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∫cos^2xdx
求不定积分
∫
x^2
cos^2xdx
答:
倍角加分步
cos^
2x=(cos2x+1)/2 原因为化为 ∫1/2*x^2dx+1/4∫x^2dsin2x =1/6x^3+1/4sin2x*x
^2
-1/
2∫
xsin2xdx =1/6x^3+1/4sin2x*x^2+1/4xcos2x-1/4
∫cos2xdx
=1/6x^3+1/4sin2x*x^2+1/4xcos2x+1/8sin2x 思路是这样,错没错不晓得 ...
∫
x
cos^2xdx
的积分怎么求?
答:
= xcosx -
∫cos^2xdx
= xcosx - ∫(1 + cos2x)/2 dx = xcosx - (x/2) - ∫cos2x/2 dx = xcosx - (x/2) - (1/4)∫(cos4x + cos2x)/2 dx = xcosx - (x/2) - (1/8)∫cos4x dx - (1/8)∫cos2x dx = xcosx - (x/2) - (1/32)sin4x - (1/16)...
∫
x
cos^2xdx
的积分如何求解?
答:
= xcosx -
∫cos^2xdx
= xcosx - ∫(1 + cos2x)/2 dx = xcosx - (x/2) - ∫cos2x/2 dx = xcosx - (x/2) - (1/4)∫(cos4x + cos2x)/2 dx = xcosx - (x/2) - (1/8)∫cos4x dx - (1/8)∫cos2x dx = xcosx - (x/2) - (1/32)sin4x - (1/16)...
如何求
∫
x
cos^2xdx
?
答:
= xcosx -
∫cos^2xdx
= xcosx - ∫(1 + cos2x)/2 dx = xcosx - (x/2) - ∫cos2x/2 dx = xcosx - (x/2) - (1/4)∫(cos4x + cos2x)/2 dx = xcosx - (x/2) - (1/8)∫cos4x dx - (1/8)∫cos2x dx = xcosx - (x/2) - (1/32)sin4x - (1/16)...
如何求
∫
x
cos^2xdx
?
答:
= xcosx -
∫cos^2xdx
= xcosx - ∫(1 + cos2x)/2 dx = xcosx - (x/2) - ∫cos2x/2 dx = xcosx - (x/2) - (1/4)∫(cos4x + cos2x)/2 dx = xcosx - (x/2) - (1/8)∫cos4x dx - (1/8)∫cos2x dx = xcosx - (x/2) - (1/32)sin4x - (1/16)...
sin^2x
cos^
2x不定积分
答:
可以降幂,答案如图所示
用凑微分法计算
cos^2xdx
答:
C为任意常数
求fcot
^2xdx
的不定积分,
答:
∫cot^2x=
∫cos^
2x/sin^2x=∫1-sin^2x/sin^2x=∫1/sin^2x-1=∫csc^2x-1=-cotx-x+c
问高数求导
∫
sin^3x
cos^2xdx
答:
∫sin^3x
cos^2xdx
=-∫sin^2xcos^2xdcosx =-∫(1-cos^2x)*cos^2xdcosx =-∫(cos^2x-cos^4x)dcosx =(1/5)*cos...
∫
sinx
cos^2xdx
=
答:
∫sinx
cos^2xdx
=-
∫cos^2xd
cosx= 1/3cos^3x
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