Question

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

Answer 1

\[y = \int\limits_{} \tan^{-1}(t) dt\]

is this the equation?

Answer 2

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

Answer 3

we might have to convert this to a trig of "u"... maybe.

I dont know any derivatives that become an arctan...

Answer 4

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

Answer 5

does that make sense so far?

Answer 6

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

Answer 7

I dont see any easy ways of doing this :)

Answer 8

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

Answer 9

do you understand piecewise functions?

Answer 10

is there an answer that you can chaeck against?

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

it is equivalent to:

[S] u sec^2(u) du tho.... and thats doable somehow

Answer 18

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 ^_^

Answer 19

lol I got you amistre :)

Answer 20

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

Answer 21

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

Answer 22

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(????)

Answer 23

\[\int\limits f'gdx = (\sec^2(u))(u) + \ln|\cos(u)| + c\]

I hope I'm right, I just followed your given lol

Answer 24

O_O what kind of table is THAT LOL

Answer 25

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

Answer 26

.-. why so complicated? LOL

Answer 27

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

Answer 28

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 :)

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

hmm:

\[\tan x = \frac{sinx}{cosx}\]

let u = cos x and du = -sinx and solve lol

Answer 34

we re-stitute for u and get:

tan^-1(t) tan(tan^-1(t)) - tan(tan^-1(t))

Answer 35

.-.

Answer 36

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

Answer 37

not convinced with the sec u amistre :(

Answer 38

can always try to derive it back ;)

Answer 39

tan x = sinx/cosx

integrating that will give you ln|cosx|

Answer 40

>_<lol

Answer 41

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

Answer 42

._.

Answer 43

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 :)

Answer 44

:) you too

Answer 45

:)

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

i used integration by parts method, not substitution

Answer 52

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

Answer 53

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

Answer 54

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)))

Answer 55

so the result?

Answer 56

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

Answer 57

n this is what u wanted?

Answer 58

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

Answer 59

since t^2+2 = sec^2(a) = sec^2(tan^-1(t)); those cancel and we are left with..

tan^-1(t)

Answer 60

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

Answer 61

hmmmm,right :)

Answer 62

did u check mine?is it wrong?

Answer 63

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

Answer 64

yup...hats off to ur patience :P