Brainlets need not apply

INTEGRATE SIN(LN(X)) NOW
NOW!

NNOOWWW!!!!!!!!

Other urls found in this thread:

uapk.org/hp-prime-pro-v-1-3-2-apk/
metadefender.com/#!/results/file/YTE3MDUxOEhKQXlscC1zZ1pyeUplSmVwV2llLQ/regular/analysis
twitter.com/SFWRedditImages

APOLOGIZE

u = log(x)

No you integrate it

> log(x)
so base 2? i'm a computer science major you can't pull the wool over my eyes

[math]
\int sin(ln(x))dx = \int sin(u)e^{u}du
[/math]
parts
[math]
w = e^{u}\\
dv = sin(u) \\
dw = e^{u} \\
v = -cos(u) \\
= -e^{u}cos(u) + \int cos(u)e^{u}du\\
dw = e^{u}\\
v = sin(u)\\
= -\frac{1}{2}e^{u}cos(u) + \frac{1}{2} e^{u}sin(u)
[/math]

forgot to resub,
-0.5*x*cos(ln(x)) + 0.5*x*sin(ln(x))

Those who use the word brainlet are even more brain-damaged than a retarded.

Another thing you could do is put put sin(x) in exponential form ie
sin(x) = (exp(ix)-exp(-ix))/2i
which simplifies the expression to
(x^i-x^(-i))/2i
the integral becomes trivial from here provided you're careful.

I am a confirmed brainlet and need to learn how to integrate complex functions.

Oh, well it's pretty simply, think about derivatives for example, the derivative of ((1-i)/2)x^(i+1) is x^i, doing the same procedure with x^(-i) and remembering to collect your terms and it's pretty easy.

sin(ln(x))dx/dx=sin(ln(x))/ln(x)=sin(x)=cos(x)

>2017
>not writing [math]\log[/math]

Use integration by parts twice with 1=du and sin(ln(x)) = v
You will get an expression like cos(ln(x)), integrate that by parts again and you end up with answer.

Sin (lnx)=Im [e^ilnx]=x^i

I hate that I know I'm smart but have no idea wtf is going on. Why is it so hard to get into it? I don't think I should have to spend hours on this. I should be able to read a couple paragraphs of shit. It's just a graph ffs

As long as I'm doing your homework for you, perhaps you would also like me to wash your dirty laundry?

t. engineer

Also a good approach! May be a little weird trying to put it back into a real form, and I'm not sure that the fact that integration rules for power functions extends to complex exponents is entirely trivial to prove, but still elegant. I like it, user.

battletoads?

link me to the apk for this pls

ln: base=e
lg: base=10

log: base=anything

uapk.org/hp-prime-pro-v-1-3-2-apk/

Here's the analysis:
metadefender.com/#!/results/file/YTE3MDUxOEhKQXlscC1zZ1pyeUplSmVwV2llLQ/regular/analysis

What's the deal with [math]*some math*[math]. Is that some /sci language or it can be put in an online calculator or sumtin to write the expression?

>[math]sin(ln(x))[/math]
>not [math]\sin(\log(x))[/math]

The brainlets are at it again.

Im [e^ilnx]=x^i

Explain this

e^(i·ln(x)) =[e^i]·[e^ln(x)] = (e^i)·x

exp(i*ln(x)) =/= exp(i)*exp(ln(x)) this is easy to see by simply noting that exp(i)*exp(ln(x)) = exp(i+ln(x)) while exp(i*ln(x)) = exp(ln(x))^i = x^i

thanks

It's how math is expressed properly in Tex, so naturally all the autistic math PhDs prefer it. Writing more complex notation like integrals is where it really becomes necessary, but it should be used for all math expressions.

i got (x/2) sin(ln(x)) (1-i)

made a couple of small mistakes. after revision i have the result:

(x/2) [sin(ln(x))-cos(ln(x)]+constant

Trolling?

ilnx=ln (x^i).

The operation you've written isn't at all valid.

ln and log are different. Just because you use retarded unspecific notation doesn't mean it's the easiest or cleanest. I bet you use sin^-1 instead of arcsin too.