Find the value of definite integration $\int\limits_0^{\dfrac{\pi }{2}} {\sin x\cos xdx} $ .

Hint-In this question, we use the concept of definite integration and also use basic trigonometric identity. We use trigonometric identity $\sin 2x = 2\sin x\cos x$ and we also use \[\int\limits_a^b {\sin n\theta = \left[ {\dfrac{{ - \cos n\theta }}{n}} \right]_a^b} \] .

Complete step-by-step answer:
Let $I = \int\limits_0^{\dfrac{\pi }{2}} {\sin x\cos xdx} $
Now, multiply by 2 on both sides of the equation.
$ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {2\sin x\cos xdx} $
We use trigonometric identity $\sin 2x = 2\sin x\cos x$
$ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\sin 2xdx} $
As we know integration \[\int\limits_a^b {\sin n\theta = \left[ {\dfrac{{ - \cos n\theta }}{n}} \right]_a^b} \]
$
   \Rightarrow 2I = \left[ {\dfrac{{ - \cos 2x}}{2}} \right]_0^{\dfrac{\pi }{2}} \\
   \Rightarrow 2I = \dfrac{{ - 1}}{2}\left[ {\cos 2x} \right]_0^{\dfrac{\pi }{2}} \\
$
$
   \Rightarrow 2I = \dfrac{{ - 1}}{2}\left[ {\cos 2 \times \dfrac{\pi }{2} - \cos 2 \times 0} \right] \\
   \Rightarrow 2I = \dfrac{{ - 1}}{2}\left[ {\cos \pi - \cos 0} \right] \\
 $
We know the value of $\cos \pi = - 1{\text{ and }}\cos 0 = 1$
$
   \Rightarrow 2I = \dfrac{{ - 1}}{2}\left[ { - 1 - 1} \right] \\
   \Rightarrow 2I = \dfrac{{ - 1}}{2} \times \left( { - 2} \right) \\
   \Rightarrow I = \dfrac{1}{2} \\
$
So the value of integration $\int\limits_0^{\dfrac{\pi }{2}} {\sin x\cos xdx} $ is $\dfrac{1}{2}$ .
Note-In such types of questions we use some important points to solve problems in an easy way. First we convert the big expression into a small expression by using trigonometric identity as mentioned in above and then apply integration because we know the integration of standard trigonometric form (like integration of sinx is cosx). Then after putting the limit we will get the required answer.