99问答网
所有问题
当前搜索:
∫cosx^3dx
∫
π/2到-π/2√cosx-
cosx^3dx
=
答:
∫π/2到-π/2√cosx-
cosx^3dx
=-2∫(0,π/2)√cosx(1-cos²x)dx =-2∫(0,π/2)sinx√cosxdx =2∫(0,π/2)√cosxdcosx =2*(2/3)(cosx)^(3/2)|(0,π/2)=-4/3
∫
xsin^3xdx 上限π 下限0 求定积分
答:
=1/3 *∫ x d(cosx)^3 = x/3 *(cosx)^3 -∫1/3 *(cosx)
^3dx
= x/3 *(cosx)^3 -∫1/3 *(cosx)^2 d(sinx)= x/3 *(cosx)^3 -∫1/3 -(sinx)^2 /3 d(sinx)= x/3 *(cosx)^3 -1/3 *sinx +1/9 *(sinx)^3 ∫-x d(cosx)= -x *cosx +
∫cosx
dx = ...
sin^3x的不定积分(cos3xsin3x的不定积分)
答:
sin^3x的不定积分为:1/3cos^3-cosx+C。解:∫sin
^3dx
=∫sin^2*sinxdx=∫)d=∫-1)dcosx=∫cos^2dcosx-∫1dcosx=1/3cos^3-cosx+C。不定积分公式:
∫cosx
dx=sinx+C、∫sinxdx=-cosx+C、∫cscxdx=-cotx+C、∫2dx=2x+C。积分中常见形式:求含有e^x的函数的积分∫x*e^xdx=∫xd...
函数f(x)=(sinx)
^3
,求定积分值
答:
=1/3 *∫ x d(cosx)^3 = x/3 *(cosx)^3 -∫1/3 *(cosx)
^3dx
= x/3 *(cosx)^3 -∫1/3 *(cosx)^2 d(sinx)= x/3 *(cosx)^3 -∫1/3 -(sinx)^2 /3 d(sinx)= x/3 *(cosx)^3 -1/3 *sinx +1/9 *(sinx)^3 ∫-x d(cosx)= -x *cosx +
∫cosx
dx = ...
(sinx)^3+(
cosx
)^3求积分
答:
方便起见,积分符号写成J J[(sinx)^3+(
cosx
)^3]dx=J(sinx)
^3dx
+J(cosx)^3dx=-J(1-cos^2x)dcosx+J(1-sin^2x)dsinx=-cosx-(cosx)^3/3+sinx-(sinx)^3/3+C
x*(sinx)
^3
的积分原函数是什么
答:
=x[(
cosx
)^3/3-cosx]-∫[(cosx)^3/3-cosx]dx =x[(cosx)^3/3-cosx]+sinx -(1/3)∫(cosx)
^3dx
=x[(cosx)^3/3-cosx]+sinx -(1/3)∫[1-(sinx)^2]dsinx =x[(cosx)^3/3-cosx]+sinx -(1/3)[sinx-(sinx)^3/3]=(2/3)sinx-xcosx+(1/3)x(cosx)^3+(1/9)(...
∫
(0,pai) x(sinx)
^3dx
=?
答:
设f(sinx) = sin³x 则∫(0~π) xsin³x dx = ∫(0~π) xf(sinx) dx = (π/2)∫(0~π) f(sinx) dx = (π/2)∫(0~π) sin³x dx = (π/2)∫(0~π) (cos²x - 1) d
cosx
= (π/2)[(1/3)cos³x - cosx] |(0~π)= (π/2)[...
∫
dx/(sin³x
cosx
)
答:
=∫[(sinx)^2+(cosx)^2]/[(sinx)^3cosx]dx =∫1/(sinxcosx)dx+
∫cosx
/(sinx)
^3dx
=2∫1/(2sinxcosx)dx+∫1/(sinx)^3d(sinx)=2∫1/sin2x dx+∫1/(sinx)^3d(sinx)=∫1/sin2x d(2x)+∫1/(sinx)^3d(sinx)=ln|tanx|-(1/2)(sinx)^(-2)+C ...
求
∫
tan^3xdx 按我的步骤做!
答:
就按楼主的步骤做 sin^3x提出一个sinx、sin^3x/cos^3x dx =1/3sin^2x/cos^3xd
cosx
=(1-cos^2x)/cos^3xdcosx=(1/cos^3x-1/cosx)dcosx =-1/4(cosx)^(-4)-ln|cosx|+C(常数)觉得好请采纳
求定积分
∫
1-(sinx)
^3dx
积分下限0 上限是派
答:
∫1-(sinx)
^3dx
=∫1+sinx-(sinx)^3-sinxdx =∫1+sinx[1-(sinx)^2]-sinxdx =∫1+sinx(
cosx
)^2-sinxdx =∫1-sinxdx+∫sinx(cosx)^2dx =∫1-sinxdx-∫(cosx)^2dcosx =x+cosx-(1/3)(cosx)^3 代入上下限,得到定积分为π-4/3 嗯,我写程序计算了一下,你的答案是正确的。
棣栭〉
<涓婁竴椤
12
13
14
15
17
18
19
20
21
涓嬩竴椤
灏鹃〉
16
其他人还搜