Question

I have to take the integral of this: x/(sqrt(9- x^2)).
The first step is to rewrite the integral as (-1/2)the integral of (9 - x^2)^(-1/2) (______x)dx. What goes in the space (_____x)?

Answer 1

\[\int\frac{x}{\sqrt{9-x^2}}dx\]

Answer 2

yep that's the integral

Answer 3

let u=9-x^2
du=-2xdx
-1/2 du=x dx

Answer 4

yep, I have that

Answer 5

straight forward u-sub. put
\[u=1-x^2\]
\[du=-2xdx\]

Answer 6

if you are answering a web question step by step it looks like you should fill in the blank with -2

Answer 7

oh, just put the u-sub in the space? with (-1/2) on the ouside?

Answer 8

oh, okay thanks

Answer 9

\[-1/2*\int\limits_{}^{}\frac{du}{\sqrt{u}}\]

Answer 10

your question was "what goes in the blank?"
it is -2

Answer 11

but why does that work?

Answer 12

the answer works

Answer 13

backwards chain rule

Answer 14

okay

Answer 15

\[\frac{d}{dx} f(u(x))=f'(u(x))u'(x)\]

Answer 16

so
\[\int f'(u(x))u'(x)dx = f(u(x))\]

Answer 17

thanks

Answer 18

yw