/sci/ - Science & Math

[Anonymous No.3785482]

/sci/ I don't understand U integration

for those who don't get what I mean like

integrate ( sin(3x)/cos(3x) dx)

I get the U = cos(3x), the Du = sin (3x)

so you end up with DU/U

but I have no idea what to do after or why I'm doing it, i've tried kahns academy but couldn't find anything helpful... so can you guys explain to me what to do after that part and why this gives the correct answer?

[Anonymous No.3785496]

if you don't understand than i see no point in telling you

[Anonymous No.3785509]

Du/U?

As in, 1/U du?

The integrate 1/U du.

[almostphysicist No.3785530]

>>3785482
You did that wrong.

U = cos(3x)
dU = -3sin(3x)dx
therefore dx = dU / -3sin(3x)

so now plug in U for cos(3x) and dU/-3sin(3x) for dx

You'll notice that you have sin(3x) on the top and bottom, they will cancel out.

Magic.

[Anonymous No.3785531]

u = cos(3x)
du/dx = 3sin(3x)
du = 3sin(3x) dx

then you have

1/3 intgeral 1/u du

[Anonymous No.3785541]

>>3785509
then you get ln U

so... the right answers ln 3sin 3x

[Anonymous No.3785549]

>>3785541
It's actually -3sin(3x). that should be a -1/3 in the derivative

[Anonymous No.3785550]

>>3785541
err ignore me

the correct answer is ln cos 3x

[Anonymous No.3785563]

fist off, your du is supposed to be -3 sin (3x). remember to use the chain rule. Then you have to to make you equation look like your sub you need to multiply the integral by a -1/3. because du/u (-3sin(3x) / cos (3x)) does not = (sin(3x)/cos(3x)). so your integral is actually -1/3 * integral of du/u. Then just by properties of ln you should know ln IxI d/dx = dx/x so the integral would becom -1/3 ln I u I -> -1/3 ln Icos(3x)I for z -> y with z being your lower bound and y being your upper.

[Anonymous No.3785602]

>>3785563
now that i have mastered math off to the arts and letter department to learn to spell!

[Anonymous No.3785606]

<span class="math">
\int{a, b}
[/spoiler]
test

[Anonymous No.3785610]

<span class="math">
\int[a,b]
[/spoiler]
test test... 123

[Anonymous No.3785622]

>>3785563
wow that is actually pretty understandable

thank you fine gent

[Anonymous No.3785640]

wow fuck /sci/ has yet to lay this question out for this shitcunt

I = int [ sin3x / cos3x]dx
u=sin3x, du/dx=3cos3x= dx=1/3(du/cos3x)

I = (1/3) int [du/cos3x (u/cos3x)] =
1/3 int u.du / 1-u^2 = 1/3 int (-2u / 1-u^2) (-1/2) = -1/6 ln (1-u^2)+c=1/6 ln(1-sin^2(3x))+c = 1/6 ln(1-(1-cos^2(3x))+c
=1/6ln(cos^2(3x))+c

or conversely much simpler ...

int sin3x/cos3x = int sin3x/cos3x (by log laws)
int f'(x)/f(x) = log_e(f(x))+c

therefore int sin3x/cos3x = -1/3(cos3x)+c = 1/3(cos3x)^-1 + c (by log laws) = 1/3 (sec3x)+c

[Anonymous No.3785648]

>>3785640
.....
where =1/6ln(cos^2(3x))+c = -2/6(cos(3x))+c (by log laws)
= -1/3cos(3x)+c=1/3(sec3x)+c

[Anonymous No.3785670]

<span class="math">
\int sin(x)cos(x) dx
<span class="math">
oh that looks hard..
but we see that if we say

u = sin(x)

then we can eliminate all x's
(we must eliminate all of the old variable)
because we know that

du/dx = cos(x)

and you must have heard "du/dx" is not division, unless it is convenient to think of it that way.
Well it is convenient.

du = cos(x)dx

Well look at that! all x's can be eliminated!

\int u du = (1/2)u^2

well that was magical
put the x's back now

(1/2)sin^2(x)

and slashmath[/spoiler][/spoiler]

[Anonymous No.3785697]

offcourse! restart!
<span class="math">
\int sin(x)cos(x) dx
[/spoiler]
oh that looks hard.. but we see that if we say
<span class="math">
u = sin(x)
[/spoiler]
then we can eliminate all x's
(we must eliminate all of the old variable) because we know that
<span class="math">
du/dx = cos(x)
[/spoiler]
and you must have heard "du/dx" is not division, unless it is convenient to think of it that way.
Well it is convenient.
<span class="math">
du = cos(x)dx
[/spoiler]
Well look at that! all x's can be eliminated!
<span class="math">
\int u du = (1/2)u^2
[/spoiler]
well that was magical put the x's back now
<span class="math">
(1/2)sin^2(x)
[/spoiler]
and slashmath

[Anonymous No.3785726]

Also,
<span class="math">
3/4 = (1/4)*3
[/spoiler]
so
<span class="math">
du/u = (1/u)du
[/spoiler]
even tho ur integrand was malformed to begin with

[Anonymous No.3785772]

<span class="math">F(u)\cdot cK \Rightarrow O(f^{f(\mathbf{u})})[/spoiler]