Thread: integration

∫ 2sin(2x+y) dx

$\displaystyle \int 2\sin(2x+y)dx$

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...