EVALUATE THE INTEGRAL sin11 theta cos5 theta d theta EVALUATE THE INTEGRAL Solution let sin(theta) = y cos(theta)^2 = 1-(y^2) theta varies from 0 to pie/2 so y varies from 0 to 1 then given function = (y^11 )*(1-y^2)^2 dy integral from 0 to 1 = integral of ( y^11)-2*(y^13)+(y^15) dy from 0 to 1 1 = [((y^12)/12) -2*((y^14)/14)+((y^16)/16) ] 0 = ( (1/12) -(2/14)+(1/16) ] = 0.00297619048.