[SPLIT] int [e^(ax)sin(bx)] dx, int [e^(ax)sin(bx)] dx

[warwick]

A couple more

21. integral of e^(ax)sin(bx)dx

= sin(bx) (1/a)e^(ax) - integral of b cos(bx)(1/a)e^(ax)dx

(2) integral of e^(ax)sin(bx)dx = [sin(bx)(1/a)e^(ax) - b cos(bx)(1/a^2)e^(ax)] (a^2/b^2)

27. integral of x^(3)e^(x^2)dx

= e^(x^2)(1/4)x^(4) - integral of 2xe^(x^2)x^(3)dx

In a situation like this, I'm not sure what to make the u and v' for the second integration by parts.

 

[galactus]

Re: A couple more

\(\displaystyle \int{e^{ax}sin(bx)dx\)

Let \(\displaystyle \L\\u=e^{ax}, \;\ dv=sin(bx)dx, \;\ du=ae^{ax}dx, \;\ v=\frac{-1}{b}cos(bx)\)

27. integral of x^(3)e^(x^2)dx

= e^(x^2)(1/4)x^(4) - integral of 2xe^(x^2)x^(3)dx

In a situation like this, I'm not sure what to make the u and v' for the second integration by parts.

You don't necessarily need parts for this one, unless you're mandated to.

It's easier to use a u subbing.

Let \(\displaystyle \L\\u=x^{2}, \;\ du=2xdx, \;\ \frac{du}{2}=xdx\)

Make the subs and get:

\(\displaystyle \L\\\frac{1}{2}\int{ue^{u}}du\)

Now, integrate and resub.

BUT, if you have to use parts:

Let \(\displaystyle \L\\u=x^{2}, \;\ dv=xe^{x^{2}}, \;\ du=2xdx, \;\ v=\frac{1}{2}e^{x^{2}}\)

 

[warwick]

Re: A couple more

So, for 21 I just used the wrong u and v' starting out?

For 27, I thought about changing the parts up, but the professor said it's a good idea to keep the u's consistent, or something like that. I now see the integral ue^udu on my handy little chart. Haha.

But.... Wait. Isn't there an extra x in your subbed integral for #27?

 

[Deleted member 4993]

Re: A couple more

Wait. Isn't there an extra x in that integral for #27?

Yes but that is taken up in dv

dv = x * e^x dx

 

[warwick]

Re: A couple more

Wait. Isn't there an extra x in that integral for #27?

Yes but that is taken up in dv

dv = x * e^x dx

Oh, I see. I was subbing the second integral I had. I'm in the Integration by Parts chapter, but the homework set says just to Find the Integral. It seems I was trying to do parts the whole time when I clearly could have done it by u substitution. Haha. The parts I chose initially must not be possible or at least easy, I suppose.

I chose

u = e^(x^2)
u' = 2xe^(x^2)
v = (1/4)x^4
v' = x^3dx

 

[galactus]

Sometimes the correct parts to use isn't so obvious. It's a matter of practice and observation.

 

[warwick]

Sometimes the correct parts to use isn't so obvious. It's a matter of practice and observation.

Is there a good way to tell if you're heading in the right direction? I know sometimes you have to use the IBP technique at least twice in some problems.

 

[galactus]

A lot of times you can tell because the wrong substitutions lead you down the primrose path and you end up with a monster of an integral for the \(\displaystyle \int{vdu}\) part. You may have to try different options until you find the 'friendly' one. No 'cast in stone' method, per se.

Like anything else in life, the more you do the better you get.