Question

How do you integrate $\int \sin 2xdx$.

Answer 1

Step- 1: Assume a variable:

The given integration is,

$\int \sin 2xdx$

Let $u=2x$.

$\therefore du=2dx$

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

Step- 2: Evaluate the integration:

Substituting this in the above equation we get,

$I=\int \sin u\frac{du}{2}$

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

$=-\frac{1}{2}\cos u+c$ $\left[\because \int \mathrm{sinxdx}=-\mathrm{cosx},\text{c be the integration constant}\right]$

$=-\frac{1}{2}\cos 2x+c$ $\left[\because u=2x\right]$

Hence, the required answer is $\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{x}\mathbf{+}\mathit{c}$.