integration

**Kvandesterren** · Jan 13th 2013, 05:15 AM

∫ 2sin(2x+y) dx

**sbhatnagar** · Jan 13th 2013, 05:28 AM

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

substitute $u=2x+y$

$\int 2\sin(2x+y)dx = \int \sin u du = -\cos u+C$

$=-\cos(2x+y)+C$

**Prove It** · Jan 13th 2013, 08:45 PM

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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