please help with math (integrals)
category: general [glöplog]
please help. anyone is able to calculate (integral) of xcos(5x) dx ?
6
(x/5)*sin(5x)+C
1/25*cos(5*x)+1/5*x*sin(5*x)
use integration by parts: http://mathworld.wolfram.com/IntegrationbyParts.html
no! use partial integration: http://mathworld.wolfram.com/PartialIntegration.html
wow, you people actually know this!
derive says:
COS(5·x)/25 + x·SIN(5·x)/5
COS(5·x)/25 + x·SIN(5·x)/5
pi
mmmmm pi
oooh derive!! old good friend! :)
Alcohol and calculs don't mix. Never drink and derive!
I demand a recount! =(
null:
You need both partial integration and the substitution method for solving this task. If you want me to explain it in more detail, please ask me. Here's the way I solved it:
Formula of partial integration:
Int u(x) v'(x) dx = u(x) v(x) - Int u'(x) v(x) dx
I choose:
v'(x) := cos(5x)
=> v(x) = 1/5 sin(5x)
u(x) := x
=> u'(x) = 1
Using substitution method, we can transform:
Int cos(5x) dx = Int cos(z) dz
by defining z := 5x
=> z' = 5 = dx/dz
=> dz = dx / 5
From which follows (using the formula for Int cos(x) dz = sin(x)):
Int cos(z) dz = 1/5 sin(5x)
Let's insert this:
Int x cos(5x) = 1/5 x sin(5x) - 1/5 Int sin(5x) dx
In order to compute Int sin(5x) dx, we use the substitution method once again, and we get:
Int sin(5x) dx = -1/5 cos(x)
So we get:
1/5 x sin(5x) - 1/5 Int sin(5x) dx = 1/5 x sin(5x) + 1/25 cos(5x)
Quod erat demonstrandum.
You need both partial integration and the substitution method for solving this task. If you want me to explain it in more detail, please ask me. Here's the way I solved it:
Formula of partial integration:
Int u(x) v'(x) dx = u(x) v(x) - Int u'(x) v(x) dx
I choose:
v'(x) := cos(5x)
=> v(x) = 1/5 sin(5x)
u(x) := x
=> u'(x) = 1
Using substitution method, we can transform:
Int cos(5x) dx = Int cos(z) dz
by defining z := 5x
=> z' = 5 = dx/dz
=> dz = dx / 5
From which follows (using the formula for Int cos(x) dz = sin(x)):
Int cos(z) dz = 1/5 sin(5x)
Let's insert this:
Int x cos(5x) = 1/5 x sin(5x) - 1/5 Int sin(5x) dx
In order to compute Int sin(5x) dx, we use the substitution method once again, and we get:
Int sin(5x) dx = -1/5 cos(x)
So we get:
1/5 x sin(5x) - 1/5 Int sin(5x) dx = 1/5 x sin(5x) + 1/25 cos(5x)
Quod erat demonstrandum.
Choose life, choose Derive.
Mathematica for life
i chose not to choose life, i chose something else.
i've chosen to love.
adok, you are damn fucking complicated.
1) null already solved his/her own problem, see the 4th entry.
2) working method. first try: x*sin(5x). that doesn't work, but almost do; so let's try to correct it: x*sin(5x)/5 + cos(5x)/25. that's all.
3) choose washing machines, cars, compact disc players and a computer with some math software, but before that, learn some math yourself
1) null already solved his/her own problem, see the 4th entry.
2) working method. first try: x*sin(5x). that doesn't work, but almost do; so let's try to correct it: x*sin(5x)/5 + cos(5x)/25. that's all.
3) choose washing machines, cars, compact disc players and a computer with some math software, but before that, learn some math yourself
Damn, I should be studying now ;P
blala: What I understood was that he didn't just need the solution, but the way of solving it.
le compte est bon!
adok: so you presented him an overcomplicated page-long argument, starting right with unmotivated "clever" choices - now he surely understand everything... if you want to teach, try instead to teach how to solve such problems in general. since general integration algorithms don't exist, i think the "try something which resembles what we want, and then try to gradually correct it" method is quite useful.
It's simple, it's a formula which you memorize and use when it's appropriate. That's it.
Memorizing formulas is generally a bad idea. The smart thing to do is to memorize and understand the ways to derive them. As an added bonus, you'll learn the formulas as well
