integral_{x=0 to pi/2} sin(x)cos(x)/[x+1] dx

first combine the sin and cos to get:

integral_{x=0 to pi/2} sin(x)cos(x)/[x+1] dx =

....................... integral_{x=0 to pi/2} sin(2x)/[2x+2] dx

Now change the variable of integration to u = 2x+2, to get:

integral_{x=0 to pi/2} sin(x)cos(x)/[x+1] dx =

....................... integral_{u=2 to pi+2} (1/2) sin(u-2)/u du

....................... integral_{u=2 to pi+2} (1/2) [sin(u)cos(2)-cos(u)sin(2)]/u du

and the result should now follow from the definitions of the sin and cosin

integrals Si and Ci.

RonL

added in explanation: you also need that for a sufficiently well behaved function:

integral_{u=a to b} f(u) du = integral_{u=0 to b} f(u) du - integral_{u=0 to a} f(u) du