anonymous
  • anonymous
find the integral.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
-5^-x/(log(5))+C
anonymous
  • anonymous
|dw:1361280612277:dw|
anonymous
  • anonymous
i have been trying to enter the site since yesterday did you all have the same problem?

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anonymous
  • anonymous
anyways there are many ways to solve this problem, one of them would be Taylor series the other one would be substitution method
anonymous
  • anonymous
@mathsmind yea ive been having problems with it to so i did the problems i know how to do until it finally works
anonymous
  • anonymous
@mathsmind i figured that one out im stuck on another one now
anonymous
  • anonymous
evaluate the intergal |dw:1361284859202:dw|
anonymous
  • anonymous
@mathsmind that one ^^^
anonymous
  • anonymous
sorry i meant i was having problems entering the website since yesterday
anonymous
  • anonymous
i know how to solve both integral in my head
anonymous
  • anonymous
@mathsmind how long are you going to be one today ?
anonymous
  • anonymous
it depends on my work, if i am busy or not
anonymous
  • anonymous
can you answer my question please?
anonymous
  • anonymous
were you able to log on to the site
anonymous
  • anonymous
@mathsmind which one ? lol
anonymous
  • anonymous
oh no i wasnt i answered it above
anonymous
  • anonymous
this website? openstudy
anonymous
  • anonymous
it wasnt loading
anonymous
  • anonymous
i felt like crying lol, with out this website i don't know what id do
anonymous
  • anonymous
ok let me reply to the last integral
anonymous
  • anonymous
the answer for the last integral would be pi/6
anonymous
  • anonymous
now by looking at the last integral you can find out from the table that it is equal to arcsin(x/2)
anonymous
  • anonymous
for the 4-x?
anonymous
  • anonymous
\[\int\limits_{0}^{1}\frac{ dx }{ \sqrt{4-x^2} }\]
anonymous
  • anonymous
is that the integral or not?
anonymous
  • anonymous
yepp
anonymous
  • anonymous
the answer is pi/6
anonymous
  • anonymous
how did you get it?
anonymous
  • anonymous
ok there are two ways to do that
anonymous
  • anonymous
one way is to look at the arctrigs integrals
anonymous
  • anonymous
you know the table of integrals right!
anonymous
  • anonymous
the other way is to go through full substitution method
anonymous
  • anonymous
lol no but ill look them up
anonymous
  • anonymous
ok do you want detailed solution then
anonymous
  • anonymous
the u sub thing ?
anonymous
  • anonymous
yep
anonymous
  • anonymous
you let x=2u
anonymous
  • anonymous
x^2 = 4u^2
anonymous
  • anonymous
okay i get it so far
anonymous
  • anonymous
can you differentiate the last bit?
anonymous
  • anonymous
dx = 2du
anonymous
  • anonymous
\[\int\limits_{0}^{1}\frac{ 2du }{\sqrt{4-4u^2} }\]
anonymous
  • anonymous
\[\int\limits\limits_{0}^{1}\frac{ du }{\sqrt{1-u^2} }\]
anonymous
  • anonymous
your a math genuis lol
anonymous
  • anonymous
thanks that was kind of you, and you are a bright student
anonymous
  • anonymous
let me give you a third way how to do that
anonymous
  • anonymous
i understand the u-sub the most
anonymous
  • anonymous
i have three more integral questions
anonymous
  • anonymous
ok go ahed
anonymous
  • anonymous
you need to know different methods to solve one problem , that for future note...
anonymous
  • anonymous
altho the differentiate part is a bit hard
anonymous
  • anonymous
i have a phone call but still submit your work
anonymous
  • anonymous
but ill look back it later because i have to get this course done to graduate
anonymous
  • anonymous
no i meant you make y =2sinx sub
anonymous
  • anonymous
different substitution
anonymous
  • anonymous
if you look at the unit triangle you can figure out that you need a sin sub
anonymous
  • anonymous
okay
anonymous
  • anonymous
find the intergal |dw:1361286908472:dw|
anonymous
  • anonymous
again you are going to get an arctan but now we need to look at the factors
anonymous
  • anonymous
just by looking at this integral we will get 4/3arctan(3x) +c
anonymous
  • anonymous
the question is what was x=? in this u sub
anonymous
  • anonymous
you need to assume that 3x=u, 9x^2=u^2, 3dx=du
anonymous
  • anonymous
\[\int\limits \frac{ 4du }{ 3(1+u^2) }=\frac{ 4 }{ 3}\arctan(u)+c\]
anonymous
  • anonymous
\[u=3x, \therefore \frac{ 4 }{ 3 }\arctan(3x)+c\]
anonymous
  • anonymous
that's your final answer
anonymous
  • anonymous
okay i just did the other one so i have one more integral question
anonymous
  • anonymous
|dw:1361287534741:dw|
anonymous
  • anonymous
ok
anonymous
  • anonymous
again you can see we will get the arctan + something
anonymous
  • anonymous
your answer should be\[\frac{ 1 }{ 2 }\arctan(\frac{ 1 }{ 2}(x-1))+c\]
anonymous
  • anonymous
\[\int\limits\limits\limits \frac{ dx }{4+(1-x)^2}\]
anonymous
  • anonymous
now 2u=1-x, 4u^2=(1-x)^2,
anonymous
  • anonymous
now one thing you need to know is how to let the 4 be a one in the denominator
anonymous
  • anonymous
how do i know that?
anonymous
  • anonymous
ok let me show, it is clear from the u-sub
anonymous
  • anonymous
remember that 2u =1-x, 2du=-dx
anonymous
  • anonymous
okay so far so good lol
anonymous
  • anonymous
\[\int\limits\frac{ 2du }{4(1+(u)^2)}\]
anonymous
  • anonymous
can you see it better this way?
anonymous
  • anonymous
\[\frac{ 1 }{ 2 }\arctan(u)+c\]
anonymous
  • anonymous
\[2u=1-x, \space u=\frac{ 1 }{ 2 }(1-x)\]
anonymous
  • anonymous
\[\frac{ 1 }{ 2 }\arctan(\frac{ 1 }{ 2 }(1-x))+c\]
anonymous
  • anonymous
that's your final answer
anonymous
  • anonymous
okay thanks ! im working on some ill let you know if i need more help
anonymous
  • anonymous
you don't sound like you got the last one?
anonymous
  • anonymous
lol i havent fully looked at it yet im finishing a problem
anonymous
  • anonymous
ok
anonymous
  • anonymous
so i read your mind well hehehe
anonymous
  • anonymous
lol yeahh i was wondering how you knew that,but im looking at it now
anonymous
  • anonymous
i still don't understand how the engagement scheme works on open study
anonymous
  • anonymous
i get that one
anonymous
  • anonymous
good
anonymous
  • anonymous
i think its by the questions you answer, and medals you receive
anonymous
  • anonymous
no i am not asking about those two, i am asking about the engagement part
anonymous
  • anonymous
the one that has learner next to it
anonymous
  • anonymous
lol i thinks its from asking questions cause thats how you learn right
anonymous
  • anonymous
okay i found two i need help on
anonymous
  • anonymous
go ahead
anonymous
  • anonymous
determine which of the integrals can be found using the basic integration formulas you have studied so far in the text.
anonymous
  • anonymous
i think you need to open a new thread because this one is getting heavy with BW
anonymous
  • anonymous
lol okay i will

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