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swat32
2 years ago
12

Prove the following integration formula:

Mathematics
2 answers:
horsena [70]2 years ago
7 0

Prove: \displaystyle \int e^a^u sin(bu)\ du = \frac{e^a^u}{a^2+b^2} (a \ sin(bu) - b \ cos(bu)) + C

Integration by parts formula: \displaystyle \int udv = uv - \int vdu

Find u, du, v, and dv for this function: \displaystyle \int e^a^u sin(bu)

  • \displaystyle u =e^a^u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = -\frac{cos(bu)}{b} \\ du=ae^a^u \ du \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = sin(bu) \ du

Plug these values into the IBP formula.

  • \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

Multiply and simplify the factors. Factor the negative out of the integral.

  • \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

Factor out a/b from the integral.

  • \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

Now we are going to apply IBP to the function: \displaystyle \int cos(bu) \ e^a^u . Find u, du, v, and dv for this function.

  • \displaystyle u =e^a^u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \frac{sin(bu)}{b} \\ du=ae^a^u \ du \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = cos(bu) \ du

Plug these values into the IBP formula.

  • \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

Multiply and simplify the factors.

  • \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

Factor out a/b from the integral.

  • \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

Notice that we have the same integral we started with. Let's plug this integral into the original IBP we did.

  • \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{]}}

Distribute a/b inside the parentheses.

  • \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

Factor 1/b out of the right side of the equation.

  • \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{]}

Multiply both sides by b to get rid of 1/b.

  • \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

Add the integral to both sides of the equation.

  • \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{)}

Factor the integral on the left side.

  • \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{)}

\displaystyle \big{(}b+\frac{a^2}{b} \big{)} = \frac{b^2+a^2}{b}, so we can multiply both sides of the equation by \displaystyle \frac{b}{a^2+b^2}.

  • \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{)}

Simplify the equation before multiplying everything by \displaystyle \frac{b}{a^2+b^2}.

  • \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{)}

Multiply the two factors together. Notice that the two b's in the denominator and numerator, respectively, cancel out. We are left with:

  • \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{)}

Factor \displaystyle e^a^u from the numerator.

  • \displaystyle \int e^a^u sin(bu) \ du  = \big{(}\frac{e^a^u( -b \ cos(bu) \ + a \ sin(bu)}{a^2+b^2} \big{)}

Split the numerator and denominator to make it appear the same as the original question.

  • \displaystyle \int e^a^u sin(bu) \ du  = \frac{e^a^u}{a^2+b^2} \big{(} a \ sin(bu) - b \ cos(bu) \big{)}

Since we are taking the integral of something, we can add a +C at the end to complete the problem.

  • \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

This is equivalent to the proof that we are given, therefore, we proved the integral correctly.

7nadin3 [17]2 years ago
6 0

Answer:

See Explanation.

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Distributive Property
  • Equality Properties

<u>Algebra I</u>

  • Combining Like Terms
  • Factoring

<u>Calculus</u>

  • Derivative 1:                  \frac{d}{dx} [e^u]=u'e^u
  • Integration Constant C
  • Integral 1:                      \int {e^x} \, dx = e^x + C
  • Integral 2:                     \int {sin(x)} \, dx = -cos(x) + C
  • Integral 3:                     \int {cos(x)} \, dx = sin(x) + C
  • Integral Rule 1:             \int {cf(x)} \, dx = c \int {f(x)} \, dx
  • Integration by Parts:    \int {u} \, dv = uv - \int {v} \, du
  • [IBP] LIPET: Logs, Inverses, Polynomials, Exponents, Trig

Step-by-step Explanation:

<u>Step 1: Define Integral</u>

\int {e^{au}sin(bu)} \, du

<u>Step 2: Identify Variables Pt. 1</u>

<em>Using LIPET, we determine the variables for IBP.</em>

<em>Use Int Rules 2 + 3.</em>

u = e^{au}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = sin(bu)du\\du = ae^{au}du \ \ \ \ \ \ \ \ \ v = \frac{-cos(bu)}{b}

<u>Step 3: Integrate Pt. 1</u>

  1. Integrate [IBP]:                                           \int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} - \int ({ae^{au} \cdot \frac{-cos(bu)}{b} }) \, du
  2. Integrate [Int Rule 1]:                                                \int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{a}{b} \int ({e^{au}cos(bu)}) \, du

<u>Step 4: Identify Variables Pt. 2</u>

<em>Using LIPET, we determine the variables for the 2nd IBP.</em>

<em>Use Int Rules 2 + 3.</em>

u = e^{au}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = cos(bu)du\\du = ae^{au}du \ \ \ \ \ \ \ \ \ v = \frac{sin(bu)}{b}

<u>Step 5: Integrate Pt. 2</u>

  1. Integrate [IBP]:                                                  \int {e^{au}cos(bu)} \, du = \frac{e^{au}sin(bu)}{b} - \int ({ae^{au} \cdot \frac{sin(bu)}{b} }) \, du
  2. Integrate [Int Rule 1]:                                    \int {e^{au}cos(bu)} \, du = \frac{e^{au}sin(bu)}{b} - \frac{a}{b} \int ({e^{au} sin(bu)}) \, du

<u>Step 6: Integrate Pt. 3</u>

  1. Integrate [Alg - Back substitute]:     \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]
  2. [Integral - Alg] Distribute Brackets:          \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
  3. [Integral - Alg] Isolate Original Terms:     \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}
  4. [Integral - Alg] Rewrite:                                (\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}
  5. [Integral - Alg] Isolate Original:                                    \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}
  6. [Integral - Alg] Rewrite Fraction:                          \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} }
  7. [Integral - Alg] Combine Like Terms:                          \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} }
  8. [Integral - Alg] Divide:                                  \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}
  9. [Integral - Alg] Multiply:                               \int {e^{au}sin(bu)} \, du = \frac{1}{a^2+b^2} [ae^{au}sin(bu) - be^{au}cos(bu)]
  10. [Integral - Alg] Factor:                                 \int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)]
  11. [Integral] Integration Constant:                     \int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)] + C

And we have proved the integration formula!

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