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