What is the integral of $int sin(2x)dx$ ?

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Answer 1 · 2016-03-17T02:56:32

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

We will use the rule:

$intsin(u)du=-cos(u)+C$

So, if we have

$intsin(2x)dx$

we should set

$u=2x" "=>" "du=2dx$

In order to have a $2dx$ in the integral expression, multiply by a $2$ on the inside and balance it with a $1/2$ on the outside:

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

Substitute:

$=1/2intsin(u)du$

Using the initial rule, this becomes:

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

Since $u=2x$ ,

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

What is the integral of int sin(2x)dxint sin(2x)dx?