
- Integration Constant C
- Integral 1: [tex]\int {e^x} \, dx = e^x + C[/tex]
- Integral 2: [tex]\int {sin(x)} \, dx = -cos(x) + C[/tex]
- Integral 3: [tex]\int {cos(x)} \, dx = sin(x) + C[/tex]
- Integral Rule 1: [tex]\int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
- Integration by Parts: [tex]\int {u} \, dv = uv - \int {v} \, du[/tex]
- [IBP] LIPET: Logs, Inverses, Polynomials, Exponents, Trig
Step-by-step Explanation:
Step 1: Define Integral
[tex]\int {e^{au}sin(bu)} \, du[/tex]
Step 2: Identify Variables Pt. 1
Using LIPET, we determine the variables for IBP.
Use Int Rules 2 + 3.
[tex]u = e^{au}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = sin(bu)du\\du = ae^{au}du \ \ \ \ \ \ \ \ \ v = \frac{-cos(bu)}{b}[/tex]
Step 3: Integrate Pt. 1
- Integrate [IBP]: [tex]\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} - \int ({ae^{au} \cdot \frac{-cos(bu)}{b} }) \, du[/tex]
- Integrate [Int Rule 1]: [tex]\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{a}{b} \int ({e^{au}cos(bu)}) \, du[/tex]
Step 4: Identify Variables Pt. 2
Using LIPET, we determine the variables for the 2nd IBP.
Use Int Rules 2 + 3.
[tex]u = e^{au}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = cos(bu)du\\du = ae^{au}du \ \ \ \ \ \ \ \ \ v = \frac{sin(bu)}{b}[/tex]
Step 5: Integrate Pt. 2
- Integrate [IBP]: [tex]\int {e^{au}cos(bu)} \, du = \frac{e^{au}sin(bu)}{b} - \int ({ae^{au} \cdot \frac{sin(bu)}{b} }) \, du[/tex]
- Integrate [Int Rule 1]: [tex]\int {e^{au}cos(bu)} \, du = \frac{e^{au}sin(bu)}{b} - \frac{a}{b} \int ({e^{au} sin(bu)}) \, du[/tex]
Step 6: Integrate Pt. 3
- Integrate [Alg - Back substitute]: [tex]\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{a}{b} [\frac{e^{au}sin(bu)}{b} - \frac{a}{b} \int ({e^{au} sin(bu)}) \, du][/tex]
- [Integral - Alg] Distribute Brackets: [tex]\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2} - \frac{a^2}{b^2} \int ({e^{au} sin(bu)}) \, du[/tex]
- [Integral - Alg] Isolate Original Terms: [tex]\int {e^{au}sin(bu)} \, du + \frac{a^2}{b^2} \int ({e^{au} sin(bu)}) \, du= \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}[/tex]
- [Integral - Alg] Rewrite: [tex](\frac{a^2}{b^2} +1)\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}[/tex]
- [Integral - Alg] Isolate Original: [tex]\int {e^{au}sin(bu)} \, du = \frac{\frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}}{\frac{a^2}{b^2} +1}[/tex]
- [Integral - Alg] Rewrite Fraction: [tex]\int {e^{au}sin(bu)} \, du = \frac{\frac{-be^{au}cos(bu)}{b^2} + \frac{ae^{au}sin(bu)}{b^2}}{\frac{a^2}{b^2} +\frac{b^2}{b^2} }[/tex]
- [Integral - Alg] Combine Like Terms: [tex]\int {e^{au}sin(bu)} \, du = \frac{\frac{ae^{au}sin(bu)-be^{au}cos(bu)}{b^2} }{\frac{a^2+b^2}{b^2} }[/tex]
- [Integral - Alg] Divide: [tex]\int {e^{au}sin(bu)} \, du = \frac{ae^{au}sin(bu) - be^{au}cos(bu)}{b^2} \cdot \frac{b^2}{a^2 + b^2}[/tex]
- [Integral - Alg] Multiply: [tex]\int {e^{au}sin(bu)} \, du = \frac{1}{a^2+b^2} [ae^{au}sin(bu) - be^{au}cos(bu)][/tex]
- [Integral - Alg] Factor: [tex]\int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)][/tex]
- [Integral] Integration Constant: [tex]\int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)] + C[/tex]
And we have proved the integration formula!
Prove: [tex]\displaystyle \int e^a^u sin(bu)\ du = \frac{e^a^u}{a^2+b^2} (a \ sin(bu) - b \ cos(bu)) + C[/tex]
Integration by parts formula: [tex]\displaystyle \int udv = uv - \int vdu[/tex]
Find u, du, v, and dv for this function: [tex]\displaystyle \int e^a^u sin(bu)[/tex]
- [tex]\displaystyle u =e^a^u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = -\frac{cos(bu)}{b} \\ du=ae^a^u \ du \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = sin(bu) \ du[/tex]
Plug these values into the IBP formula.
- [tex]\displaystyle \int e^a^u sin(bu) \ du = e^a^u \cdot \frac{-cos(bu)}{b} - \int -\frac{cos(bu)}{b} \cdot ae^a^u \ du[/tex]
Multiply and simplify the factors. Factor the negative out of the integral.
- [tex]\displaystyle \int e^a^u sin(bu) \ du = -\frac{e^a^u cos(bu)}{b} +\int \frac{a \ cos(bu) \ e^a^u}{b} \ du[/tex]
Factor out a/b from the integral.
