∫ 2sin(2x+y) dx
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$\displaystyle \int 2\sin(2x+y)dx$ substitute $\displaystyle u=2x+y$ $\displaystyle \int 2\sin(2x+y)dx = \int \sin u du = -\cos u+C$ $\displaystyle =-\cos(2x+y)+C$
Originally Posted by Kvandesterren ∫ 2sin(2x+y) dx sbhatnagar's answer is correct if y is considered to be independent of x (so you are really doing partial integration, which is the opposite of partial differentiation). However it's incorrect if y is a function of x...
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