# How do you integrate int sin 2x dx?

We have to integrate sin2xdx

### Solution

$\int \sin 2x dx = \frac{1}{2}\int \sin (2x)2dx$ —————(i)

Let us assume u = 2x and du = 2.dx

We know that $\int \sin x= -\cos x + C$

Hence on substituting eq (i) becomes

$\int \sin 2xdx = \frac{1}{2}\int sin(u)du$

=$\int \sin 2xdx = \frac{1}{2} – cosu.du$ + C

= $-\frac{1}{2} cos(2x)+C$

$\int sin 2x dx =-\frac{1}{2} cos(2x)+C$