Question

Amistre64 you are here?

Answer 1

i am :)

Answer 2

well thank God! I will send you a handy-dandy screen shot if I may : )

Answer 3

as long as its rateg 'g' :)

Answer 4

you kill me! I laugh at calculus when you're around!

Answer 5

don't know the inverse secant formula. Isn't it something to do with cosine?

Answer 6

gonna have ta review this a little bit

Answer 7

pls do! I will await you patiently! ; )

Answer 8

they still telling you to do integration by parts right?

Answer 9

yep. same thin as last night!

Answer 10

the fun just never ends, right?

Answer 11

i know that the sec inverse function gives us an angle back ..... not that that helps much at the moment :)

Answer 12

the secant of 2 is 60 degrees which is pi/3. Now correct me if I'm wrong, but isn't that the inverse of 1/2? The cosine of 60 is 1/2, so the inverse of 1/2 is two. Now, the cosine of 30 degrees is sqrt3 over 2; so wouldn't the inverse secant of that be 2/sqrt 3?

Answer 13

it would, but that can be written a few ways; like 2sqrt(3)/3

Answer 14

http://www.wolframalpha.com/input/?i=int%287t+sec^-1%28t%29%29dt+from+%282%2Fsqrt%283%29%29+to+2

this is what wolfram gives me on the original problem; but doesnt give the steps :)

Answer 15

you typed in 7 and it's 17. I';m sure tha changes things?

Answer 16

it does; i used the original problem instead of the 'example' problem :)

Answer 17

the clue is to use sec-1 as the deriving part; and 17t as the inegrating part

Answer 18

i see that! so sorry!

Answer 19

sec-1(t) derives to: \(\frac{1}{t \sqrt{t^2 -1}}\)

Answer 20

so that answer in that attachment is the answer to that whole mess of a problem in the orginal problem?

Answer 21

and 17t integrates up to: \(17t^2\over 2\)

Answer 22

yes :)

Answer 23

let me see if I can table it for youin this thing.....

Answer 24

\[\left|
\begin{array}c
.&.&intV
\\.&dU&17t
\\+&sec^{-1}(t)&\frac{17t^2}{2}
\\-&\frac{1}{t\sqrt{t^2-1}}& \frac{17t^3}{6}
\end{array}
\right|\]

Answer 25

im not following the second part; whatd I mess up?

Answer 26

what second part? And btw, if I want to just use that website to get the answers, how do I type in the equation the right way?

Answer 27

int(equation stuff) dx from this# to that#

is what i do

Answer 28

i tried that with the 17 one and it just gave me the number value of 2/sqrt3. What did I do wrong?

Answer 29

int(17t sec^-1(t))dt from (2/sqrt(3)) to 2

Answer 30

got it! I just used that calculator to get the answer for that problem, and I really don't know that I will understand that, so i am going to just leave it. If I can get the right answer with that calculator I don't need to know how to work it. I know thaqt's a chincy way to do it. But that is way way too hard for me!

Answer 31

:) its just a matter of following the same steps as before; just with more confusing terms :)

Answer 32

so you fill in the values for that problem where there is a t. For t you use 2 and then the 2/sqrt3? But it's the sqrt of 3 that throws me!

Answer 33

in other words, how the heck do you find the inverse secant of 2/sqrt 3?

Answer 34

{S} u dv = uv - {S} v du

u = sec-1(t) du = 1/(tsqt(t^2-1))
v = 17t {S}v = (17t^2)/2

Answer 35

\[\int_{} 17t.sec^{-1}(t).dt=\frac{17t^2}{2}sec^{-1}(t)\] - \int_{} \frac{17t^2}{2}.\frac{1}{t\sqrt{t^2-1}}
I understand that part

Answer 36

well I would have if it had formatted right lol

Answer 37

\[\int_{}17t.sec^{−1}(t).dt=\left(\frac{17t^2}{2}.sec^{−1}(t)\right)- \left(\int_{} \frac{17t^2}{2}.\frac{1}{t\sqrt{t^2-1}}\right)\]

Answer 38

ok, but how do you find what all that equalss to?

Answer 39

power went out :) if your here laterlet me know ;)

Answer 40

after integrating by parts we see that the integral of on the right there is easier to operate on.

\[\int_{}\frac{17t^2}{2t\sqrt{t^2-1}}dt \iff \frac{17}{2}\int_{}\frac{2}{2}*\frac{t}{\sqrt{t^2-1}}\]

Answer 41

\[\frac{17}{2}\int_{}\frac{2t}{2\sqrt{t^2-1}} \implies \frac{17}{2}(\sqrt{t^2-1})\]

Answer 42

to put it all together we get:

\[\int_{}17t.sec^{-1}(t) \implies \frac{17t^2}{2}sec^{-1}(t) - \frac{17}{2}(\sqrt{t^2-1}); [_{\frac{2}{\sqrt{3}}}^{2}\]