Question

∫R（sinx,cosx)dx=∫R[2u/(1+u^2),(1-u^2)/(1+u^2)]*[2/(1+u^2)]du

Answer 1

令U=tan(x/2)

sinx=2u/(1+u^2)

cosx=(1-u^2)/(1+u^2)

dx=2/(1+u^2)du

我就举一个例子 sinx=sinx/1 (*) (sinx=2sin(x/2)cos(x/2) 1=sin(x/2)^2+cos(x/2)^2

带入（*）后,分子分母同除以cos(x/2)^2 就得到sinx=2u/(1+u^2)