Mathematics, 16.12.2020 06:50 srosrguezbracho

Prove the following integration formula: $\displaystyle \int e^{au}\sin(bu)\, du=\frac{e^{au}}{a^2+b^2}(a\sin(bu)-b\cos(bu))+C$

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Answer from: macybarham

See Explanation.

General Formulas and Concepts:

Pre-Algebra

Distributive PropertyEquality Properties

Algebra I

Combining Like TermsFactoring

Calculus

Derivative 1: $\frac{d}{dx} [e^u]=u'e^u$ Integration Constant CIntegral 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:

Step 1: Define Integral

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

Step 2: Identify Variables Pt. 1

Using LIPET, we determine the variables for IBP.

Use Int Rules 2 + 3.

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

Step 3: Integrate Pt. 1

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

Step 4: Identify Variables Pt. 2

Using LIPET, we determine the variables for the 2nd IBP.

Use Int Rules 2 + 3.

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

Step 5: Integrate Pt. 2

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

Step 6: Integrate Pt. 3

Integrate [Alg - Back substitute]: $\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{a}{b} [\frac{e^{au}sin(bu)}{b} - \frac{a}{b} \int ({e^{au} sin(bu)}) \, du]$ [Integral - Alg] Distribute Brackets: $\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2} - \frac{a^2}{b^2} \int ({e^{au} sin(bu)}) \, du$ [Integral - Alg] Isolate Original Terms: $\int {e^{au}sin(bu)} \, du + \frac{a^2}{b^2} \int ({e^{au} sin(bu)}) \, du= \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}$ [Integral - Alg] Rewrite: $(\frac{a^2}{b^2} +1)\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}$ [Integral - Alg] Isolate Original: $\int {e^{au}sin(bu)} \, du = \frac{\frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}}{\frac{a^2}{b^2} +1}$ [Integral - Alg] Rewrite Fraction: $\int {e^{au}sin(bu)} \, du = \frac{\frac{-be^{au}cos(bu)}{b^2} + \frac{ae^{au}sin(bu)}{b^2}}{\frac{a^2}{b^2} +\frac{b^2}{b^2} }$ [Integral - Alg] Combine Like Terms: $\int {e^{au}sin(bu)} \, du = \frac{\frac{ae^{au}sin(bu)-be^{au}cos(bu)}{b^2} }{\frac{a^2+b^2}{b^2} }$ [Integral - Alg] Divide: $\int {e^{au}sin(bu)} \, du = \frac{ae^{au}sin(bu) - be^{au}cos(bu)}{b^2} \cdot \frac{b^2}{a^2 + b^2}$ [Integral - Alg] Multiply: $\int {e^{au}sin(bu)} \, du = \frac{1}{a^2+b^2} [ae^{au}sin(bu) - be^{au}cos(bu)]$ [Integral - Alg] Factor: $\int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)]$ [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!

Answer from: faithforlife3573

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.

Answer from: Quest

No a triangle cannot be formed because when you add the two smallest number together (which would be 11) the have to be bigger than the number it is "against" ( which would be 12)

Answer from: Quest

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice

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