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