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)?
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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
\[u=1-x^2\]
\[du=-2xdx\]
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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
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OpenStudy (anonymous):
but why does that work?
OpenStudy (anonymous):
the answer works
OpenStudy (anonymous):
backwards chain rule
OpenStudy (anonymous):
okay
OpenStudy (anonymous):
\[\frac{d}{dx} f(u(x))=f'(u(x))u'(x)\]
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