大一高数不定积分求解 求解大一高数不定积分问题
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∫(cosx)^2/(sinx)^3*dx
=∫cosx/(sinx)^3*d(sinx)
=-2∫cosxd[1/(sinx)^2]
=-2cosx/(sinx)^2+2∫1/(sinx)^2*d(cosx)
=-2cosx/(sinx)^2+2∫1/[1-(cosx)^2]*d(cosx)
=-2cosx/(sinx)^2+2*(1/2)ln|(1+cosx)/(1-cosx)|+C
=-2cosx/(sinx)^2+ln[(1+cosx)/(1-cosx)]+C
绛旓細瑙ｏ細鈭1/[x(x^10+1)]dx =鈭(x^9)/[x^10路(x^10+1)]dx 銆愪护t=x^10锛屽垯dt=10x^9 dx銆=(1/10)路鈭1/[t(t+1)] dt =(1/10)路鈭玔1/t-1/(t+1)]dt =(1/10)路[lnt-ln(t+1)]+C =(1/10)路ln[t/(t+1)]+C =(1/10)路ln[(x^10)/(x^10+1)]+C 鎴...
绛旓細=鈭玠y/鈭歔(Cy²+1)/y²]=鈭珁*dy/鈭(Cy²+1)=1/(2C) * 鈭2C*dy/鈭(Cy²+1)=1/(2C) * 鈭玠(Cy²+1)/鈭(Cy²+1)=鈭(Cy²+1) + C'
绛旓細(44)x^3+1 = x(x^2+1) - x +1 鈭(x^3+1)/(x^2+1)^2 dx =鈭 x/(x^2+1) dx - 鈭(x-1)/(x^2+1)^2 dx =鈭 x/(x^2+1) dx - 鈭玿/(x^2+1)^2 dx + 鈭玠x/(x^2+1)^2 =(1/2)ln|x^2+1| + 1/[2(x^2+1)] + 鈭玠x/(x^2+1)^2 let x...
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绛旓細鈭玔xe^(arctanx)/(1+x^2)^(3/2)]dx=鈭玔xe^(arctanx)/(1+x^2)^(3/2)]d(arctanx)=鈭玔x/(1+x^2)^(1/2)]d[e^(arctanx)]=[x/(1+x^2)^(1/2)]e^(arctanx)-鈭玔e^(arctanx)/(1+x^2)^(1/2)]-[e^(arctanx)*x^2/(1+x^2)^(3/2)]dx =[x/(1+...
绛旓細浠=sinu,dt=cosudu 鈭垰(1-t^2)dt =鈭(cosu)^2du =1/2鈭(1+cos2u)du =1/2u+1/4sin2u+C =1/2arcsint+1/2t鈭(1-t^2)+C
绛旓細浠も垰x=t,鍒檟=t2,dx=2tdt 鍘熷紡鍙寲涓猴細鈭玸in2(2t)/t*2tdt =鈭2sin2(2t)dt =鈭(1-cos4t)dt =t-1/4sin4t+C 灏嗏垰x=t浠ｅ叆锛屽緱锛氣垰x-1/4sin4鈭歺+C
绛旓細鏈涢噰绾
绛旓細sin3x * cos4x =1/2 * [sin(3x +4x) - sin(3x -4x)]=1/2 * ( sin7x + sinx)鎵浠ワ紝涓婇潰鐨绉垎灏卞彲浠ュ彉鎹负锛=1/2 *鈭(sin7x + sinx)*dx =1/2 *鈭玸in7x *dx + 1/2 * 鈭玸inx *dx =1/14 * 鈭玸in7x * d(7x) + 1/2 * 鈭玸inx *dx =1/14 * [-cos7x] + ...
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