Question

What is $\sin 2x$ integration?

Answer 1

Integration of trigonometric function:

$\int \sin 2x\mathrm{dx}$.

$$\begin{gathered} \mathrm{Let}u=2x \\ \Rightarrow du=2dx \\ \Rightarrow dx=\frac{1}{2}du \end{gathered}$$

Put the above values in $\int \sin 2x\mathrm{dx}$:

$$\begin{gathered} \therefore \int \sin 2xdx=\frac{1}{2}\int \sin udu \\ \Rightarrow \int \sin 2xdx=\frac{1}{2}(-\cos u)+C\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{\int }\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{}\mathbf{dx}\mathbf{=}\mathbf{-}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{]} \\ \Rightarrow \int \sin 2xdx=-\frac{1}{2}\cos 2x+C\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{u}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{]} \end{gathered}$$

Hence, the required integral value is $\mathbf{\int }\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{}\mathbf{dx}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{C}$