求不定积分详细过程 第一换元法求不定积分

老朋友,,,,帮你搞定1 令tant=x (-π/2<x<π/2) 原式=∫ cost dt/ (tant)^4 =∫ (sec t)^2 d t/ (tant)^4 (sec t)^3 =∫ (cost)^4 d sint / (sint)^4 = ∫ (1-(sint)^2)^2 d sint / (sint)^4 = ∫ (1-2(sint)^2+(sint)^4) d sint / (sint)^4 然后将他拆开 后面的很简单的 自己做吧 前面的也不知道搞错没 方法是没错的 有些运算自己看有没错

The L.C.M. of 1/3 and 1/2 is 1/6So let x = t^6 then dx = (6t^5)dt and we have∫ x^(1/3) / [x(√x + x^(1/3))] dx= ∫ t² / [t^6(t³+t²)] * (6t^5)dt= 6∫ t/(t³+t²) dt= 6∫ [(1+t)-t] / [t(1+t)] .

the l.c.m. of 1/3 and 1/2 is 1/6so let x = t^6 then dx = (6t^5)dt and we have∫ x^(1/3) / [x(√x + x^(1/3))] dx= ∫ t² / [t^6(t³+t²)] * (6t^5)dt= 6∫ t/(t³+t²) dt= 6∫ [(1+t)-t] / [t(1+t)] dt= .

解:积分域:d={(x,y)|x^2+y^2≤x+y+1}={(x,y)|x^2+y^2-x-y-1≤0}={(x,y)|(x-1/2)^2+(y-1/2)^2≤3/2} 积分域为圆心为(1/2,1/2)半径为√6/2的圆的内部(包括边界) 转换成极坐标:.

x=t³,dx=3t²dt∫dx/(1+x^1/3)=∫3t²dt/(1+t)=3∫t(t+1-1)/(1+t)dt=3∫dt-3∫(t+1-1)/(1+t)dt=3(t²/2)-3∫dt+3∫dt/(1+t)=(3/2)t²-3t+3ln|1+t|+C=(3/2)x^(2/3)-3x^(1/3)+3ln|1+x^(1/3)|+C

积分化为(省略积分号)=-(cosxdcosx)/(1+cos^2x),如果你学积分时间不长,可做个代换:u=cosx更清楚=-udu/(1+u^2)=-(1/2)d(1+u^2)/(1+u^2)=-(1/2)ln(1+u^2)+C,再将u换回cosx,在积分熟练后,这个代换就不必做,而直接计算

原式=∫[(x³+2x²)-(2x²+4x)+(7x+14)-12]/(x+2)dx =∫[x²(x+2)-2x(x+2)+7(x+2)-12]/(x+2)dx =∫x²-2x+7-12/(x+2) dx =1/3*x³-x²+7x-12ln|x+2|+C望采纳

cos²xsin⁴x=¼(2sinxcosx)²·sin²x=¼sin²(2x)sin²x=¼[sin(2x)sinx]²=¼[(-½)(cos(3x)-cosx)]²=(1/16)[cos²(3x)+cos²x-2cosxcos(3x)]=(1/32)[1+cos(6x)+1+cos(2x)-2.

令x=t^6,则dx=6t^5原式=∫6t^5/(t^3+t^2)dt=6∫t^3/(t+1)dt=6∫(t^3+1-1)/(t+1)dt=6∫(t^3+1)/(t+1)dt-6∫1/(t+1)dt=6∫(t^2+t+1)dt-6ln(t+1)=2t^3+3t^2+6t-6ln(t+1)+C代入t=x^(1/6),原式=2*x^(1/2)+3*x^(1/3)+6*t^(1/6)-6ln[x^(1/6)+1]+C

(sinx)^4 = (sinx^2)^2 = ((1 - cos2x)/2)^2 = (1 - 2cos2x + (cos2x)^2)/4= 0.25 - 0.5cos2x + 0.125(1 + cos4x)= (cos4x)/8 - (cos2x)/2 + 3/8 ∫ (sinx)^4dx = ∫ ((cos4x)/8 - (cos2x)/2 + 3/8)dx= ∫ ((cos4x)/8)dx - ∫ ((cos2x)/2)dx + ∫ (3/8)dx= (1/32)∫ cos4xd4x - (1/4)∫ cos2xd2x + (3x/8)= (sin4x)/32 - (sin2x)/4 + (3x/8) + c