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

|dw:1361280612277:dw|

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

@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?

yepp

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 ?

yep

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

okay

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

|dw:1361287534741:dw|

ok

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

ok

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

good

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

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

lol okay i will