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∫Cos²(x/2)Dx=?

∫cos²(x/2)dx =∫[(1+cosx)/2]dx =1/2∫dx+1/2∫cosxdx =(x+sinx)/2+C

cos²x=(1+cos2x)/2 所以∫cos²xdx=∫1/2dx+1/2*∫cos2xdx =x/2+1/4*∫cos2xd(2x) =x/2+1/4*sin2x =(2x+sin2x)/4 定积分就不加常数C了,你把积分的上下限代入即可

cos2x=2cos^2x-1 所以 1+cos2x=2cos²x

cosx = 1 - 2sin²(x/2) 2cosx = 2 - 4sin²(x/2) ∫ dx/(1 + 2cosx) = ∫ dx/[3 - 4sin²(x/2)] = ∫ dx/[3sin²(x/2) + 3cos²(x/2) - 4sin²(x/2)] = ∫ dx/[3cos²(x/2) - sin²(x/2)] = ∫ sec²(x/2)/[3 -...

(d/dx)∫[cosx, x²](x-t)costdt = (d/dx){x*∫[cosx, x²]costdt-∫[cosx, x²]tcostdt} = ∫[cosx, x²]costdt+x*{coscosx*(-sinx)-cosx²*2x} -[cosx*coscosx*(-sinx)-x²*cosx²*2x] = ……。

半角公式:sin²(x/2)=(1-cosx)/2

=cos(πcos²x)×(cosx)'-cos(πsin²x)×(sinx)' =cos(πcos²x)×(-sinx)-cos(πsin²x)×cosx, 继续整理的话, =cos(π-πsin²x)×(-sinx)-cos(πsin²x)×cosx =cos(πsin²x)×sinx-cos(πsin²x)×cosx =cos(πcos&#...

由1+cosx=2cos²(x/2)得 ∫(1/1+cosx)² dx =∫(1/2cos²(x/2))² dx =1/4∫sec⁴(x/2) dx =1/2∫sec⁴(x/2) d(x/2) =1/2∫sec²(x/2) dtan(x/2) =1/2∫[tan²(x/2)+1] dtan(x/2) =(1/6)tan³(x/2)+(1/2)tan(x/...

令x=2sina dx=2cosada 原式=∫2cosa*2cosada =2∫(1+cos2a)/2 d(2a) =2a+sin2a+C =2a+2sinacosa+C =2arcsin(x/2)+x*√(4-x²)/2+C

cosx+3/xdx=1/12x24x254cosx+C ∫x212x22xsinx+3/12x2dx3cos3x-1/2xsinx-3/12x2dcos3x =-3/dx=1/4x212x23cosx34∫x218xsin3x+1/4x2cosx+3/4∫cosxdx24x2∫x2cos3x-1/4∫x23∫sinx32∫xcosxdx+1/sinx3sin3=-1/4x2 =-3/4∫x218xsin3x-1/18∫sin3xdx =-3/sin...

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