anonymous
  • anonymous
Integral of (1/(1+y^2)^(3/2))dy???
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Well do u wanna use u subsitution?
anonymous
  • anonymous
I don't really know. I tried to integrate by parts, but apparently it's wrong.. or I did it wrong.
anonymous
  • anonymous
well wrong according to what?

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More answers

anonymous
  • anonymous
according to wolframalpha lol
anonymous
  • anonymous
u can get diff answers depending on the way u integrate but it is in reality all the sam answer
anonymous
  • anonymous
I think subsitution is easier
anonymous
  • anonymous
ohhh maybe not
anonymous
  • anonymous
Well, they put y=tanu and then they did something, but I don't understand it :/
anonymous
  • anonymous
ohh they used trig subsitution
anonymous
  • anonymous
but i guess u didnt learn that
anonymous
  • anonymous
ya that wld be the simplest
anonymous
  • anonymous
did u learn trig subsitution yet?
anonymous
  • anonymous
Nope, never learned it. this is for my DE class :(
anonymous
  • anonymous
lol that is weird
anonymous
  • anonymous
k give me a sec to figure it out
anonymous
  • anonymous
ok thanks
razor99
  • razor99
|dw:1327991590145:dw|
razor99
  • razor99
59% correct iguess
anonymous
  • anonymous
lol that is wrong
razor99
  • razor99
why
anonymous
  • anonymous
I am getting an ugly one
anonymous
  • anonymous
oh sal can u help?
anonymous
  • anonymous
you have to use trig substitution... take a look at this.
anonymous
  • anonymous
ya but he never learnt that yet
anonymous
  • anonymous
is this for hmw? or you are just doing problems for fun?
anonymous
  • anonymous
yah, i read that.. but how does sec^3(tan^-1) = cos??
anonymous
  • anonymous
i have an exam tomoroww lol
anonymous
  • anonymous
ok I will show u what i got. u = 1+y^2 du=2ydy (1/2)du=ydy
anonymous
  • anonymous
y=sqrt(u-1)
anonymous
  • anonymous
so (1/2) integral (u)^-(2/3)(u-1)^(1/2)du
anonymous
  • anonymous
sorry i just dont know how to use latex
anonymous
  • anonymous
rld did you get a solution in the end using ur way?
anonymous
  • anonymous
ya but an ugly one
anonymous
  • anonymous
well look at the solution you know how y=tanu , the solution says that at the begining? you substitute y= tanu, then u=tan-1y
anonymous
  • anonymous
ya sal's method will give u a nice answer and it is the simplest
anonymous
  • anonymous
and \[\sin^2u+\cos^2u=1\] divide all by \[\cos^2u\] you will get, \[\tan^2u+1=\sec^2u\] you gotta know these identities
anonymous
  • anonymous
lol the guy disappeared
anonymous
  • anonymous
lol
anonymous
  • anonymous
U scared him off
anonymous
  • anonymous
amma disappear too...
anonymous
  • anonymous
lol
anonymous
  • anonymous
who?
anonymous
  • anonymous
the poster
anonymous
  • anonymous
my screen frozed. i suppose i will mull it over. or just hope there will be no trig sub integrals.
anonymous
  • anonymous
pray hard :D
anonymous
  • anonymous
lol, yeh man just pray hard. go get some sleep so u can write the exam 2morrow
anonymous
  • anonymous
hahaha thanks
anonymous
  • anonymous
np
razor99
  • razor99
http://www.numberempire.com/integralcalculator.php
razor99
  • razor99
this might help ya linhong10
anonymous
  • anonymous
lol razor u r the best
razor99
  • razor99
why?
anonymous
  • anonymous
cuz u r always so helpful
razor99
  • razor99
thax i love to help people even i dont know it.
razor99
  • razor99
by the way i alway help my friends
anonymous
  • anonymous
coool
anonymous
  • anonymous
keep up the great work =)
razor99
  • razor99
rld u might be tired
anonymous
  • anonymous
hehe never tired
razor99
  • razor99
wow ur so patient i hope .
razor99
  • razor99
someday i might be like u
razor99
  • razor99
i mean a being very patient
anonymous
  • anonymous
hehe
anonymous
  • anonymous
cool free notes :D
anonymous
  • anonymous
ya sals notes
razor99
  • razor99
bye guyz wont see till friday okay
anonymous
  • anonymous
k bb
anonymous
  • anonymous
hf
razor99
  • razor99
cya around later guyz the school has finally started
anonymous
  • anonymous
kk

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