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joHn eDwArd's question at Yahoo! Answers regarding an indefinite integral

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MarkFL

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Feb 24, 2012
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Here is the question:

How to get the integral of (x^3 + 2x + sin x) e^(3x) dx?

please include the complete solution. thanks
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello joHn eDwArd,

We are given to evaluate:

\(\displaystyle \int\left(x^3+2x+\sin(x) \right)e^{3x}\,dx\)

I would write:

\(\displaystyle y=\int\left(x^3+2x+\sin(x) \right)e^{3x}\,dx\)

Now, differentiating with respect to $x$, we obtain the ODE:

\(\displaystyle \frac{dy}{dx}=\left(x^3+2x+\sin(x) \right)e^{3x}\)

We see the homogeneous solution is:

\(\displaystyle y_h(x)=c_1\)

and, using the method of undetermined coefficients, we should expect a particular solution of the form:

\(\displaystyle y_p(x)=\left(Ax^3+Bx^2+Cx+D+E\cos(x)+F\sin(x) \right)e^{3x}\)

Differentiating with respect to $x$ and substituting into the ODE, we obtain after dividing though by $e^{3x}$:

\(\displaystyle 3Ax^3+(3A+3B)x^2+(2B+3C)x+(C+3D)+(3F-E)\sin(x)+(3E+F)\cos(x)=x^3+2x+\sin(x)\)

Equating coefficients, we find:

\(\displaystyle 3A=1\,\therefore\,A=\frac{1}{3}\)

\(\displaystyle 3A+3B=0\,\therefore\,B=-A=-\frac{1}{3}\)

\(\displaystyle 2B+3C=2\,\therefore\,C=\frac{8}{9}\)

\(\displaystyle C+3D=0\,\therefore\,D=-\frac{C}{3}=-\frac{8}{27}\)

\(\displaystyle 3F-E=1\)

\(\displaystyle 3E+F=0\,\therefore\,E=-\frac{1}{10},\,F=\frac{3}{10}\)

And so we have:

\(\displaystyle y_p(x)=\left(\frac{1}{3}x^3-\frac{1}{3}x^2+\frac{8}{9}x-\frac{8}{27}-\frac{1}{10}\cos(x)+\frac{3}{10}\sin(x) \right)e^{3x}\)

\(\displaystyle y_p(x)=\frac{e^{3x}}{270}\left(90x^3-90x^2+240x-80-27\cos(x)+81\sin(x) \right)\)

And so by superposition, we have:

\(\displaystyle y(x)=y_h(x)+y_p(x)\)

Hence, we may conclude:

\(\displaystyle \int\left(x^3+2x+\sin(x) \right)e^{3x}\,dx=\frac{e^{3x}}{270}\left(90x^3-90x^2+240x-80-27\cos(x)+81\sin(x) \right)+C\)
 

Prove It

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Jan 26, 2012
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Here is the question:



I have posted a link there to this topic so the OP can see my work.
Here's how I would do the problem:

\(\displaystyle \displaystyle \begin{align*} \int{ \left[ x^3 + 2x + \sin{(x)} \right] e^{3x}\,dx} = \int{ \left( x^3 + 2x \right) e^{3x} \,dx} + \int{ \sin{(x)}e^{3x}\,dx} \end{align*}\)

So now we can evaluate each integral by using Integration by Parts.

[tex]\displaystyle \begin{align*} \int{ \left( x^3 + 2x \right) e^{3x}\,dx} &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \int{ \frac{1}{3} \left( 3x^2 + 2 \right) e^{3x}\,dx } \\ &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{3} \left[ \frac{1}{3} \left( 3x^2 + 2 \right) e^{3x} - \int{ \frac{1}{3} \left( 6x \right) e^{3x} \,dx } \right] \\ &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{9} \left( 3x^2 + 2 \right) e^{3x} + \frac{2}{3} \int{ x\,e^{3x}\,dx} \\ &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{9} \left( 3x^2 + 2 \right) e^{3x} + \frac{ 2}{3} \left[ \frac{1}{3} x\,e^{3x} - \int{ \frac{1}{3} e^{3x}\,dx } \right] \\ &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{9} \left( 3x^2 + 2 \right) e^{3x} + \frac{2}{9}x\,e^{3x} - \frac{2}{9} \int{ e^{3x}\,dx} \\ &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{9} \left( 3x^2 + 2 \right) e^{3x} + \frac{2}{9}x\,e^{3x} - \frac{2}{27}e^{3x} \end{align*}[/tex]

and

[tex]\displaystyle \begin{align*} I &= \int{\sin{(x)}e^{3x}\,dx} \\ I &= \frac{1}{3}\sin{(x)}e^{3x} - \int{ \frac{1}{3}\cos{(x)}e^{3x}\,dx } \\ I &= \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{3} \int{ \cos{(x)}e^{3x}\,dx} \\ I &= \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{3} \left[ \frac{1}{3}\cos{(x)}e^{3x} - \int{ -\frac{1}{3}\sin{(x)}e^{3x}\,dx } \right] \\ I &= \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{9}\cos{(x)}e^{3x} - \frac{1}{9}\int{\sin{(x)}e^{3x}\,dx} \\ I &= \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{9}\cos{(x)}e^{3x} - \frac{1}{9} I \\ \frac{10}{9}I &= \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{9}\cos{(x)}e^{3x} \\ I &= \frac{9}{10} \left[ \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{9}\cos{(x)}e^{3x} \right] \\ I &= \frac{3}{10}\sin{(x)}e^{3x} - \frac{1}{10}\cos{(x)}e^{3x} \end{align*}[/tex]

Therefore [tex]\displaystyle \begin{align*} \int{ \left[ x^3 + 2x + \sin{(x)} \right] e^{3x}\,dx}\end{align*}[/tex]

[tex]\displaystyle \begin{align*} = \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{9} \left( 3x^2 + 2 \right) e^{3x} + \frac{2}{9}x\,e^{3x} - \frac{2}{27}e^{3x} + \frac{3}{10}\sin{(x)}e^{3x} - \frac{1}{10}\cos{(x)}e^{3x} + C \end{align*}[/tex]
 
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MarkFL

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Feb 24, 2012
13,775
Here's how I would do the problem:...
I truly appreciate you taking the time to present this so nicely! The OP is probably expected to use IBP here, but I got "lazy" and decided to treat it as an ODE instead. (Wasntme)
 

Prove It

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Jan 26, 2012
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I truly appreciate you taking the time to present this so nicely! The OP is probably expected to use IBP here, but I got "lazy" and decided to treat it as an ODE instead. (Wasntme)
Don't worry Mark, your solution is quite beautiful, and you weren't the only lazy one. I could not be arsed taking out common factors and collecting like terms (if there are any) hahaha.