Question

What is integral of $\sin 2x$?

Answer 1

Step 1: Substitute $u$ in the place $2x$ in the given integration

Let $I=\int \sin 2xdx$ ….(1)

and $2x=u$

Now differentiate the equation $2x=u$ with respect to $x$

$2dx=du$

$\Rightarrow$$dx=\frac{1}{2}du$

Now putting $dx=\frac{1}{2}du$in equation (1)

$I=\frac{1}{2}\int \sin udu$

Step 2: Apply the formula $\int \sin xdx=-\cos x+c$

$I=\frac{1}{2}\int \sin udu$

$\Rightarrow$$I=-\frac{1}{2}\cos u+c$ … where $c$ is an integration constant

Putting $u=2x$

$\Rightarrow$$I=-\frac{1}{2}\cos 2x+c$

Hence the integral of $\cos 2x$ is $-\frac{1}{2}\cos 2x+c$, where $c$ is an integration constant.