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December 22, 2014

December 22, 2014

Posted by **abhi** on Friday, October 12, 2012 at 11:38am.

- math -
**Steve**, Friday, October 12, 2012 at 11:43am∫sin√x dx

let

u = √x

du = 1/(2√x) dx

dx = 2√x du = 2u du

∫sin√x dx = ∫sinu (2u du)

= 2∫u sinu du

now use integration by parts to get

2sinu - 2u cosu

and convert back to x's

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