大一高数不定积分例题
答:是错的,没有分类讨论 √1-sinx的积分如图所示
答:原式=1/2∫1/(1+sin^4x) dsin²x =1/2 arctan(sin²x)+c
答:方法如下,请作参考:
答:8.(1)∫x^3dx/(x+3)=∫(x^3+27-27)dx/(x+3)=∫(x^2-3x+9-27/(x+3))dx =x^3/3-3/2x+9x-27ln|x+3|+C (2)∫dx/sin^2xcosx 令t=sinx,x=arcsint,dx=dt/√(1-t^2)原式=∫1/[t^2√(1-t^2)]*dt/√(1-t^2)=∫dt/[t^2(1-t^2)]=∫[1/t^2+1...
答:等式两边同时乘以1/5,有dX=1/5d(5X)(3)d(X^2+1)=2XdX,等式两边同时乘以1/2,有XdX=1/2d(X^2+1)(5)d(√X-2)=1/2(1/√X)dX,等式两边同时乘以2,有dX/√X=2d(√X-2)(7)d(arctan2X)=2/(1+4X^2)dX,等式两边同时乘以1/2,有dX/(1+4X^2)=1/2d(arctan2X)...
答:设√(1-e^x)=t x=ln(t^2+1)原式=∫2t^2/(t^2+1)dt =∫2-2/(t^2+1)dt =2t-2arctant+C =2√(1-e^x)-2arctan√(1-e^x)+C
答:∫cos(√x)dx 令√x=u,则dx/2√x=du,dx=2(√x)du=2udu,原式=2∫ucosudu =2∫ud(sinu)=2[usinu-∫sinudu]=2(usinu+cosu)+C =2[(√x)sin(√x)+cos(√x)]+C ~~~∫√x(x+1)^2dx 令√x=t, 则dx=2tdt,带入 =∫t(t^2+1)^2*2tdt =∫2t^6+4t^4+2t^2dt =...
答:(44)x^3+1 = x(x^2+1) - x +1 ∫(x^3+1)/(x^2+1)^2 dx =∫ x/(x^2+1) dx - ∫(x-1)/(x^2+1)^2 dx =∫ x/(x^2+1) dx - ∫x/(x^2+1)^2 dx + ∫dx/(x^2+1)^2 =(1/2)ln|x^2+1| + 1/[2(x^2+1)] + ∫dx/(x^2+1)^2 let x...
答:∫e^√(x+1) dx let y=√(x+1)dy = dx/[2√(x+1)]dx= 2ydy ∫e^√(x+1) dx =2∫ye^y dy =2∫yde^y =2ye^y -2∫e^y dy =2ye^y -2e^y +C =2√(x+1) .e^√(x+1) -2e^√(x+1) + C
答:如图所示
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鲜罡18824489979:
大学数学不定积分的题目1、∫1/√(x∧2+4)∧3 dx2、∫x^2/√(4 - x∧2)dx3、∫In√x dx4、∫x^2arctanx dx5、∫e^ - 2x cosx dx -
2406翟促
:[答案] 先化简 然后分布积分.懒得算
鲜罡18824489979:
一道高数积分题目,求lnsinx对x的不定积分 -
2406翟促
:[答案] ∫lnsinx dx=xlnsinx-∫x d(lnsinx)=xlnsinx-∫x*1/sinx*cosx dx=xlnsinx-∫xcotx dx基本上∫xcotx dx是无法用初等函数解决的,可利用复数形式解但∫xcotx dx=xln[1-e^(2ix)]-1/2*i{x²+Li_2 [e^(2ix)]}∴原式=xl...
