Question

calc help, intergral (2-x)/sqrt(4-x^(2))
help step by step please

Answer 1

We have
\[\int \frac{2-x}{\sqrt{4-x^2}}dx\]
\[\int \frac{2}{\sqrt{4-x^2}}-\frac{x}{\sqrt{4-x^2}}dx\]
Do you get this step?

Answer 2

yeah

Answer 3

Do you know the derivative of \(\sin^{-1}x\)?

Answer 4

thats 1/sqrt(1-x^(2))

Answer 5

i believe

Answer 6

the first set i recieved 2arcsin(x/2) but i dont remember how to do the one with x as the numerator

Answer 7

good:)
second is easy

Answer 8

\[\int-\frac{x}{\sqrt{4-x^2}}dx\]
put
\[4-x^2=t\]
so
\[-2x dx=dt\]
or
\[\frac{dt}{-2x}=dx\]
now our integral \[\int-\frac{x}{\sqrt{t}}\frac{dt}{-2x}\]
we get
\[\int \frac{dt}{2\sqrt t}dt\]
I think you can do this, can' t you?

Answer 9

yup then i got 1/2 ln | u or 4-x^(2)| + c

Answer 10

i just add that with the first one and im fine correct.

Answer 11

It's not 1/t, you won't get ln t
it's
\[\int \frac{1}{2 \underline {\sqrt t}}dt\]

Answer 12

oh, ok i see that but now im stuck lol

Answer 13

it wouldnt be 1/2arcsin(x/2) would it?

Answer 14

The first one is almost correct
it'd be
\[\arcsin \frac x 2\]

Answer 15

how is the first equation not 2arcsin(x/2)?

Answer 16

Sorry you're correct with the first one
it'd be
\[2\arcsin \frac x2\]

Answer 17

o, ok and the second one would be 1/2arcsin(x/2)?

Answer 18

use the following for second
\[\int x^n dx=\frac{x^{n+1}}{n+1}\]
here we have
\[\int \frac{1}{\sqrt t} dt\]
here n=\(-\frac 12\)

Answer 19

nope, please read the post for the second integral again

Answer 20

ohhh im had a dumb moment, i can bring the 1/sqrt(t) up and make it t^(-1)dt that becomes t^(1/2) / (1/2) or 2t(1/2)

Answer 21

if you bring it up, it'll be \(\large t^{-\frac 1 2}\)

Answer 22

^

Answer 23

oops, after integration
you're right then
\[2\sqrt t\]

Answer 24

you got it?

Answer 25

yeah my final looks like 2arcsin(x/2) - sqrt(u) or sqrt (4-x^(2)) + c

Answer 26

good work:D
Sorry to confuse you in between:(

Answer 27

lol its alright i appreciate the help.

Answer 28

thanks:)