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

- anonymous

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

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- amistre64

\[y = \int\limits_{} \tan^{-1}(t) dt\]
is this the equation?

- anonymous

yes but it is a definite integral 1/x to 4x

- 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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## More answers

- 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

does that make sense so far?

- amistre64

i wonder if we can decompose this......

- amistre64

I dont see any easy ways of doing this :)

- anonymous

i think i'm just making it more difficult than it needs to be in some way

- anonymous

do you understand piecewise functions?

- amistre64

is there an answer that you can chaeck against?

- amistre64

i know if piecewise functions; but havenet come across a way to integrate them

- anonymous

mmk and there isn't an answer to check :/ thanks for trying though!

- anonymous

.-. I don't think that's right amistre though I didn't try it myself lol >_<

- anonymous

wh are you letting tan^-1(t) = u? ._.

- amistre64

this is prolly an integration by parts method; but I dont know much about when it goes into multiple integration by parts :)

- amistre64

becasue there is no function that derives down to a tan^-1

- amistre64

it is equivalent to:
[S] u sec^2(u) du tho.... and thats doable somehow

- 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

lol I got you amistre :)

- amistre64

int by parts is still a mystery to me for some functions :)

- 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

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

\[\int\limits f'gdx = (\sec^2(u))(u) + \ln|\cos(u)| + c\]
I hope I'm right, I just followed your given lol

- anonymous

O_O what kind of table is THAT LOL

- amistre64

its an integration by parts table; but it lines everything up as tho it was multiple integration by parts :)

- anonymous

.-. why so complicated? LOL

- anonymous

you can do it my way :) much easier ^_^

- 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

>_< lol, I've learned something new today! thank you :)

- amistre64

:) all I need is an integral for tan(u) and I got it solved lol

- amistre64

the table of integration says: tan(x) ints to ln(sec(x)) cool :)

- amistre64

u tan(u) - ln(sec(u))

- anonymous

hmm:
\[\tan x = \frac{sinx}{cosx}\]
let u = cos x and du = -sinx and solve lol

- amistre64

we re-stitute for u and get:
tan^-1(t) tan(tan^-1(t)) - tan(tan^-1(t))

- anonymous

.-.

- amistre64

well...... ln(sec(tan^-1(t))) lol got excited ;)

- anonymous

not convinced with the sec u amistre :(

- amistre64

can always try to derive it back ;)

- anonymous

tan x = sinx/cosx
integrating that will give you ln|cosx|

- anonymous

>_

- anonymous

\[\tan ^{-1}t=u, tanu = t\]
\[dt=\sec ^{2}u du=(1 + \tan ^{2}u)d u\]

- 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

:) you too

- anonymous

:)

- anonymous

\[\int\limits_{?}^{?}\tan ^{-1}u=\int\limits_{?}^{?}1.\tan ^{-1}u\]

- anonymous

integrating by parts
\[\tan ^{-1}u.u-\int\limits_{?}^{?}u.1/(1+u ^{2})d u\]

- anonymous

\[utan ^{-1}u -2 \int\limits_{?}^{?}2u/(1+u ^{2})\]

- amistre64

if I did it right; mine derives back to tan^-1(t)....

- anonymous

\[utan ^{-1}u - 2 \ln (1+u ^{2})\]

- anonymous

i used integration by parts method, not substitution

- 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

\[\frac{\tan(\tan^{-1}(t)).\sec^2 (\tan^{-1}(t)).\sec(\tan^{-1}(t))+\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))-\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))}{\sec(\tan^{-1}(t)).(t^2+1)}\]

- amistre64

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

so the result?

- amistre64

\[\frac{\tan^{-1}(t).\sec(^2(\tan^{-1}(t))}{t^2+1}\]

- anonymous

n this is what u wanted?

- anonymous

i guess i should terminate at Tan^(-1)t?

- 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

...since t^2 + 1.....typoed it :)

- anonymous

hmmmm,right :)

- anonymous

did u check mine?is it wrong?

- amistre64

i had enough trouble proofing mine lol...my computer is bogging with all thisequation editor stuff....lol

- anonymous

yup...hats off to ur patience :P

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