Question

1/(sinxcosx)^3的原函数

Answer 1

∫1/(sinxcosx)^3dx=8∫1/(sin2x)^3dx=-4∫1/(sin2x)^4dcos2x=-4∫1/[1-（cos2x)^2]^2dcos2x

设cos2x=y

上式=-4∫1/[1-y^2]^2dy=-8∫(1/(1-y^2)+1/(1+y^2)]dy=-8arctany+16ln|(y+1)/(y-1)|+c

∫1/(sinxcosx)^3dx==-8arctan(cos2x)+16ln|(cos2x+1)/(cos2x-1)|+c