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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[y = \int\limits_{} \tan^{-1}(t) dt\] is this the equation?
yes but it is a definite integral 1/x to 4x
we might have to convert this to a trig of "u"... maybe. I dont know any derivatives that become an arctan...

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
does that make sense so far?
i wonder if we can decompose this......
I dont see any easy ways of doing this :)
i think i'm just making it more difficult than it needs to be in some way
do you understand piecewise functions?
is there an answer that you can chaeck against?
i know if piecewise functions; but havenet come across a way to integrate them
mmk and there isn't an answer to check :/ thanks for trying though!
.-. I don't think that's right amistre though I didn't try it myself lol >_<
wh are you letting tan^-1(t) = u? ._.
this is prolly an integration by parts method; but I dont know much about when it goes into multiple integration by parts :)
becasue there is no function that derives down to a tan^-1
it is equivalent to: [S] u sec^2(u) du tho.... and thats doable somehow
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 ^_^
lol I got you amistre :)
int by parts is still a mystery to me for some functions :)
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
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(????)
\[\int\limits f'gdx = (\sec^2(u))(u) + \ln|\cos(u)| + c\] I hope I'm right, I just followed your given lol
O_O what kind of table is THAT LOL
its an integration by parts table; but it lines everything up as tho it was multiple integration by parts :)
.-. why so complicated? LOL
you can do it my way :) much easier ^_^
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 :)
>_< lol, I've learned something new today! thank you :)
:) all I need is an integral for tan(u) and I got it solved lol
the table of integration says: tan(x) ints to ln(sec(x)) cool :)
u tan(u) - ln(sec(u))
hmm: \[\tan x = \frac{sinx}{cosx}\] let u = cos x and du = -sinx and solve lol
we re-stitute for u and get: tan^-1(t) tan(tan^-1(t)) - tan(tan^-1(t))
.-.
well...... ln(sec(tan^-1(t))) lol got excited ;)
not convinced with the sec u amistre :(
can always try to derive it back ;)
tan x = sinx/cosx integrating that will give you ln|cosx|
>_
\[\tan ^{-1}t=u, tanu = t\] \[dt=\sec ^{2}u du=(1 + \tan ^{2}u)d u\]
._.
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 :)
:) you too
:)
\[\int\limits_{?}^{?}\tan ^{-1}u=\int\limits_{?}^{?}1.\tan ^{-1}u\]
integrating by parts \[\tan ^{-1}u.u-\int\limits_{?}^{?}u.1/(1+u ^{2})d u\]
\[utan ^{-1}u -2 \int\limits_{?}^{?}2u/(1+u ^{2})\]
if I did it right; mine derives back to tan^-1(t)....
\[utan ^{-1}u - 2 \ln (1+u ^{2})\]
i used integration by parts method, not substitution
\[D[\tan^{-1}(t).\tan(\tan^{-1}(t)) - \ln(\sec(\tan^{-1}(t)))]\] \[\frac{\tan^{-1}(t)\sec^2(\tan^{-1}(t))}{t^2 +1}+\frac{\tan(\tan^{-1}(t))}{t^2+1}- \frac{\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))}{\sec(\tan^{-1}(t)).(t^2+1)}\]
\[\frac{\tan(\tan^{-1}(t)).\sec^2 (\tan^{-1}(t)).\sec(\tan^{-1}(t))+\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))-\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))}{\sec(\tan^{-1}(t)).(t^2+1)}\]
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)))
so the result?
\[\frac{\tan^{-1}(t).\sec(^2(\tan^{-1}(t))}{t^2+1}\]
n this is what u wanted?
i guess i should terminate at Tan^(-1)t?
since t^2+2 = sec^2(a) = sec^2(tan^-1(t)); those cancel and we are left with.. tan^-1(t)
1 Attachment
...since t^2 + 1.....typoed it :)
hmmmm,right :)
did u check mine?is it wrong?
i had enough trouble proofing mine lol...my computer is bogging with all thisequation editor stuff....lol
yup...hats off to ur patience :P

Not the answer you are looking for?

Search for more explanations.

Ask your own question