求∫cosx/sin^2xdx; ∫sec5xdx; ∫1/(√x+3√x)dx; ∫[√﹙x-1﹚/x]dx; ∫sin√xdx;

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 07:36:33
求∫cosx/sin^2xdx;∫sec5xdx;∫1/(√x+3√x)dx;∫[√﹙x-1﹚/x]dx;∫sin√xdx;求∫cosx/sin^2xdx;∫sec5xdx;∫1/(√x+3√x)dx

求∫cosx/sin^2xdx; ∫sec5xdx; ∫1/(√x+3√x)dx; ∫[√﹙x-1﹚/x]dx; ∫sin√xdx;
求∫cosx/sin^2xdx; ∫sec5xdx; ∫1/(√x+3√x)dx; ∫[√﹙x-1﹚/x]dx; ∫sin√xdx;

求∫cosx/sin^2xdx; ∫sec5xdx; ∫1/(√x+3√x)dx; ∫[√﹙x-1﹚/x]dx; ∫sin√xdx;
1、∫cosxdx/sin²x=∫cscxcotxdx=-cscx + c
2、∫sec5xdx=(1/5)∫sec5xd(5x)=(1/5)ln|sec5x+tan5x| + c
3、∫dx/[x^(1/2)+x^(1/3)],u=x^1/6,x=u^6,dx=6u^5du
= 6∫u^5du/(u³+u²)
= 6∫u³du/(u+1)
= 6∫[u²-u-1/(u+1)+1]du
= 6[u³-u²-ln|u+1|+u] + c
= 2x^(1/2)-3x^(1/3)+6x^(1/6)-6ln|1+x^(1/6)| + c
4、∫√(x-1)dx / x,u=√(x-1),du=dx/[2√(x-1)]
= 2∫u²du/(1+u²)
= 2∫[1-1/(1+u²)]du
= 2(u-arctanu) + c
= 2√(x-1) - 2arctan√(x-1) + c
5、∫sin√xdx
= ∫ (2√x)sin√xdx / (2√x)
= 2∫√xsin√xd√x
= -2∫√xdcos√x
= - 2√xcos√x + 2∫cos√xd√x
= 2sin(√x) - 2(√x)cos(√x) + c

1)∫sec5xdx=1/5∫sec5xd5x=1/5ln|sec5x+tan5x|+c
2)∫sin√xdx;
令√x=t,x=t^2,dx=2tdt,
∫sin√xdx= 2∫tsintdt = -2∫tdcost = -2(tcost-∫costdt) = -2(tcost-sint)+c=-2(√xcos√x-sin√x)+c
3)∫[√﹙x-1﹚/...

全部展开

1)∫sec5xdx=1/5∫sec5xd5x=1/5ln|sec5x+tan5x|+c
2)∫sin√xdx;
令√x=t,x=t^2,dx=2tdt,
∫sin√xdx= 2∫tsintdt = -2∫tdcost = -2(tcost-∫costdt) = -2(tcost-sint)+c=-2(√xcos√x-sin√x)+c
3)∫[√﹙x-1﹚/x]dx
令√﹙x-1﹚=t, x=t^2+1, dx=2tdt, ∫[√﹙x-1﹚/x]dx =∫t*2t/(t^2+1)*dt=2∫[1-1/(t^2+1)]dt
= 2(t-arctant)+c = 2[√﹙x-1﹚-arctan√﹙x-1﹚]+c

收起