Question

What is integration of $\sin x\cos x$ using $\sin 2x$ formula?

Answer 1

Integration of $\mathbf{sinx}\mathbf{}\mathbf{cosx}$ using $\mathbf{sin}\mathbf{2}\mathbf{x}$ formula.

We know, $\sin 2x=2\sin x\cos x$

Now,

$\begin{array}{rcl}\int \sin x\cos xdx& =& \frac{1}{2}\int 2\sin x\cos xdx\\ & =& \frac{1}{2}\int \sin 2xdx\\ & =& \frac{1}{2}\times \frac{-\cos 2x}{2}+c\mathbf{}\left[\because \int sinaxdx=-\frac{1}{a}cos\left(ax\right)\right]\\ & =& -\frac{1}{4}\cos 2x+c\end{array}$

Thus, using the formula $\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{x}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{x}$ ,$\mathbf{\int }\mathbf{sinx}\mathbf{}\mathbf{cosx}\mathbf{}\mathbf{dx}\begin{array}{rcl}& =& \end{array}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{c}$.