三角函數(shù)萬能公式
令tanx/2=t
(secx/2)^2dx/2=dt
dx=2dt/(t^2+1)
∫dx/(sinx+tanx)
=∫2dt/{(t^2+1)*[2t/(1+t^2)+2t/(1-t^2)]}
=∫dt/{(1+t^2)*2t/[(1+t^2)(1-t^2)]}
=∫(1-t^2)/(2t)dt
=1/2∫(1/t-t)dt
=1/2lnt-t^2/4+c
=1/2ln[tan(x/2)]-[tan(x/2)]^2/4+c