Quick integration question (1 Viewer)

youngminii · Feb 26, 2009

Integrate
x^2/sqrt(1 - x^2)

Thankyedesu.

hermand · Feb 26, 2009

$\frac{x^{2}}{\left (1-x^{2} \right )^{\frac{1}{2}}}$
$\frac{dy}{dx}=\frac{vu'-uv'}{v^{2}}$
$u=x^{2}$ $u'=2x$
$v=(1-x^{2})^{\frac{1}{2}}$
$v'=\frac{-x}{\left (1-x^{2} \right )^{\frac{1}{2}}}$
$\therefore \frac{dy}{dx}=\frac{(1-x^{2})^{\frac{1}{2}}2x-x^{2}(\frac{-x}{(1-x^{2})^{\frac{1}{2}}})}{\left [(1-x^{2})^{\frac{1}{2}} \right ]^{2}}$
taking a common denominator in the numerator then will transfer to the denominator of the entire fraction.
$\therefore \frac{dy}{dx}=\frac{\left [(1-x^{2})^{\frac{1}{2}} \right ]^{2}2x+x^{3}}{(1-x^{2})\sqrt{1-x^{2}}}$
$\therefore \frac{dy}{dx}=\frac{(1-x^{2})2x+x^{3}}{(1-x^{2})\sqrt{1-x^{2}}}$
$\therefore \frac{dy}{dx}=\frac{2x-x^{3}}{(1-x^{2})\sqrt{1-x^{2}}}$

EDIT; shit, i differentiated.
sorry!

lolokay · Feb 26, 2009

$\frac{x^2}{\sqrt{1-x^2}}\\ = \frac{-(1-x^2)}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-x^2}}\\ = -\sqrt{1-x^2} + \frac{1}{\sqrt{1-x^2}}\\ \\$
integrate first part with trig substitution

trailblazer · Feb 26, 2009

hermand said:
$\frac{x^{2}}{\left (1-x^{2} \right )^{\frac{1}{2}}}$
$\frac{dy}{dx}=\frac{vu'-uv'}{v^{2}}$
$u=x^{2}$ $u'=2x$
$v=(1-x^{2})^{\frac{1}{2}}$
$v'=\frac{-x}{\left (1-x^{2} \right )^{\frac{1}{2}}}$
$\therefore \frac{dy}{dx}=\frac{(1-x^{2})^{\frac{1}{2}}2x-x^{2}(\frac{-x}{(1-x^{2})^{\frac{1}{2}}})}{\left [(1-x^{2})^{\frac{1}{2}} \right ]^{2}}$
taking a common denominator in the numerator then will transfer to the denominator of the entire fraction.
$\therefore \frac{dy}{dx}=\frac{\left [(1-x^{2})^{\frac{1}{2}} \right ]^{2}2x+x^{3}}{(1-x^{2})\sqrt{1-x^{2}}}$
$\therefore \frac{dy}{dx}=\frac{(1-x^{2})2x+x^{3}}{(1-x^{2})\sqrt{1-x^{2}}}$
$\therefore \frac{dy}{dx}=\frac{2x-x^{3}}{(1-x^{2})\sqrt{1-x^{2}}}$

EDIT; shit, i differentiated.
sorry!

LOL, thought you were doing some awesome method, before i realised it was the quotient rule.

Manboobs · Feb 26, 2009

you could use the quotient rule that way if you wanted. It's similar to the parts method.

u an v are functions of x . Integrating the quotient rule.

Rearranging

So just using a substitution and the answer will pop out. Albeit being pretty neat, using this method is only useful in certain cases.

youngminii · Feb 26, 2009

Oh cool, thanks heaps everyone~

Trebla · Feb 26, 2009

The "integration by reverse quotient rule" is exactly the same as integration by parts, except far more complicated. You're better off integrating by parts because the quotient rule is basically the product rule applied on u x 1/v.
However, the method lolokay suggested is by far the most efficient method that I know of, which is basically a "partial fractions" method by splitting the numerator.

Quick integration question (1 Viewer)

youngminii

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hermand

je t'aime.

lolokay

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trailblazer

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Manboobs

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youngminii

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Trebla

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