How do you integrate $int sin2x dx$ ?

Question

147711 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-17T12:54:55

$intsin(2x)dx=-1/2cos(2x)+C$

Using integration by substitution together with the known integral $intsin(x)dx = -cos(x)+C$ , we first let $u = 2x => du = 2dx$ . Then

$intsin(2x)dx = 1/2intsin(2x)2dx$

$=1/2intsin(u)du$

$=1/2(-cos(u))+C$

$=-1/2cos(2x)+C$

How do you integrate int sin2x dxint sin2x dx?