xsin+4xdx
答:∫xcos4x dx =(1/4)∫xdsin4x =(1/4)xsin4x-(1/4)∫sin4xdx =(1/4)xsin4x+(1/16)cos4x ∫x(sinx)^4dx =(1/8)[(3/2)x^2-∫ 4xcos2xdx +∫xcos4x dx ]=(1/8)[(3/2)x^2-2xsin2x -cos2x +(1/4)xsin4x+(1/16)cos4x ] + C ...
答:xsin²x是奇函数,所以 它的积分为0 而 sin^4x是偶函数,所以 原式=2∫(0,π)sin^4xdx =4∫(0,π/2)sin^4xdx =4×3/4×1/2×π/2 =3π/4
答:这个不用具体求,这是奇函数在对称区间上得定积分,结果就是0,这是定积分基本性质之一
答:不会,解出来后给你上图
答:原式=∫cos²x(1-sin²x)dx =∫cos²xdx-∫cos²xsin²xdx =∫cos²xdx-1/4∫sin²2xdx =∫(1+cos2x)/2 dx-1/4∫(1-cos4x)/2 dx =(2x+sin2x)/4+(8x-sin4x)/32+C =(8sin2x-sin4x+24x)/32+C ...
答:公式:∫[0→π] xf(sinx) dx = (π/2)∫[0→π] f(sinx) dx ∫[0→π] x(sinx)⁶(cosx)⁴ dx 由公式:=(π/2)∫[0→π] (sinx)⁶(cosx)⁴ dx =(π/2)∫[0→π/2] (sinx)⁶(cosx)⁴ dx + (π/2)∫[π/2→π] (sinx)...
答:计算过程:∫cos^2x=∫(1+cos2x)/2dx =1/2∫(1+cos2x)dx =1/2(x+sin2x/2)+C =1/2x+1/4sin2x+C 原函数的定义 primitive function已知函数f(x)是一个定义在某区间的函数,如果存在可导函数F(x),使得在该区间内的任一点都有dF(x)=f(x)dx,则在该区间内就称函数F(x)为函数f(...
答:=a^π/4)3.这个题为了表达清楚,我干脆先求不定积分,最后把上下限一带就可以!原式=∫x*[(1-cos2x)/2]*dx =∫xdx/2 - (1/2)*∫xcos2xdx =x^/4 - (1/2)*∫(x/2)*d(sin2x)=x^/4 - (1/4)*∫xd(sin2x)=x^/4 - (1/4)*xsin2x + (1/4)*∫sin2xdx =x^/...
答:∫xsin4xdx =(-1/4)∫xdcos4x =(-1/4)xcos4x+(1/4)∫cos4xdx =(-1/4)xcos4x+(1/16)sin4x+C
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xsin4xdx求不定积分 -
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:[答案] ∫xsin4xdx =(-1/4)∫xdcos4x =(-1/4)xcos4x+(1/4)∫cos4xdx =(-1/4)xcos4x+(1/16)sin4x+C
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xsin4xdx求不定积分 -
737孙欧
: ∫xsin4xdx =(-1/4)∫xdcos4x =(-1/4)xcos4x+(1/4)∫cos4xdx =(-1/4)xcos4x+(1/16)sin4x+C
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不定积分x²cos³xsinxdx怎么算? -
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: ^不定积分结果=-(1/8)x^2cos2x+(1/8)xsin2x+(1/16)cos2x-(1/32)x^2cos4x+(1/64)xsin4x+(1/256)cos4x+C过程如下:x^2(cosx)^2cosxsinxdx=(1/4)x^2(1-cos2x)2cosxsinxdx=(1/4)x^2(sin2x-cos2xsin2x)dx=(1/4)x^2sin2xdx-(1/8)x^2sin4xdx(1/4)x^2...
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∫xsin∧4(x)dx= ?急求 谢谢啦! -
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: ∫x(sinx)^4dx=(1/4)∫x(1-cos2x)^2dx=(1/4)∫[x-2xcos2x +x(cos2x)^2 ]dx=(1/8)∫[2x-4xcos2x +x(1+cos4x) ]dx=(1/8)∫[3x-4xcos2x +xcos4x ]dx=(1/8)[(3/2)x^2-∫ 4xcos2xdx +∫xcos4x dx ] consider ∫4xcos2xdx=2∫xdsin2x=2xsin2x - 2∫sin2x dx=2xsin2x +cos2x ...
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∫cos5xsin4xdx. -
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:[答案] ∫ cos5x sin4xdx=∫ (1-sin2x)2 sin4xdsinx =∫ 1-2u2+u4 u4du=∫(u-4-2u-2+1)du =- 1 3u-3+2u-1+u+C =- 1 3sin3x+ 2 sinx+sinx+C.
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∫cos^3xsin^4xdx 用第一类换元法,谢谢给个解题过程 -
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: ∫ (cosx)³(sinx)⁴dx=∫ (cosx)²(sinx)⁴dsinx=∫ (1-sin²x)(sinx)⁴dsinx=∫ [(sinx)⁴-(sinx)⁶] dsinx=(1/5)(sinx)^5 - (1/7)(sinx)^7 + C 【数学之美】团队为您解答,若有不懂请追问,如果解决问题请点下面的“选为满意答案”.
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xe^xsinx的不定积分 -
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: ∫xe^xsinxdx=[-xe^xcosx+(cosx*e^x+sinx*e^x)/2+xe^xsinx-(sinx*e^x-cosx*e^x)/2]/2+C 解题过程如下: ∫xe^xsinxdx =-∫xe^xdcosx =-xe^xcosx+∫cosxdxe^x =-xe^xcosx+∫cosx(e^x+x*e^x)dx =-xe^xcosx+∫cosx*e^xdx+∫cosx*x*e^xdx ∫cosx*e^xdx=∫cosxde^...
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∫x^2sin(4x)dx 怎么算啊 -
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: ∫x^2sin(4x)dx =1/4∫x^2sin(4x)d4x =-1/4∫x^2dcos(4x)=-1/4*x^2cos4x+1/4∫cos4xdx^2=-1/4*x^2cos4x+1/2∫xcos4xdx=-1/4*x^2cos4x+1/8∫xcos4xd4x=-1/4*x^2cos4x+1/8∫xdsin4x=-1/4*x^2cos4x+1/8*xsin4x-1/8∫sin4xdx=-1/4*x^2cos4x+1/8*xsin4x-1/32∫sin4xd4x=-(1/4)x^2cos4x+(1/8)xsin4x+(1/32)cos4x+C
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∫sin2x/sin^4x+cos^4xdx??//求详细步骤..... -
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: ∫sin2x/sin^4x+cos^4xdx=∫sin2x/[(sin²x+cos²x)²-2sin²xcos²x]dx=∫sin2x/[1-1/2sin²2x]dx=2∫sin2x/[2-sin²2x]dx=2∫sin2x/(1+cos²2x)dx=-∫1/(1+cos²2x)dcos2x=-arctancos2x+c
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xsinx积分怎么算 -
737孙欧
: xsinx积分是-xcosx+sinx+C. 分部积分法:∫udv=uv-∫vdu ∫ xsinx dx = - ∫ x d(cosx) =-xcosx+∫ cosx dx =-xcosx+sinx+C 所以xsinx积分是-xcosx+sinx+C. 扩展资料: 1、不定积分的公式 (1)∫ a dx = ax + C,a和C都是常数 (2)∫ x^a dx = [x^(a + 1)]/...