- [tex]\displaystyle \int e^a^u sin(bu) \ du = -\frac{e^a^u cos(bu)}{b} + \frac{a}{b} \int cos(bu) \ e^a^u \ du[/tex]
Now we are going to apply IBP to the function: [tex]\displaystyle \int cos(bu) \ e^a^u[/tex] . Find u, du, v, and dv for this function.
- [tex]\displaystyle u =e^a^u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \frac{sin(bu)}{b} \\ du=ae^a^u \ du \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = cos(bu) \ du[/tex]
Plug these values into the IBP formula.
- [tex]\displaystyle \int e^a^u cos(bu) \ du = e^a^u \cdot \frac{sin(bu)}{b} - \int \frac{sin(bu)}{b} \cdot ae^a^u \ du[/tex]
Multiply and simplify the factors.
- [tex]\displaystyle \int e^a^u cos(bu) \ du = \frac{e^a^u \ sin(bu)}{b} - \int \frac{a \ sin(bu)\ e^a^u}{b} \ du[/tex]
Factor out a/b from the integral.
- [tex]\displaystyle \int e^a^u cos(bu) \ du = \frac{e^a^u \ sin(bu)}{b} - \frac{a}{b} \int sin(bu)\ e^a^u} \ du[/tex]
Notice that we have the same integral we started with. Let's plug this integral into the original IBP we did.
- [tex]\displaystyle \int e^a^u sin(bu) \ du = -\frac{e^a^u cos(bu)}{b} + \frac{a}{b} \big{[ }\frac{e^a^u sin(bu)}{b} - \frac{a}{b} \int sin(bu) \ e^a^u \ du \big{]}}[/tex]
Distribute a/b inside the parentheses.
- [tex]\displaystyle \int e^a^u sin(bu) \ du = -\frac{e^a^u cos(bu)}{b} + \frac{a \ e^a^u sin(bu)}{b^2} - \frac{a^2}{b^2} \int sin(bu) \ e^a^u \ du[/tex]
Factor 1/b out of the right side of the equation.
- [tex]\displaystyle \int e^a^u sin(bu) \ du = \frac{1}{b}\big{[} -e^a^u cos(bu)+ a \big{(}\frac{e^a^u sin(bu)}{b} \big{)} - \frac{a^2}{b} \int sin(bu) \ e^a^u \ du \big{]}[/tex]
Multiply both sides by b to get rid of 1/b.
- [tex]\displaystyle b \int e^a^u sin(bu) \ du = -e^a^u cos(bu)+ a \big{(}\frac{e^a^u sin(bu)}{b} \big{)} - \frac{a^2}{b} \int sin(bu) \ e^a^u \ du[/tex]
Add the integral to both sides of the equation.
- [tex]\displaystyle b \int e^a^u sin(bu) \ du + \frac{a^2}{b} \int sin(bu) \ e^a^u \ du= -e^a^u cos(bu)+ a \big{(}\frac{e^a^u sin(bu)}{b} \big{)}[/tex]
Factor the integral on the left side.
- [tex]\displaystyle \int e^a^u sin(bu) \ du \ \big{(} b + \frac{a^2}{b} \big{)}= -e^a^u cos(bu)+ a \big{(}\frac{e^a^u sin(bu)}{b} \big{)}[/tex]
[tex]\displaystyle \big{(}b+\frac{a^2}{b} \big{)} = \frac{b^2+a^2}{b}[/tex], so we can multiply both sides of the equation by [tex]\displaystyle \frac{b}{a^2+b^2}[/tex].
- [tex]\displaystyle \int e^a^u sin(bu) \ du = -e^a^u cos(bu)+ a \big{(}\frac{e^a^u sin(bu)}{b} \big{)} \big{(} \frac{b}{a^2+b^2} \big{)}[/tex]
Simplify the equation before multiplying everything by [tex]\displaystyle \frac{b}{a^2+b^2}[/tex].
- [tex]\displaystyle \int e^a^u sin(bu) \ du = \big{(}\frac{-e^a^u cos(bu) \ b + a \ e^a^u sin(bu)}{b} \big{)} \big{(} \frac{b}{a^2+b^2} \big{)}[/tex]
Multiply the two factors together. Notice that the two b's in the denominator and numerator, respectively, cancel out. We are left with:
- [tex]\displaystyle \int e^a^u sin(bu) \ du = \big{(}\frac{-e^a^u cos(bu) \ b + a \ e^a^u sin(bu)}{a^2+b^2} \big{)}[/tex]
Factor [tex]\displaystyle e^a^u[/tex] from the numerator.
- [tex]\displaystyle \int e^a^u sin(bu) \ du = \big{(}\frac{e^a^u( -b \ cos(bu) \ + a \ sin(bu)}{a^2+b^2} \big{)}[/tex]
Split the numerator and denominator to make it appear the same as the original question.
- [tex]\displaystyle \int e^a^u sin(bu) \ du = \frac{e^a^u}{a^2+b^2} \big{(} a \ sin(bu) - b \ cos(bu) \big{)}[/tex]
Since we are taking the integral of something, we can add a +C at the end to complete the problem.
- [tex]\displaystyle \int e^a^u sin(bu) \ du = \frac{e^a^u}{a^2+b^2} \big{(} a \ sin(bu) - b \ cos(bu) \big{)} + C[/tex]
This is equivalent to the proof that we are given, therefore, we proved the integral correctly.
