原式=∫1/[√2sin(x+π/4)]² dx
=1/2∫dx/sin²(x+π/4)
=1/2∫csc²(x+π/4)d(x+π/4)
=-1/2∫-csc²(x+π/4)d(x+π/4)
=-1/2*cot(x+π/4)+C
1/((sinx+cosx)^2)的积分怎么算?
原式=∫1/[√2sin(x+π/4)]² dx
=1/2∫dx/sin²(x+π/4)
=1/2∫csc²(x+π/4)d(x+π/4)
=-1/2∫-csc²(x+π/4)d(x+π/4)
=-1/2*cot(x+π/4)+C