1.y=∫(sint+costsint)dt=∫(sint(1+cost)dt=-∫(1+cost)dcost=-1/2(cosx)^2-cosx=-1/2 (cosx+1)^2+1/2
所以,当cosx=-1时,y有最大值1/2.
2.f(x)=(1-tanx)*0+1*(1+sin2x+cos2x)=1+sin2x+cos2x 定义域为R
f(x+π/8)=1+sin(2x+π/4)+cos(2x+π/4)=1+(根号下2)/2 *sin2x+(根号下2)/2 *cos2x=根号下2÷5
所以 sin2x+cos2x=2/5-2*根号下2 即 f(x)=7/5-2*根号下2