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anonymous
 5 years ago
find dy/dx for y=integral(arctan)tdt from (1/x,4x)
anonymous
 5 years ago
find dy/dx for y=integral(arctan)tdt from (1/x,4x)

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[y = \int\limits_{} \tan^{1}(t) dt\] is this the equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but it is a definite integral 1/x to 4x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we might have to convert this to a trig of "u"... maybe. I dont know any derivatives that become an arctan...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0does that make sense so far?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i wonder if we can decompose this......

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I dont see any easy ways of doing this :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think i'm just making it more difficult than it needs to be in some way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you understand piecewise functions?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0is there an answer that you can chaeck against?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i know if piecewise functions; but havenet come across a way to integrate them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mmk and there isn't an answer to check :/ thanks for trying though!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0.. I don't think that's right amistre though I didn't try it myself lol >_<

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wh are you letting tan^1(t) = u? ._.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this is prolly an integration by parts method; but I dont know much about when it goes into multiple integration by parts :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0becasue there is no function that derives down to a tan^1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it is equivalent to: [S] u sec^2(u) du tho.... and thats doable somehow

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integration 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
 5 years ago
Best ResponseYou've already chosen the best response.0lol I got you amistre :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0int by parts is still a mystery to me for some functions :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it'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
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits f'gdx = (\sec^2(u))(u) + \ln\cos(u) + c\] I hope I'm right, I just followed your given lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0O_O what kind of table is THAT LOL

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its an integration by parts table; but it lines everything up as tho it was multiple integration by parts :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0.. why so complicated? LOL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can do it my way :) much easier ^_^

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its 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
 5 years ago
Best ResponseYou've already chosen the best response.0>_< lol, I've learned something new today! thank you :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0:) all I need is an integral for tan(u) and I got it solved lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the table of integration says: tan(x) ints to ln(sec(x)) cool :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0u tan(u)  ln(sec(u))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm: \[\tan x = \frac{sinx}{cosx}\] let u = cos x and du = sinx and solve lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we restitute for u and get: tan^1(t) tan(tan^1(t))  tan(tan^1(t))

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well...... ln(sec(tan^1(t))) lol got excited ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not convinced with the sec u amistre :(

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0can always try to derive it back ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tan x = sinx/cosx integrating that will give you lncosx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\tan ^{1}t=u, tanu = t\] \[dt=\sec ^{2}u du=(1 + \tan ^{2}u)d u\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0students 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{?}^{?}\tan ^{1}u=\int\limits_{?}^{?}1.\tan ^{1}u\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integrating by parts \[\tan ^{1}u.u\int\limits_{?}^{?}u.1/(1+u ^{2})d u\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[utan ^{1}u 2 \int\limits_{?}^{?}2u/(1+u ^{2})\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if I did it right; mine derives back to tan^1(t)....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[utan ^{1}u  2 \ln (1+u ^{2})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i used integration by parts method, not substitution

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[D[\tan^{1}(t).\tan(\tan^{1}(t))  \ln(\sec(\tan^{1}(t)))]\] \[\frac{\tan^{1}(t)\sec^2(\tan^{1}(t))}{t^2 +1}+\frac{\tan(\tan^{1}(t))}{t^2+1} \frac{\sec(\tan^{1}(t)).\tan(\tan^{1}(t))}{\sec(\tan^{1}(t)).(t^2+1)}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\tan(\tan^{1}(t)).\sec^2 (\tan^{1}(t)).\sec(\tan^{1}(t))+\sec(\tan^{1}(t)).\tan(\tan^{1}(t))\sec(\tan^{1}(t)).\tan(\tan^{1}(t))}{\sec(\tan^{1}(t)).(t^2+1)}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0all 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)))

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\tan^{1}(t).\sec(^2(\tan^{1}(t))}{t^2+1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0n this is what u wanted?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i guess i should terminate at Tan^(1)t?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0since t^2+2 = sec^2(a) = sec^2(tan^1(t)); those cancel and we are left with.. tan^1(t)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0...since t^2 + 1.....typoed it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did u check mine?is it wrong?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i had enough trouble proofing mine lol...my computer is bogging with all thisequation editor stuff....lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yup...hats off to ur patience :P
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