Mathematics 48 Online
OpenStudy (anonymous):

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)?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yep that's the integral

myininaya (myininaya):

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

OpenStudy (anonymous):

yep, I have that

OpenStudy (anonymous):

straight forward u-sub. put $u=1-x^2$ $du=-2xdx$

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

oh, okay thanks

myininaya (myininaya):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

but why does that work?

OpenStudy (anonymous):

OpenStudy (anonymous):

backwards chain rule

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so $\int f'(u(x))u'(x)dx = f(u(x))$

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

yw

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