Question

How do you integrate $\int \sin x\cos xdx$.

Answer 1

Step-1: Assume a variable:

The given integration is,

$\int \sin x\cos xdx=\frac{1}{2}\int 2\sin x\cos xdx$

$=\frac{1}{2}\int \sin 2xdx$ $\left[\because 2sinxcosx=sin2x\right]$

Let $u=2x$.

$\therefore du=2dx$

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

Step-2: Evaluate the integration:

Substituting this in the above equation we get,

$I=\frac{1}{2}\int \sin u\frac{du}{2}$

$=\frac{1}{4}\int \sin udu$

$=-\frac{1}{4}\cos u+c$ $\left[\because \int \mathrm{sinxdx}=-\mathrm{cosx},\text{c be the integration constant}\right]$

$=-\frac{1}{4}\cos 2x+c$ $\left[\because u=2x\right]$

Hence, the required answer is $\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}\mathbf{+}\mathit{c}$.