How do you integrate int sin 2x dx?

We have to integrate sin2xdx


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

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


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

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