Question

integral of x/sqrt(1-x^2) from (-1/8) to 0

Answer 1

\[\int\limits_{{-1/8}}^{0} \frac{x}{\sqrt{1-x^2}}\]

Answer 2

now normally it should be arcsin, but that x is on top, we cannot throw it infront of the integral though right?

Answer 3

you can either use u-sub or trig-sub. which one do you want to use?

Answer 4

u sub i guess..

Answer 5

u = x
du = 1

Answer 6

u on top? doesn't work...

Answer 7

no..the bottom is \(1-x^2\) then let u =1-x^2 => du/dx = -2x

Answer 8

I figured it would be trig but yeah that works too.

-1/2du = xdx

Answer 9

so then we have a u'/u

Answer 10

you dont change it to -1/2du=xdx it wont work

Answer 11

why';s that? Don't we want to get the x on top as our u'?

Answer 12

hmm I guess that doesnt' work with the sqrt..

Answer 13

\[\int\limits\frac{x}{\sqrt{1-x^{2}}} dx \]
\[\int\limits\frac{x}{\sqrt{u}} *-\frac{du}{-2x} => \int\limits\frac{\cancel{x}}{\sqrt{u}} *-\frac{du}{\cancel{-2x}} \]

Answer 14

do you understand? you have to cancel the \(x\)

Answer 15

which leads to
\[\frac{1}{2} \int\limits\frac{1}{\sqrt{u}} du\]

Answer 16

hmm never have seen that as du/something.

Answer 17

then trig sub? lol

Answer 18

i gotta go sorry

Answer 19

it's fine, thanks for the help thus far.

Answer 20

I guess when it's du = something dx then it's du/ that something?

Answer 21

or actually it's like I was showing before except instead of 1/2du it's 1/2x du..I got it.