anonymous
  • anonymous
find dy/dx for y=integral(arctan)tdt from (1/x,4x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
\[y = \int\limits_{} \tan^{-1}(t) dt\] is this the equation?
anonymous
  • anonymous
yes but it is a definite integral 1/x to 4x
amistre64
  • amistre64
we might have to convert this to a trig of "u"... maybe. I dont know any derivatives that become an arctan...

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amistre64
  • amistre64
if we let u = tan^-1(t) then; t = tan(u) ; dt = sec^2(u) du which gives us... [S] u sec^2(u) du ; to solve
amistre64
  • amistre64
does that make sense so far?
amistre64
  • amistre64
i wonder if we can decompose this......
amistre64
  • amistre64
I dont see any easy ways of doing this :)
anonymous
  • anonymous
i think i'm just making it more difficult than it needs to be in some way
anonymous
  • anonymous
do you understand piecewise functions?
amistre64
  • amistre64
is there an answer that you can chaeck against?
amistre64
  • amistre64
i know if piecewise functions; but havenet come across a way to integrate them
anonymous
  • anonymous
mmk and there isn't an answer to check :/ thanks for trying though!
anonymous
  • anonymous
.-. I don't think that's right amistre though I didn't try it myself lol >_<
anonymous
  • anonymous
wh are you letting tan^-1(t) = u? ._.
amistre64
  • amistre64
this is prolly an integration by parts method; but I dont know much about when it goes into multiple integration by parts :)
amistre64
  • amistre64
becasue there is no function that derives down to a tan^-1
amistre64
  • amistre64
it is equivalent to: [S] u sec^2(u) du tho.... and thats doable somehow
anonymous
  • anonymous
integration by parts is something like this:\[\int\limits f'gdx = fg - \int\limits fg'dx\] where f' = sec^2(u) and g = u then find f and g' and plug it in the equation ^_^
anonymous
  • anonymous
lol I got you amistre :)
amistre64
  • amistre64
int by parts is still a mystery to me for some functions :)
anonymous
  • anonymous
it's alright ^_^ all you have to do is find f and g' and then plug it in the equation, with that, you have solved the problem. g' = 1 and f = tan x then you'll get : \[\int\limits f'gdx = (\sec^2(u))(u) - \int\limits \tan(u) du\] then just find the integral lol
amistre64
  • amistre64
i usually do a table like this: | v up ----------- u dn | sec^2 ----------- +u | tan ----------- -1 | ?? ---------- +0 | N/A ---------- And so I get for the integral: u tan(u) - 1(????)
anonymous
  • anonymous
\[\int\limits f'gdx = (\sec^2(u))(u) + \ln|\cos(u)| + c\] I hope I'm right, I just followed your given lol
anonymous
  • anonymous
O_O what kind of table is THAT LOL
amistre64
  • amistre64
its an integration by parts table; but it lines everything up as tho it was multiple integration by parts :)
anonymous
  • anonymous
.-. why so complicated? LOL
anonymous
  • anonymous
you can do it my way :) much easier ^_^
amistre64
  • amistre64
its easier when doing multiple integration by parts; which is all the table amounts to. You shift between + and - your u derives down and your v suits up.... its the same process :) just a different format to keep everything in line :)
anonymous
  • anonymous
>_< lol, I've learned something new today! thank you :)
amistre64
  • amistre64
:) all I need is an integral for tan(u) and I got it solved lol
amistre64
  • amistre64
the table of integration says: tan(x) ints to ln(sec(x)) cool :)
amistre64
  • amistre64
u tan(u) - ln(sec(u))
anonymous
  • anonymous
hmm: \[\tan x = \frac{sinx}{cosx}\] let u = cos x and du = -sinx and solve lol
amistre64
  • amistre64
we re-stitute for u and get: tan^-1(t) tan(tan^-1(t)) - tan(tan^-1(t))
anonymous
  • anonymous
.-.
amistre64
  • amistre64
well...... ln(sec(tan^-1(t))) lol got excited ;)
anonymous
  • anonymous
not convinced with the sec u amistre :(
amistre64
  • amistre64
can always try to derive it back ;)
anonymous
  • anonymous
tan x = sinx/cosx integrating that will give you ln|cosx|
anonymous
  • anonymous
>_
anonymous
  • anonymous
\[\tan ^{-1}t=u, tanu = t\] \[dt=\sec ^{2}u du=(1 + \tan ^{2}u)d u\]
anonymous
  • anonymous
._.
anonymous
  • anonymous
students have faced such a question last semester, thankfully I wasn't there lol , anyhow, thanks for your time amistre and uzma ^_^ but I gotta study communications , take care both :)
anonymous
  • anonymous
:) you too
anonymous
  • anonymous
:)
anonymous
  • anonymous
\[\int\limits_{?}^{?}\tan ^{-1}u=\int\limits_{?}^{?}1.\tan ^{-1}u\]
anonymous
  • anonymous
integrating by parts \[\tan ^{-1}u.u-\int\limits_{?}^{?}u.1/(1+u ^{2})d u\]
anonymous
  • anonymous
\[utan ^{-1}u -2 \int\limits_{?}^{?}2u/(1+u ^{2})\]
amistre64
  • amistre64
if I did it right; mine derives back to tan^-1(t)....
anonymous
  • anonymous
\[utan ^{-1}u - 2 \ln (1+u ^{2})\]
anonymous
  • anonymous
i used integration by parts method, not substitution
amistre64
  • amistre64
\[D[\tan^{-1}(t).\tan(\tan^{-1}(t)) - \ln(\sec(\tan^{-1}(t)))]\] \[\frac{\tan^{-1}(t)\sec^2(\tan^{-1}(t))}{t^2 +1}+\frac{\tan(\tan^{-1}(t))}{t^2+1}- \frac{\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))}{\sec(\tan^{-1}(t)).(t^2+1)}\]
amistre64
  • amistre64
\[\frac{\tan(\tan^{-1}(t)).\sec^2 (\tan^{-1}(t)).\sec(\tan^{-1}(t))+\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))-\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))}{\sec(\tan^{-1}(t)).(t^2+1)}\]
amistre64
  • amistre64
all that work and it cuts it short lol...; that last part is zeros the middle part to get:\[\frac{\tan^{-1}(t).\sec^2(\tan^{-1}(t)).\sec(\tan^{-1}())}{\sec(\tan^{-1}(t)).(t^2+1)}\] had to fix a typo in that second round... it aint a tan(tan^-1(t)))
anonymous
  • anonymous
so the result?
amistre64
  • amistre64
\[\frac{\tan^{-1}(t).\sec(^2(\tan^{-1}(t))}{t^2+1}\]
anonymous
  • anonymous
n this is what u wanted?
anonymous
  • anonymous
i guess i should terminate at Tan^(-1)t?
amistre64
  • amistre64
since t^2+2 = sec^2(a) = sec^2(tan^-1(t)); those cancel and we are left with.. tan^-1(t)
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amistre64
  • amistre64
...since t^2 + 1.....typoed it :)
anonymous
  • anonymous
hmmmm,right :)
anonymous
  • anonymous
did u check mine?is it wrong?
amistre64
  • amistre64
i had enough trouble proofing mine lol...my computer is bogging with all thisequation editor stuff....lol
anonymous
  • anonymous
yup...hats off to ur patience :P

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