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find the integral.

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i have been trying to enter the site since yesterday did you all have the same problem?

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Other answers:

anyways there are many ways to solve this problem, one of them would be Taylor series the other one would be substitution method
@mathsmind yea ive been having problems with it to so i did the problems i know how to do until it finally works
@mathsmind i figured that one out im stuck on another one now
evaluate the intergal |dw:1361284859202:dw|
@mathsmind that one ^^^
sorry i meant i was having problems entering the website since yesterday
i know how to solve both integral in my head
@mathsmind how long are you going to be one today ?
it depends on my work, if i am busy or not
can you answer my question please?
were you able to log on to the site
@mathsmind which one ? lol
oh no i wasnt i answered it above
this website? openstudy
it wasnt loading
i felt like crying lol, with out this website i don't know what id do
ok let me reply to the last integral
the answer for the last integral would be pi/6
now by looking at the last integral you can find out from the table that it is equal to arcsin(x/2)
for the 4-x?
\[\int\limits_{0}^{1}\frac{ dx }{ \sqrt{4-x^2} }\]
is that the integral or not?
the answer is pi/6
how did you get it?
ok there are two ways to do that
one way is to look at the arctrigs integrals
you know the table of integrals right!
the other way is to go through full substitution method
lol no but ill look them up
ok do you want detailed solution then
the u sub thing ?
you let x=2u
x^2 = 4u^2
okay i get it so far
can you differentiate the last bit?
dx = 2du
\[\int\limits_{0}^{1}\frac{ 2du }{\sqrt{4-4u^2} }\]
\[\int\limits\limits_{0}^{1}\frac{ du }{\sqrt{1-u^2} }\]
your a math genuis lol
thanks that was kind of you, and you are a bright student
let me give you a third way how to do that
i understand the u-sub the most
i have three more integral questions
ok go ahed
you need to know different methods to solve one problem , that for future note...
altho the differentiate part is a bit hard
i have a phone call but still submit your work
but ill look back it later because i have to get this course done to graduate
no i meant you make y =2sinx sub
different substitution
if you look at the unit triangle you can figure out that you need a sin sub
find the intergal |dw:1361286908472:dw|
again you are going to get an arctan but now we need to look at the factors
just by looking at this integral we will get 4/3arctan(3x) +c
the question is what was x=? in this u sub
you need to assume that 3x=u, 9x^2=u^2, 3dx=du
\[\int\limits \frac{ 4du }{ 3(1+u^2) }=\frac{ 4 }{ 3}\arctan(u)+c\]
\[u=3x, \therefore \frac{ 4 }{ 3 }\arctan(3x)+c\]
that's your final answer
okay i just did the other one so i have one more integral question
again you can see we will get the arctan + something
your answer should be\[\frac{ 1 }{ 2 }\arctan(\frac{ 1 }{ 2}(x-1))+c\]
\[\int\limits\limits\limits \frac{ dx }{4+(1-x)^2}\]
now 2u=1-x, 4u^2=(1-x)^2,
now one thing you need to know is how to let the 4 be a one in the denominator
how do i know that?
ok let me show, it is clear from the u-sub
remember that 2u =1-x, 2du=-dx
okay so far so good lol
\[\int\limits\frac{ 2du }{4(1+(u)^2)}\]
can you see it better this way?
\[\frac{ 1 }{ 2 }\arctan(u)+c\]
\[2u=1-x, \space u=\frac{ 1 }{ 2 }(1-x)\]
\[\frac{ 1 }{ 2 }\arctan(\frac{ 1 }{ 2 }(1-x))+c\]
that's your final answer
okay thanks ! im working on some ill let you know if i need more help
you don't sound like you got the last one?
lol i havent fully looked at it yet im finishing a problem
so i read your mind well hehehe
lol yeahh i was wondering how you knew that,but im looking at it now
i still don't understand how the engagement scheme works on open study
i get that one
i think its by the questions you answer, and medals you receive
no i am not asking about those two, i am asking about the engagement part
the one that has learner next to it
lol i thinks its from asking questions cause thats how you learn right
okay i found two i need help on
go ahead
determine which of the integrals can be found using the basic integration formulas you have studied so far in the text.
i think you need to open a new thread because this one is getting heavy with BW
lol okay i will

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