Working from the general form integral

[dejet]

Homework Statement

lot of work to go around here so need to see if i did it right. thanks!
$$\int e^2xsin3xdx$$

Homework Equations

$$\int uav=uv-\int vdu$$

The Attempt at a Solution

$$\int\frac{-e^{2x}}{3}cos3x+\frac{2}{3}\int(e^{2x})cos3xdx$$

parts work: let u=e^2x, dv= sin3xdx
so: du=2e^2x, and v= $$\frac{-1}{3}cos3x$$

next:$$\int(e^{2x}cos3xdx)=\frac{e^{2x}}{3}sin3x-\frac{-2}{3}\int(e^{2x}sin3xdx)$$

work:let u=$$\frac{-e^{2x}}{3}, dv= cos3xdx, du= 2e^{2x} v= \frac{1}{3}sin3x$$

now: $$\int e^{2x}sin3xdx= \frac{-e^{2x}}{3}cos3x+\frac{2}{3}[\frac{e^{2x}}{3}sin3x-\frac{2}{3}\int e^{2x}sin3xdx]$$

" " "= $$\int e^{2x}sin3xdx+ \frac{-4}{9} \int e^{2x}sin3xdx= \frac{-e^{2x}}{3}cos3x+\frac{2}{9}e^{2x}sin3x$$

$$\frac{9}{13}\frac{13}{9}\int e^{2x}sin3xdx= [\frac{-e^{2x}}{3}cos3x+\frac{2}{9}e^{2x}sin3x]\frac{9}{13}$$

$$\int e^{2x}sin3xdx = \frac{-3}{13}e^{2x}cos3x+ \frac{2}{13}e^{2x}sin3x+C = e^{2x} [ \frac{2}{13}sin3x- \frac{3}{13}cos3x+C$$

ok now last: a=2 b=3

$$\int e^{2x}sin3xdx= \frac{1}{2^{2}+3^{2}}e^{2x}[2sin3x-3cos3x]+c= e^{2x}[\frac{2}{13}sin3x-\frac{3}{13}cos3x]+C$$

done. i think. we learned this a few days back and was given it to see if we can do it. it is extra work so it wont kill my grade or anything for getting it wrong, but it will help me if i can get it right. i looked it over and to me it looks fine just want to see if anyone can see if i overlooked something.

also i hope it looks right this is the 1st time i am using this sight(well used it few days back but not for anything like this)

 

[dynamicsolo]

I agree with your result, working from the general form. (This is one of those integrations-by-parts where it is very important to track your layers of integration correctly to make sure you pick up all the minus signs and multiplicative constants...)

 

[Defennder]

I can't figure out what you are writing here:[

$$\int\frac{-e^{2x}}{3}cos3x+\frac{2}{3}\int(e^{2x})cos3xdx$$

Where did this come from? How is this related to the original integral? If you meant to break up the original integral into 1/3 and 2/3, I should ask, why is that done?

next:$$\int(e^{2x}cos3xdx)=\frac{e^{2x}}{3}sin3x-\frac{-2}{3}\int(e^{2x}sin3xdx)$$

Should be $$-\frac{e^{2x}}{3}sin3x$$ instead if I'm reading it properly.

now: $$\int e^{2x}sin3xdx= \frac{-e^{2x}}{3}cos3x+\frac{2}{3}[\frac{e^{2x}}{3}sin3x-\frac{2}{3}\int e^{2x}sin3xdx]$$

It should be $$-\frac{1}{3}e^{2x}cos(3x) + \frac{2}{3} \int e^{2x}cos(3x) dx$$. Right?

The rest is kind of too messy to follow. Probably they are erroneous since the above working is. Double check them.

 

[dynamicsolo]

Yes, there are a number of mistakes in the TeX lines, but dejet seems to have done the math correctly, but the result for the anti-derivative is right. (A grader, though, would certainly have taken some few points off for this...)