Question

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

Answer 1

Hint: Use formula $\sin 2x = 2\sin x\cos x.$
Let the value of integration be $I.$
$ \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\sin x\cos xdx} ,$
Dividing and multiplying by $2$ on right hand side, we’ll get:
$ \Rightarrow I = \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {2\sin x\cos xdx} ,$
We know that $2\sin x\cos x = \sin 2x$, using this we’ll get:
$ \Rightarrow I = \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\sin 2xdx} ,$
And we also know that, \[\int {\sin 2xdx = } - \frac{{\cos 2x}}{2} + C\], using this we will get:
\[
   \Rightarrow I = \frac{1}{2}\left[ { - \frac{{\cos 2x}}{2}} \right]_0^{\frac{\pi }{2}}, \\
   \Rightarrow I = - \frac{1}{4}\left[ {\cos 2x} \right]_0^{\frac{\pi }{2}} \\
\]
Putting limit of integration, we’ll get:
$
   \Rightarrow I = - \frac{1}{4}\left[ {\cos \pi - \cos 0} \right], \\
   \Rightarrow I = - \frac{1}{4}( - 1 - 1) = - \frac{1}{4} \times ( - 2), \\
   \Rightarrow I = \frac{1}{2}. \\
$
Thus, the value of integration is $\frac{1}{2}.$
Note: We will ignore the constant of integration while putting the limit because it is a definite integration.