**sovlved**

[mdchesne]

Integral of [ sin(3x) dx
is this just [-cos(3x)]/3 ?

Integral of [arctan(7x)]
??

 

[shadduphoe]

Originally posted by: mdchesne
Integral of [ sin(3x) dx
is this just [-cos(3x)]/3 ?

Integral of [arctan(7x)]
??

easy,


.1 /1 .11 1.11 .111

 

[djpolstee]

linky

 

[mdchesne]

Originally posted by: djpolstee
linky

perfect! wow, love that link. bookmarking that immediately.

Originally posted by: shadduphoe

easy,


.1 /1 .11 1.11 .111

WTF? stop posting crap

 

[david46675]

Originally posted by: djpolstee
linky

holy rusted metal batman!

 

[mdchesne]

yea, i thought it was awesome too, but it doesn't like solving integrals by parts.
Like, the integral of xln(x) is
1/2ln(x)x^2 - 1/4x^2 + c
but mathematica only gives
(x^2*x*ln)/2

so... yea... it's kinda weak. or else I'm not pluggin it in right. oh well, nice link though. i have to play with it more once i have time tonight

 

[Kyteland]

Originally posted by: mdchesne
Integral of [ sin(3x) dx
is this just [-cos(3x)]/3 ?

Integral of [arctan(7x)]
??

The formulas you need are these:
?sin(x) dx = -cos(x)+C
?arctan(x) dx = x*arctan(x)-ln(1+x^2)/2+C

To actually do the integration, you need touse substitutions. ?sin(3x) dx doesn't quite match the generic rule you have above, so you need to manipulate it until it does.
3x = u
3 dx = du
dx = du/3

Now you can substitute in to the rule.
?sin(3x) dx
=?sin(u) du/3
=(1/3)*?sin(u) du
=(1/3)*(-cos(u))+C
=cos(3x)/3+C

The other one is similar:
7x = u
7 dx = du
dx = du/7

So repeat the step from above:
?arctan(7x) dx
= ?arctan(u) du/7
= (1/7)*?arctan(u) du
=(1/7)*(u*arctan(u)-ln(1+u^2)/2+C)
=(1/7)*(7x*arctan(7x)-ln(1+(7x)^2)/2+C)
=x*arctan(7x)-ln(1+49*x^2)/14+C

 

[Kyteland]

Originally posted by: mdchesne
yea, i thought it was awesome too, but it doesn't like solving integrals by parts.
Like, the integral of xln(x) is
1/2ln(x)x^2 - 1/4x^2 + c
but mathematica only gives
(x^2*x*ln)/2

so... yea... it's kinda weak. or else I'm not pluggin it in right. oh well, nice link though. i have to play with it more once i have time tonight

You have to give it the proper input. Take a look at the botton "How to enter input" below the input box. x*Log[x] gives the answer you're looking for.

 

[mdchesne]

i thought the arctan needed to be solved for by integration of parts. where it would be 1 * arctan(u) where u = 7x. or did i just completely miss what you did and you actually have it in there (i'm tired so I can't concentrate now, lol)?

 

[Kyteland]

Your substitution is u=7x. Taking the derivative of both sides gives you du=7 dx. When you start with ?arctan(7x) dx you need to substitute the pieces you solved earlier. You know 7x=u and dx=du/7 which gives ?arctan(u) du/7. You know the general rule to solve ?arctan(u) du, but this still leaves everything in terms og u and you need them in terms of x. Since you know u=7x you can make the simple substitution. to get the final answer.


Edit: Maybe I'm forgetting what integration by parts is, but isn't that ?u dv = uv - ?v du? I don't see why that would apply to this problem? That's more useful for stuff like ?x*arctan(x) dx.

I'll have to dig out my calc book when I get home tonight....

 

[DaShen]

That link is kind of cool. The parser must be written well.