Question

integrate ((x)/(squareroot(6x-x^2)))dx

......if you look at the photo that is how far I got....I know the answer will have arcsin in it but I always get stuck at this point!

Answer 1

I think it should be u = x-3

Answer 2

now take the 9 out of the square root
and substitute u=sinx/3

Answer 3

that should help :)

Answer 4

yes I'm using trig substitution

Answer 5

Let us take a look at the denominator first huh.
\[\sqrt{6x-x^2}\]
In order to have positive value of x, we need to take away the negative sign right? Therefore it should be\[\sqrt{- (x^2-6)}\]

Answer 6

I mean positive coefficient

Answer 7

oh snap it is x-3

thank you!

Answer 8

Do you know what to do next? @megannicole51

Answer 9

did u look on my picture? u just do basically u substitution and solve from there....

Answer 10

NO ,, WRIRE NUMERATOR AS (X-3)+3

Answer 11

I think you missed something. Isn't it that be u = asin(theta)
where u = x-3
and a = 3

Answer 12

THEN SPLIT INTO TWO PARTS AND INTEGRATE SEPERATELY

Answer 13

6X-X^2=3^2-(X-3)^2

Answer 14

Therefore,
dx = 3cos(theta) d(theta)

Answer 15

ONE PART WILL BE

Answer 16

AND OTHER PART USING SINTITA

Answer 17

CLEAR

Answer 18

Before I forgot, when you are doing trigonometric substitution. You must provide a new expression in terms of theta for each term. Thefore, you must provide 3 expression in terms of theta to substitute for x, dx and sqrt(6x-x^2) Did you get it @megannicole51 ?

Answer 19

okay so.....

x-3=3sin(theta)
dx=3cos(theta)d(theta)

which means....

the integral of ((dx)/(sqrt3sin^2(x)+3)) ??? I feel like this part is wrong

@Yttrium

Answer 20

\[\frac{ xdx }{ \sqrt{6x-x^2} } = \frac{ xdx }{ \sqrt{9 - (x-3)^2} }\]
now
u = x-3
du = dx
\[\frac{ u+3 }{ \sqrt{9 - u^2}\ }du\ = \frac{ u }{ \sqrt{9 - u^2}\ }du\ + \frac{ 3 }{ \sqrt{9 - u^2}\ }du\ \]

Answer 21

the first can be solved by another substitution
and the second is indeed arcsin.

Answer 22

@Coolsector I dont understand how you got u+3 in the numerator...

Answer 23

because u=x-3
we had xdx
so
x=u+3
and dx=du

xdx = (u+3)du

Answer 24

okay so then what do u do when you split up the equation?

Answer 25

With what we've discussed earlier, you already have value of dx which is 3cos(theta)d(theta)
And as I said, you need to find substitutes for all of the terms and you have two remaining: the x and the sqrt(6-x^2)

Using the equation x-3 = 3sin(theta)
You can therefore conclude that x = 3sin(theta) + 3

And using cos function you can conclude that sqrt(6-x^2) = (x+3)cos(theta)

Answer 26

\[\frac{ u+3 }{ \sqrt{9 - u^2}\ }du\ = \frac{ u }{ \sqrt{9 - u^2}\ }du\ + \frac{ 3 }{ \sqrt{9 - u^2}\ }du\ \]

now we can use for the first one
\[\frac{ u }{ \sqrt{9 - u^2}\ }du\]
the substitution:
t=9-u^2
dt = -2udu
so
udu = -dt/2
and it becomes:
\[\frac{ u }{ \sqrt{9 - u^2}\ }du\ =\frac{ -1 }{ 2\sqrt{t}\ }dt\ \]
and you should know how to integrate it

and the second :
\[\frac{ 3 }{ \sqrt{9 - u^2}\ }du = \frac{ 1 }{ \sqrt{1 - u^2/9}\ }du \]
now you should use the sub
p = u/3
so
du = 3dp
\[\ \frac{ 3 }{ \sqrt{1 - p^2}\ }dp \]
which is 3arcsin(p)

Answer 27

hope i didnt make mistakes here
so check me.

Answer 28

im lost...

Answer 29

okay can we start over with the original problem and using the same variables

Answer 30

ok

Answer 31

\[\frac{ xdx }{ \sqrt{6x - x^2}\ } = \frac{xdx }{ \sqrt{9 -(x-3)^2}\ } \]
agree?

Answer 32

yes im with you

Answer 33

now what happens when you sub
u = x-3

Answer 34

if we do it ( change var to u) we have to express everything using u.
so
the x in the numerator becomes
x = u+3
and
dx =du
and
9-(x-3)^2=9-u^2

ok?

Answer 35

okay here is where im getting mixed up with ur method and mine.... im using...

x-3=3sin(theta)
dx=3cos(theta)d(theta)

and i would just plug 3sin(theta) into the numerator where x-3 is...

Answer 36

no, if you do it you have to plug into the numerator
x = 3sin(theta) + 3
ant multiply it by 3cos(theta)d(theta)

Answer 37

but why would you do that

Answer 38

yes....because that is how it was taught to me

Answer 39

Because it is also applicable @Coolsector

Answer 40

i know it is but it is longer..
ok so this is what you get ?
\[\frac{ (3sin(\theta) + 3)3cos(\theta)d\theta) }{ \sqrt{9 - (3sin(\theta))^2} }\]

Answer 41

ignore the ) after the d(theta)

Answer 42

yes i know exactly how you got that

Answer 43

can you just finish my way and then show me yours

Answer 44

@megannicole51 ?

Answer 45

@megannicole51 can u read my inbox I sent u