MHB JoHn eDwArd's question at Yahoo Answers regarding an indefinite integral

AI Thread Summary
The discussion centers on solving the integral of (x^3 + 2x + sin x) e^(3x) dx. Two main methods are presented: one using the method of undetermined coefficients to derive a particular solution and another employing integration by parts (IBP). The final solution combines results from both approaches, yielding a complex expression involving exponential, sine, and cosine functions. Participants express appreciation for the clarity of the solutions and acknowledge the different strategies used. The conversation highlights the flexibility in solving integrals and the collaborative nature of mathematical problem-solving.
MarkFL
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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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Hello joHn eDwArd,

We are given to evaluate:

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

I would write:

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

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

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

We see the homogeneous solution is:

$$y_h(x)=c_1$$

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

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

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

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

Equating coefficients, we find:

$$3A=1\,\therefore\,A=\frac{1}{3}$$

$$3A+3B=0\,\therefore\,B=-A=-\frac{1}{3}$$

$$2B+3C=2\,\therefore\,C=\frac{8}{9}$$

$$C+3D=0\,\therefore\,D=-\frac{C}{3}=-\frac{8}{27}$$

$$3F-E=1$$

$$3E+F=0\,\therefore\,E=-\frac{1}{10},\,F=\frac{3}{10}$$

And so we have:

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

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

And so by superposition, we have:

$$y(x)=y_h(x)+y_p(x)$$

Hence, we may conclude:

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

[math]\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*}[/math]

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

\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*}

and

\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*}

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

\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*}
 
Prove It said:
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)
 
MarkFL said:
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.
 
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