Integration: int ( x / sqrt[8-2x^2] ) dx

sepoto · Postby **sepoto** » Tue Dec 17, 2013 6:47 am

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In regards to:
$\int\frac{x}{\sqrt{8-2x^2}}dx$

$u=8-2x^2$
$du=-4x$

So the solutions manual says:
$\int u^{1/2}(-1/4)du$

I'm wondering is that a typo? Is it not supposed to be:
$\int u^{-1/2}(-1/4)du$

Thank You...

nona.m.nona · Postby **nona.m.nona** » Tue Dec 17, 2013 1:19 pm

sepoto wrote: $\int\frac{x}{\sqrt{8-2x^2}}dx$

$u=8-2x^2$
$du=-4x$

So the solutions manual says:
$\int u^{1/2}(-1/4)du$

I'm wondering is that a typo?

If one reviews the second solution method -- the "guess and check" -- one notes that the displayed differentiation does not result in the desired form. I agree with your noted change of sign.

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Integration: int ( x / sqrt[8-2x^2] ) dx

Integration: int ( x / sqrt[8-2x^2] ) dx

Re: Integration: int ( x / sqrt[8-2x^2] ) dx