Help me mathfags
Help me mathfags
Nathan Sanders
Other urls found in this thread:
Dylan Reed
74
Dominic Gonzalez
Math me helpfags
Camden Watson
Equals sqrt(Pi/2).
Asher Butler
Last two digits of my post in the answer
Julian Ortiz
-e^(cot(x)) (-1 + cot(x))
Jeremiah Stewart
First thoughts: rewrite cot as cos/sin; let u=sin;???;profit?
Joseph Ramirez
I'm too drunk to help, but use these resources.
Khan Academy
PatrickJMT
ProfRobBob
Professor Leonard
>Symbolab only as a last resort
Andrew Turner
Have you not heard of wolframalpha?
Nolan Fisher
holy shitballs OP. This is fucking nasty. With no pencil/paper on hand, the only thing I can say is go to Wolfram Alpha
Luis Nelson
Either way, the answer is -e^[cot(x)] + C.
Just look at the function and its super obvious.
Benjamin Richardson
...
Isaiah Myers
this. use u-sub if not ibp
Ayden Wright
Gabriel Barnes
The indefinite integral of e to the power of any non-polynomial function cannot be solved for real values. Meaning there's gonna be an i there somewhere.
Wyatt Murphy
-e^cot(x) * (cot(x) - 1) + c
...I think
Dylan Powell
> ProfRobBob
Mein Neiger! BAM!
Also fuck calculus II, wait till you get to decomposition. Shits bananas.
Adam Johnson
let u be cot you fucking moron
Lucas Hughes
Your fucking stupid
Samuel Lewis
>-e^[cot(x)] + C
I think you forgot the first part, which just goes to cot(x) - 1 + C after substituting integrating
Aiden Flores
Op use trig identities to transform cot(x)/sin^2(x) into cot(x)*csc^2(x)
then use u substitution on cot(x), and you get -u*e^u du. Then int by parts and bobs your uncle.
Chase Martin
>Indefinite integral of e to any non-polynomial function can't be solved for real values
That is bullshit
You are wrong
Michael Scott
or you can go to the Sup Forums message boards and have a bunch of faggots and shitlords to the thinking for you. Works just as well
James Edwards
Kek, his loss.
Wyatt Garcia
-(e^cotx)*(cotx-1)+C
First you substitute u=cotx, since -du=[(cscx)^2]dx
Then you use integration by parts.
James Morgan
what shit are you smoking?
Leo Harris
88 Jiggawatts
Kayden Hall
Thank you im new to both the u-substitution and the cot/csc trigs. Much appreciated
Julian Ross
cot(x)e^(cot(x))-e^(cot(x))+c
Thomas Bennett
...
Michael Rodriguez
here OP i solved it for you the same way i got through it in school
Nolan Morales
...
Ian Ross
42
James Edwards
(you have to simplify the answer because there's no built-in cot function, like i replaced it in the input)
((-(cos(x)-sin(x)))*e^(1/(tan(x))))/(sin(x))
which can be rewritten as -e^(cot(x)) (cot(x) - 1)