Question

integration help

Answer 1

\[\int\limits_{}^{} dx \div x \sqrt{x ^{2}-a ^{2}}\]

Answer 2

dang amistre, you became a moderator? i didnt know they even had mods on here haha

Answer 3

:) nah .. i just got moldy

Answer 4

that appears to be one of the inverse trig integrals

Answer 5

i dont get it.... but i wanna be a mod -.- anyways, my professor told me the answer is 1/a arcsec of x/a +c

Answer 6

it is an arctrig haha, only problem is that in all my proofs, i dont get that first 1/a, i only get arcsec(x/a)

Answer 7

\[\int \frac{1}{x\sqrt{x^2-a^2}}dx\]

Answer 8

how you get the fraction ? i can only get a divy symbol

Answer 9

well; since a is considered a filler for a constant; lets use a=3

\frac{top}{bottom}

Answer 10

lets try to work this backwards and see what we can get; sec-1 derived

Answer 11

sec-1(x) = y

x = sec(y)

dx/dx = sec(y)tan(y) dy/dx

1 = sec(y)tan(y) y'

y' = \(\frac{1}{sec(y)tan(y)}\) right?

Answer 12

yeahp

Answer 13

since y = sec-1(x); lets plug it back in and see where it leads

Answer 14

1
----------------------
sec(sec-1(x)) tan(sec-1(x))


1
------------ ; a triangle to see the movements of the tan(sec-1)
x tan(sec-1(x)) might be helpful next

Answer 15

too complicated O.o we dont need to prove the inverse of it
i just need to use basic integrationt echniques to make it lead to a perfect arctrig integral

Answer 16

to see how it derives helps me see how to put it back together

Answer 17

from this we get the derivative from sec-1 as:

\[\frac{1}{x\sqrt{x^2-1}}\]

Answer 18

the 1 is just a place holder and can be substituted for a^2

Answer 19

recall from trig now that:

tan^2 + 1 = sec^2

tan^2 = sec^2 - 1

tan = sqrt(sec^2 - 1)

if we substitute x = sec; we can reform this integral right?

Answer 20

my computer jsut randomly restarted...

Answer 21

right but when trying to find a simplified integral for any given a, it should be that the integral is (1/a)arcsec(x/a)+c

Answer 22

however when i try to simplify to that using this process:

Answer 23

x = a sec gives us

x.sqrt[(asec)^2 - a^2]

x.sqrt[a^2 sec^2 - a^2]

x.sqrt[(sec^2 - 1()a^2)]

x.sqrt(tan^2 . a^2)

x.sqrt(tan^2).sqrt(a^2)

x.tan.a

Answer 24

factor out a^2 from the square root, which gives you \[ax \sqrt{\frac{x ^{2}}{a ^{2}}+1}\] as the bottom of the fraction
then i take the a out as a constant, and use usub where u = (x/a) but then that cancells out the a i took out as a constant which leaves me with no more a's and just a arcsec(x/a)+c

Answer 25

the 1/a accounts for the chainrule

Answer 26

sec(x/a) is not only the derivative of a sec; but also the innards x/a

Answer 27

arcsec that is :)

Answer 28

i can get that from a derivitive standpoint
but i cant get the 1.a from my integration

Answer 29

1/a***

Answer 30

you have to include it as a form of 1 in order to integrate appropriately; give me the steps you took again

Answer 31

iight gimmy a sec to type it all up

Answer 32

so integrate
dx/ [x*sqrt(xx-aa)]
factor out aa so i get
dx/ [ax*sqrt{(x/a)^2 -1}]
then take out the a as a constant:
1/a * integral dx/[x*sqrt{(x/a)^2-1}]
u sub on x/a so u=x/a du=dx/a so dx= adu
thus:
1/a * integral adu/[x*sqrt{uu-1}]
pull out the "a" from adu as a constant
\[\frac{a}{a} \int\limits_{}^{}\frac{du}{x \sqrt{u ^{2}-1}}\]
which is the same as
\[\int\limits_{}^{}\frac{du}{x \sqrt{u ^{2}-1}}\]
which integrated is just
\[arcsec(\frac{x}{a})+C\]

my problem is that i still have an x after usub, which i forgot how to deal with

Answer 33

your u-sub choice can be manipulated to redefine x right?

Answer 34

u = x/a ; therefore ua = x

Answer 35

true, but define it as what?

Answer 36

oh... damn -.-

Answer 37

thats no basic calculus technique, hahaha no wonder i forgot about redeining u in terms of x, thanks so much amistre

Answer 38

youre welcome :)

Answer 39

ill be on later, most likely to help out, but ill deffinitely be back with questions.

I just finished freshman year of HS but im taking calc2 over the summer to get the credit for all the extra work i did for fun. cept highschool calc2 isint the same as college >< its so much harder! and i got my first college test on friday :/