Question

hey guys , pls help me here

i'll post the integration question in a sec

Answer 1

find the integration for:

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

I already know the answer , but when I checked using u substitution the final answer should be tan^-1 (ax) +c ,, but in my notes it says it's 1/a tan^-1(ax)+c

Answer 2

2- find the integration for :

\[\frac{ 1 }{ \sqrt{-5+6x-x^2}}\]

when we tried to take 6x-x^2 as u , please explain how to find u and du here.

Answer 3

2) do you know the method of completing the squares ?

Answer 4

you mean (a+b)^2 = a^2 +2ab + b^2 ?

Answer 5

-5 +6x -x^2
= - (x^2 -6x + ????) -5 +????

what should replace ???? so that
(x^2 -6x +????) becomes a perfect square.

Answer 6

sorry sometimes I won't recall such things because these things I took it before college " was studying in my main language :D " so the names might be different

Answer 7

I think since we're squaring b we take 3 and it becomes 9 inside the ()

Answer 8

well the other part that is the one that I couldn't understand :D

Answer 9

9 is correct :)
-5 +6x -x^2
= - (x^2 -6x + 9) -5 +9

= 4 - (x-3)^2
= 2^2 - (x-3)^2

so your integral has the denominator in the form of sqrt (A^2 -X^2)

can you use standard formulas for integration ?

Answer 10

sure , but wait please :D

why did we put -5 then +9 ?

Answer 11

the formula that we're going to use here is 1 over square root of 1-x^2 which is sin^-1

Answer 12

ok, since we have the expression as -5 +6x -x^2
we MUST NOT change it.
so if i am subtracting 9 from this to make a perfect square, i must also add 9 to it so that the whole expression remains unchanged.

Answer 13

ah , thank you sir ^^

the rest is easy no need to solve it :D

how about the first question , could you please help ^^

Answer 14

please post the question for 1) again.

Answer 15

the first question in the notes it says 1/a in the final answer " look at what I posted " but when I solved it it should be tan^-1 u instead of what I posted

Answer 16

sure

Answer 17

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

Answer 18

wait

it's 1/1+a^2x^2 , sorry

Answer 19

there's 1 before a^2x^2

Answer 20

i thought so

Answer 21

:D

Answer 22

write that as
\(\Large \dfrac{1}{a^2 }\dfrac{1}{(1/a)^2 +x^2}\)

Answer 23

(1/a^2) [ a arctan (ax) ] +c

Answer 24

1/a tan^-1(ax)+c

did you see your error ?

Answer 25

wait please , here he took ax as u

Answer 26

so we took 1/a^2 out side of the integration and we remain with 1/1/a^2 + x^2

Answer 27

oh, you want to solve it bys substitution ?

Answer 28

we want to get rid of 1/a^2

Answer 29

our prof solved it using u substitution :D

Answer 30

1/(1+ a^2x^2) dx

let u = ax
du = a dx
(du)/a =dx

------------> 1/ (1+u^2) (du/a)

Answer 31

so since du = a we put 1/a out side so we can get rid of a ?

Answer 32

du = a?????
no...


we took, u = ax , right ?

diff. w.r.t x, we get
du/dx = a

du/a = dx

so i am plugging in du/a in place of dx

Answer 33

so the final form will be 1/1+(ax^2) adx

and we want to get rid of a ? here's what I couldn't understand :D

Answer 34

idk whay you always wanna "get rid of a"

did you understand the substitution ?
did u get how
dx = du/a first ?

Answer 35

since du = adx then dx = 1/a du

Answer 36

yes
good.
so our new integral becomes
int (1/a ) (1 / (1+u^2) ) du

right ?

Answer 37

I think 1/a is out side of the int

Answer 38

so int 1/1+u^2 is tan^-1u

Answer 39

but why did we put 1/a out side of the int ?

Answer 40

yes

because dx was du/a

= (1/a) du

Answer 41

ah thank you :D

Answer 42

sorry for wasting your time :(

Answer 43

it was not a waste since you understood the problem :)

Answer 44

and you're most welcome ^_^

Answer 45

Hope this post helps who couldn't understand similar problems beside me :) , thanks again ^^

Answer 46

since you're learning integration, this might be helpful....may go through it
http://openstudy.com/users/hartnn#/updates/50960518e4b0d0275a3ccfba

Answer 47

Yeah , this semester the course will be all about integration , I'll look at what you have posted , thanks alot , you really helped ^^