I=∫(π:0)xdx∫(x:0)cos(x+y)dy

=∫(π:0)[(sin(x+y)|(x:0)]xdx

=∫(π:0)(xsin2x-xsinx)dx

=-x(cos2x)/2+(sin2x)/4+xcosx-sinx |(π:0)

=-3π/2

Supplement to the question:

Omit the upper and lower limits for the sake of form.

∫(xsin 2x-xsinx)dx =∫(xsin 2x)dx-∫(xsinx)dx

Integration by parts: (refer to page 352 (1). This book should be used by the landlord.

The previous part:

∫(xsin 2x)dx =-x(cos2x)/2-∫(- 1/2)cos2xdx =-x(cos2x)/2+(sin2x)/4

The second half

∫(xsin x)dx =-xcos x-∫cosx dx =-xcos x+sinx

The process in the book is more troublesome. I usually use this method:

Divide xsinx by (1)*(2)

sinx( 1).....x(2)

-cosx(3).... 1(4)

(1) to (3) take the integral and (2) to (4) take the derivative.

∫(xsinx)dx =(2)*(3)-(3)(4)dx = ... (omitted)