Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\int_0^1 {x\cos xy} } dydx \cr
& = \int_0^{\pi /2} {\left[ {\int_0^1 {x\cos xy} dy} \right]} dx \cr
& {\text{solve the inner integral}}{\text{, treat }}x{\text{ as a constant}} \cr
& \int_0^1 {x\cos xy} dy = \left[ {\sin xy} \right]_0^1 \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \sin x\left( 1 \right) - \sin 0 \cr
& {\text{simplifying}} \cr
& = \sin x \cr
& \cr
& = \int_0^{\pi /2} {\left[ {\int_0^1 {x\cos xy} dy} \right]} dx = \int_0^{\pi /2} {\sin x} dx \cr
& {\text{integrating}} \cr
& = \left( { - \cos x} \right)_0^{\pi /2} \cr
& = \left( {\cos x} \right)_{\pi /2}^0 \cr
& {\text{evaluate}} \cr
& = \cos \left( 0 \right) - \cos \left( {\frac{\pi }{2}} \right) \cr
& = 1 - 0 \cr
& = 1 \cr} $$
