作业帮1/(sinxcosx)^3的原函数
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/09 13:15:36
1/(sinxcosx)^3的原函数
1/(sinxcosx)^3的原函数
1/(sinxcosx)^3的原函数
∫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
sin x=sin x*sinx =(1-cos x)sinx 原式=∫(1-cos x)sinxdx =∫(cos x-1)dcosx =cos x/3-cosx+C ARCSIN 3次根号下X sinx ^3
∫1/(sinxcosx)^3dx=8∫1/(sin2x)^3dx=-4∫1/(sin2x)^4dcos2x=-4∫1/[1-（cos2x)^2]^2dcos2x
设cos2x=y
1/(1-y^2)^2=(Ay+B)/(1+y)^2+(Cy+D)/(1-y)^2这一步
展开(Ay+B)(1+y^2+2y)+(Cy+D)(1+y^2-2y)=1
全部展开
∫1/(sinxcosx)^3dx=8∫1/(sin2x)^3dx=-4∫1/(sin2x)^4dcos2x=-4∫1/[1-（cos2x)^2]^2dcos2x
设cos2x=y
1/(1-y^2)^2=(Ay+B)/(1+y)^2+(Cy+D)/(1-y)^2这一步
展开(Ay+B)(1+y^2+2y)+(Cy+D)(1+y^2-2y)=1
计算得A=1/4,B=1/2,C=-1/4,D=1/2.
所以1/(1-y^2)^2=(1/4y+1/2)/(1+y)^2+(-1/4y+1/2)/(1-y)^2,
原式=-∫[(2-y)/(1-y)^2]dy-∫[(y+2)/(1+y)^2]dy
=-1/2∫d(y^2-2y+1)/(y^2-2y+1)-∫dy/(1-y)^2-1/2∫d(y^2+2y+1)/(y^2+2y+1)-∫dy/(1+y)^2
=ln(y-1)+1/(y-1)-ln(y+1)+1/(1+y)+C
收起