Question

${\int }_0^{\frac{\pi }{2}}\sin 2x.\cos x$ $dx$

Answer 1

We have,

${\int }_0^{\frac{\pi }{2}}\sin 2x\cos xdx$

${\int }_0^{\frac{\pi }{2}}2\sin x\cos x\cos xdx$

$\Rightarrow 2{\int }_0^{\frac{\pi }{2}}\sin x{\cos }^{2}xdx$

Let

$\cos x=t$

$\sin xdx=dt$

Therefore,

$2{\int }_1^0{t}^{2}dt$

On integrating and we get,

$=2{{\left[\frac{t^3}{3}\right]}_1}^0$

Then,

$=2\left[\frac{0^3-1^3}{3}\right]$

$=\frac{-2}{3}$

Hence, this is the answer.