Integration Question (1 Viewer)

Eddykungfu · May 6, 2009

From my 3u mid years I had this question...

How do you do it??

Drongoski · May 6, 2009

Eddykungfu

$\100dpi \int xcosxdx\ \ =\ \ \int xdsinx\ \ =\ \ xsinx\ -\ \int sinxdx\ \ =\ \ xsinx+\ cosx\ +\ C$

This was integration by parts, a 4U topic.

Eddykungfu · May 6, 2009

I don't get how that works

. whered the xsinx come from

Drongoski · May 6, 2009

In integration by parts we use the formula:

$\100dpi \ \ \ \ \ \int udv\ \ =\ \ uv\ -\ \int vdu\ \ \ \ \ (u\ &\ v\ being\ functions\ of\ x)$

gurmies · May 6, 2009

$\frac{d(x\sin x)}{dx}=\sin x + x\cos x$

$x\cos x = \frac{d(x\sin x)}{dx}-\sin x$

$\int (x\cos x)dx = \int [\frac{d(x\sin x)}{dx}-\sin x]dx$

$= x\sin x + \cos x + C$

jet · May 6, 2009

gurmies said:
$\frac{d(x\sin x)}{dx}=\sin x + x\cos x$

$x\cos x = \frac{d(x\sin x)}{dx}-\sin x$

$\int (x\cos x)dx = \int [\frac{d(x\sin x)}{dx}-\sin x]dx$

$= x\sin x + \cos x + C$

The proper 3u way of doing it. This is integration by parts in an implicit form, though approaches it from a perspective whereby you don't need to understand it.

Integration Question (1 Viewer)

Eddykungfu

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Drongoski

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Eddykungfu

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Drongoski

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gurmies

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jet

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