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∫sin^3x Cosx/³√sinx%Cosx Dx

可以停止了,因为原函数不初等

∫x²sin³xdx=1/4∫x²(3sinx-sin3x)dx =3/4∫x²sinxdx-1/4∫x²sin3xdx =-3/4∫x²dcosx+1/12∫x²dcos3x =-3/4x²cosx+3/4∫cosxdx²+1/12x²cos3x-1/12∫cos3xdx² =-3/4x²cosx+3/2∫xcosxdx+1/12...

解:∵(cosx)^4是偶函数,(sinx)^3是奇函数 ∴∫(cosx)^4dx=2∫(cosx)^4dx ∫(sinx)^3dx=0 故 ∫((cosx)^4+(sinx)^3)dx =∫(cosx)^4dx+∫(sinx)^3dx =2∫(cosx)^4 =(1/2)∫[3/2+2cos(2x)+cos(4x)/2]dx (应用倍角公式) =(1/2)[3x/2+sin(2x)+sin(4x)/8]│ =(1...

猜[2ysin(x/y)+ 3xcos(x/y)]dy-3ycos(x/y)dx=0, 两边都除以y,得[2sin(x/y)+(3x/y)cos(x/y)]dy-3cos(x/y)dx=0, 设x=uy,则dx=ydu+udy,上式变为 [2sinu+3ucosu]dy-2cosu(ydu+udy)=0, 整理得(2sinu+ucosu)dy=2ycosudu, 分离变量得dy/y=2cosudu/(2s...

sinx-cosx=½ sin²x-2sinxcosx+cos²x=¼ ∴sinxcosx=⅜ sin³x-cos³x=(sinx-cosx)(sin²x+sinxcosx+cos²x) =½(1+⅜) =11/16

原式=sinx-(1/3)sin3x ;区间为-π/2 到π/2 =sin(π/2)-(1/3)sin3π/2-[sin(-π/2)-(1/3)sin(-3π/2)] =1+1/3-(-1-1/3) =2+2/3 =2又2/3

求不定积分∫dx/(sin³xcosx) 解:原式=∫(sin²x+cos²)dx/(sin³xcosx)=∫dx/(sinxcosx)+∫cosxdx/sin³x =∫d(2x)/sin(2x)+∫d(sinx)/sin³x=ln∣tanx∣-1/(2sin²x)+C

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