find the integral.

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- anonymous

find the integral.

- schrodinger

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- anonymous

-5^-x/(log(5))+C

- anonymous

|dw:1361280612277:dw|

- anonymous

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

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## More answers

- 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

@mathsmind yea ive been having problems with it to so i did the problems i know how to do until it finally works

- anonymous

@mathsmind i figured that one out im stuck on another one now

- anonymous

evaluate the intergal |dw:1361284859202:dw|

- anonymous

@mathsmind that one ^^^

- anonymous

sorry i meant i was having problems entering the website since yesterday

- anonymous

i know how to solve both integral in my head

- anonymous

@mathsmind how long are you going to be one today ?

- anonymous

it depends on my work, if i am busy or not

- anonymous

can you answer my question please?

- anonymous

were you able to log on to the site

- anonymous

@mathsmind which one ? lol

- anonymous

oh no i wasnt i answered it above

- anonymous

this website? openstudy

- anonymous

it wasnt loading

- anonymous

i felt like crying lol, with out this website i don't know what id do

- anonymous

ok let me reply to the last integral

- anonymous

the answer for the last integral would be pi/6

- anonymous

now by looking at the last integral you can find out from the table that it is equal to arcsin(x/2)

- anonymous

for the 4-x?

- anonymous

\[\int\limits_{0}^{1}\frac{ dx }{ \sqrt{4-x^2} }\]

- anonymous

is that the integral or not?

- anonymous

yepp

- anonymous

the answer is pi/6

- anonymous

how did you get it?

- anonymous

ok there are two ways to do that

- anonymous

one way is to look at the arctrigs integrals

- anonymous

you know the table of integrals right!

- anonymous

the other way is to go through full substitution method

- anonymous

lol no but ill look them up

- anonymous

ok do you want detailed solution then

- anonymous

the u sub thing ?

- anonymous

yep

- anonymous

you let x=2u

- anonymous

x^2 = 4u^2

- anonymous

okay i get it so far

- anonymous

can you differentiate the last bit?

- anonymous

dx = 2du

- anonymous

\[\int\limits_{0}^{1}\frac{ 2du }{\sqrt{4-4u^2} }\]

- anonymous

\[\int\limits\limits_{0}^{1}\frac{ du }{\sqrt{1-u^2} }\]

- anonymous

your a math genuis lol

- anonymous

thanks that was kind of you, and you are a bright student

- anonymous

let me give you a third way how to do that

- anonymous

i understand the u-sub the most

- anonymous

i have three more integral questions

- anonymous

ok go ahed

- anonymous

you need to know different methods to solve one problem , that for future note...

- anonymous

altho the differentiate part is a bit hard

- anonymous

i have a phone call but still submit your work

- anonymous

but ill look back it later because i have to get this course done to graduate

- anonymous

no i meant you make y =2sinx sub

- anonymous

different substitution

- anonymous

if you look at the unit triangle you can figure out that you need a sin sub

- anonymous

okay

- anonymous

find the intergal |dw:1361286908472:dw|

- anonymous

again you are going to get an arctan but now we need to look at the factors

- anonymous

just by looking at this integral we will get 4/3arctan(3x) +c

- anonymous

the question is what was x=? in this u sub

- anonymous

you need to assume that 3x=u, 9x^2=u^2, 3dx=du

- anonymous

\[\int\limits \frac{ 4du }{ 3(1+u^2) }=\frac{ 4 }{ 3}\arctan(u)+c\]

- anonymous

\[u=3x, \therefore \frac{ 4 }{ 3 }\arctan(3x)+c\]

- anonymous

that's your final answer

- anonymous

okay i just did the other one so i have one more integral question

- anonymous

|dw:1361287534741:dw|

- anonymous

ok

- anonymous

again you can see we will get the arctan + something

- anonymous

your answer should be\[\frac{ 1 }{ 2 }\arctan(\frac{ 1 }{ 2}(x-1))+c\]

- anonymous

\[\int\limits\limits\limits \frac{ dx }{4+(1-x)^2}\]

- anonymous

now 2u=1-x, 4u^2=(1-x)^2,

- anonymous

now one thing you need to know is how to let the 4 be a one in the denominator

- anonymous

how do i know that?

- anonymous

ok let me show, it is clear from the u-sub

- anonymous

remember that 2u =1-x, 2du=-dx

- anonymous

okay so far so good lol

- anonymous

\[\int\limits\frac{ 2du }{4(1+(u)^2)}\]

- anonymous

can you see it better this way?

- anonymous

\[\frac{ 1 }{ 2 }\arctan(u)+c\]

- anonymous

\[2u=1-x, \space u=\frac{ 1 }{ 2 }(1-x)\]

- anonymous

\[\frac{ 1 }{ 2 }\arctan(\frac{ 1 }{ 2 }(1-x))+c\]

- anonymous

that's your final answer

- anonymous

okay thanks ! im working on some ill let you know if i need more help

- anonymous

you don't sound like you got the last one?

- anonymous

lol i havent fully looked at it yet im finishing a problem

- anonymous

ok

- anonymous

so i read your mind well hehehe

- anonymous

lol yeahh i was wondering how you knew that,but im looking at it now

- anonymous

i still don't understand how the engagement scheme works on open study

- anonymous

i get that one

- anonymous

good

- anonymous

i think its by the questions you answer, and medals you receive

- anonymous

no i am not asking about those two, i am asking about the engagement part

- anonymous

the one that has learner next to it

- anonymous

lol i thinks its from asking questions cause thats how you learn right

- anonymous

okay i found two i need help on

- anonymous

go ahead

- anonymous

determine which of the integrals can be
found using the basic integration formulas you have studied
so far in the text.

- anonymous

i think you need to open a new thread because this one is getting heavy with BW

- anonymous

lol okay i will

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