Question

Calculus integration question

Answer 1

Yesh? I can help

Answer 2

\[\int\limits 1/\sqrt{6x-x^2}dx\]

Answer 3

ok so i completed the square and got the correct answer which is arcsin((x-3)/3) + C

Answer 4

but then the next part of the question asks to find the integral by making the substitution \[u = \sqrt{x}\]

Answer 5

im sort of stuck on how to do this

Answer 6

u=sqrt(x) so x=u^2

Answer 7

integral of 1/sqrt(6x-x^2) dx is then equal to?

Answer 8

yes i did that but im getting wrong answer

Answer 9

im assuming you take out the x from the inside of the radical

Answer 10

\[1/\sqrt{x}\sqrt{6-x}\]

Answer 11

then you do the u substitution and it gets rid of that

Answer 12

square root of x

Answer 13

Its the same thing lol

Answer 14

If you plugged in u^2 and then took out sqrt(u^2) from the radical, it gives you the same thing. But yes, you can do that.

Answer 15

1/sqrt(x)*(sqrt(6-x)) is correct.

Answer 16

ok so i do the u substitution but that leaves me with a left over 2 on the top

Answer 17

\[du = 1/2\sqrt{x} dx\]

Answer 18

im assuming the sqrt(x) is in the denominator yes?

Answer 19

yes

Answer 20

that is correct

Answer 21

but you have to solve for dx

Answer 22

\[\int\limits 2/\sqrt{6-u^2} du\]

Answer 23

dx=2du

Answer 24

so its actually int of 2/u*sqrt(6-u^2) du

Answer 25

aka you forgot to multiply a u in your denominator.

Answer 26

hmmm but i thought the square root of x got taken care of after i did the u substitution

Answer 27

\[\int\limits 2\sqrt{x}/\sqrt{x}\sqrt{6-x}\]

Answer 28

so the square roots of x would cancel correct?

Answer 29

no theres no sqrt(x) in the top of the integral.

Answer 30

You know how you solved for du?

Answer 31

but doesn't dx = 2sqrt(x)du

Answer 32

remember how we said that

Answer 33

u=sqrt(x) and x=u^2?

Answer 34

if you take the derivative of x=u^2 you get dx=2udu

Answer 35

ok so i would just plug in the u for the sqrt of x

Answer 36

gotcha

Answer 37

wait nvm im high, its the same thing.

Answer 38

so yes integral of 2/sqrt(6-x) would be correct.

Answer 39

oh ok so i do know what im doing lol

Answer 40

yesh

Answer 41

so once you have that you should plug in your u^2

Answer 42

yeah so int of 2/sqrt(6-u^2) du

Answer 43

if i integrate that then it would be 2arcsin(sqrt(x)/sqrt(6))

Answer 44

but that doesn't match up to what i got when i completed the square

Answer 45

hm lets do this step by step

Answer 46

you can take out the 6 from the denominator to get

Answer 47

2/sqrt(6)*sqrt(1-(u^2)/(sqrt(6))^2)

Answer 48

take out 2/sqrt(6) from the integral and you get

Answer 49

why take out the 6? isn't the point to get it into a^2 - x^2 form so i can get the arcsin

Answer 50

(2/sqrt(6))*arcsin(u/sqrt(6)

Answer 51

u=sqrt(x) so it would be (2/sqrt(6))*arcsin(sqrt(x)/sqrt(6) if i didnt do my math wrong...

Answer 52

this still doesnt match up does it...

Answer 53

maybe i should write it out...typing it is a little confusing...ill brb

Answer 54

what i did is i made the 6 into sqrt(6)^2 so i wouldn't have to mess with it and then it would be in a^2 - x^2 for which would make it easy to integrate it into arcsin

Answer 55

hm im still getting (2/sqrt(6))*arcsin(sqrt(x/6))

Answer 56

i graphed your answer and mine and mine looks the same as when i completed the square but shifted down like 1 point something

Answer 57

hm i could write out what i did and take a picture for you

Answer 58

sure that would be great

Answer 59

im pretty sure there is supposed to be a sqrt(6) before the arc sin tho...

Answer 60

Is it readable?

Answer 61

From what I got there is a sqrt(6) before the arcsin

Answer 62

yes it readable but what you did im pretty sure is not valid because the u^2 cannot have a coefficient infront of it, if you do it your way then it has that 1/sqrt(6) infront

Answer 63

what i did was I factored out the 6

Answer 64

from the sqrt

Answer 65

that's why i did sqrt(6-u^2) = sqrt(sqrt(6)^2 - u^2)

Answer 66

http://www.wolframalpha.com/input/?i=integral+of+1%2Fsqrt%281-%28x%2F2%29%5E2%29+dx

Answer 67

yes i understand you factored it out but that makes it so u^2 is divided by sqrt(6) which you dont want

Answer 68

http://www.wolframalpha.com/input/?i=integral+of+1%2Fsqrt%284-x%5E2%29+dx

Answer 69

fk im confusing myself

Answer 70

lol

Answer 71

Why arent these integrals equal da fk

Answer 72

yeah ik dude it doesn't make sense to me either

Answer 73

OH

Answer 74

if you graph my answer and the original answer they are the freaking same but my answer is shifted down

Answer 75

they are equal

Answer 76

how?

Answer 77

Lets pretend that 6-u^2 is the same as 4-x^2 in wolfram

Answer 78

ok

Answer 79

http://www.wolframalpha.com/input/?i=integral+of+1%2Fsqrt%284-x%5E2%29+dx

Answer 80

wait hrm

Answer 81

but that just means... my answer shouldve been the exact same as ur answer...

Answer 82

which is 2*arcsin(u/sqrt(6)

Answer 83

fk my bad

Answer 84

and 2arcsin(sqrt(x)/sqrt(6)) doesnt match up to the complete the square answer...

Answer 85

why not...

Answer 86

i have no idea!!!

Answer 87

give me a moment... sorry

Answer 88

it's ok take your time

Answer 89

if anyone else has any ideas you are more than welcome to jump in

Answer 90

completing the square gives sqrt(9-(x-3)^2) correct?

Answer 91

-9 but yeah

Answer 92

you sure -9?

Answer 93

-9 would mean that the bottom is imaginary

Answer 94

well yeah because you have to get rid of the 9 if you try to convert back to the original expression

Answer 95

\[\sqrt{-(x-3)^2-9}\]

Answer 96

factor out a negative 1 from that... you have an imaginary number lol