 QUESTION

# 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 $\int\limits_a^b {\sin n\theta = \left[ {\dfrac{{ - \cos n\theta }}{n}} \right]_a^b}$ .

Let $I = \int\limits_0^{\dfrac{\pi }{2}} {\sin x\cos xdx}$
$\Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {2\sin x\cos xdx}$
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 $\dfrac{1}{2}$ .