鲜罡18824489979:
大一高数~求不定积分∫sin3xsin5xdx -
2406翟促
:[答案] 须知公式:sinAsinB = (1/2)[cos(A-B)-cos(A+B]∫sin(3x)sin(5x) dx= (1/2)∫[cos(-2x)-cos(8x)] dx= (1/2)∫cos2x dx - (1/2)∫cos8x dx= (1/4)∫cos2x d(2x) - (1/16)∫cos8x d(8x)= (1/4)sin2x - (1/16)sin8x + C
鲜罡18824489979:
请教一高数不定积分题1/(1+根号x - 1)dx 的不定积分 -
2406翟促
:[答案] 设 t = √(x-1) 则 x = t²+1,dx = 2tdt∫1/[1+√(x-1)]dx =∫1/[1+ t] 2tdt= 2∫(t+1-1)/(1+ t)dt= 2t -2∫1/(1+ t)dt= 2t - 2ln(1+t) + C= 2√(x-1) - 2ln(1+√(x-1))+ C
鲜罡18824489979:
大一高数不定积分∫1/(sinx+cosx) dx 求详解 -
2406翟促
:[答案] 因为sinx+cosx=2sin(x/2)cos(x/2)+cos²(x/2)-sin²(x/2) 设t=tan(x/2) 则dt=1/2sec²xdx 代入原式可得 ∫1/(sinx+cosx)dx =∫2/(1+2t-t²)dt =∫2/[2-(t-1)²]d(t-1) =√2/2∫[1/[√2+(t-1)]+1/[√2-(t-1)]]d(t-1) =√2/2ln|(t-1+√2)/(t-1-√2)|+C
鲜罡18824489979:
高数不定积分题一枚,∫ (arcsin√x)/(√x(1 - x))dx 注:分母中x(1 - x)均在根号内 -
2406翟促
:[答案] 令x^0.5=t 则积分对象变为:arcsint/(t*(1-t^2)^0.5)*d(t^2)=2arcsint/(1-t^2)^0.5*dt 令p=arcsint,则t=sinp,积分对象变为: 2p/cosp*cosp*dp=2p*dp=d(p^2) 所以积分结果为p^2+C=(arcsin(x^0.5))^2+C
鲜罡18824489979:
求高数不定积分的题目,∫(a^x)(e^x)dx 和 ∫(secx)^4 (tanx)^2 dx,说得详细点,如果说得好再加10分 -
2406翟促
:[答案] ∫(a^x)(e^x)dx =∫(ae)^xdx ,用不定积分公式∫(A^x)dx ,其中A=ae即可.∫(secx)^4 (tanx)^2 dx=∫(secx)^2 (tanx)^2(secx)^2dx,注意(secx)^2dx=dtanx=∫(secx)^2 (tanx)^2dtanx=∫[1+(tanx)^2] (tanx)^2dtanx=∫[(ta...
鲜罡18824489979:
高数不定积分问题:设f(x)的一个原函数arcsinx,则不定积分∫ xf'(x)dx= , -
2406翟促
:[答案] 由于f(x)的一个原函数arcsinx 所以∫ f(x)dx = arcsinx + C f(x)= (arcsinx)' = 1/根号(1-x²) ∫ xf'(x)dx = ∫ xd(f(x)) =xf(x) - ∫ f(x)dx =xf(x) + arcsinx + C =x/根号(1-x²) + arcsinx + C
鲜罡18824489979:
高数2求不定积分的题目例题是这样的:(1) {1/x^2+x - x*dx={1/(x - 1)(x+2)*dx但它是怎么变成 1/3{(1/x - 1 - 1/x+2)dx的而最后变成 1/3ln!x - 1/x+2!+C的(2) {dx/x(x^... -
2406翟促
:[答案] 1/(x-1)-1/(x+2)=[(x+2)-(x-1)]/(x-1)(x+2)=3/(x-1)(x+2)注意到分子是3所以为了使等式两边相等必须要乘1个(1/3)使等式两边相等因此=1/3{(1/(x-1)-1/(x+2))dx然后因为1/(x-1)的积分是ln(x-1),1/(x+2)的积分是ln(x+2...
鲜罡18824489979:
一个高数不定积分的题,∫{x^2e^x/(x+2)^2}dx=?这是高等数学的题,分母是(x+2)^2 -
2406翟促
:[答案] ∫(x^2e^x/(2+x)^2 dx ∫[x^2*e^x/(x+2)^2]dx 换元积分=-∫(x^2*e^x)d[1/(x+2)] 分部积分=-{x^2*e^x/(x+2)-∫[1/(x+2)]d(x^2*e^x)} =[-x^2*e^x/(x+2)]+∫[1/(x+2)]*(2x*e^x+x^2*e^x)dx =[-x^2*e^x/(x+2)]+∫[1/(x+2)...