考研真题 不定积分与定积分.docx
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考研真题不定积分与定积分
arcsinxdx
0
(2020数二)
⎰af
(x)dx=
2⎰a[f
(x)+f
(a+b-x)]dx
⎰arcsinx
dx=1⎰
⎛arcsinx
+
arcsin1-x
⎫dx
1
020⎝
x(1-x)(1-x)x⎭
α=arcsin
1-x
1
=2⎰0
π2dx
β=arcsinx
=π⎰1
1d(2x-1)
40
1
=πarcsin(2x-1)
40
=π2
4
x(1-x)=
1-⎛çx-1⎫⎪
2
4⎝2⎭
⎰
arcsinxdx
0
(2020数二)简化
拓展⎰
arcsinxdx
arcsinx
1
0
dx=⎰0
arcsint⋅2tdt
令x=
t⇒x
=t2
=⎰0
2arcsintdt
x:
0→1
t:
0→1
1
=⎰0
2arcsintdarcsint
巧妙不适用
=arcsin2t1
=π2
4
拓展⎰
arcsinxdx
⎰af
(x)dx=
2⎰a[f
(x)+f
(a+b-x)]dx
⎰arcsinxdx=1⎰
⎛arcsinx+
arcsin1-x
⎫dx
20⎝
x(1-x)+1
(1-x)x+1⎭
=1⎰π2
5⎛1⎫
dx20x(1-x)+1
x(1-x)+1=
-çx-⎪
4⎝2⎭
=π⎰1
1d⎛ç2
x-1⎫⎪
40⎝55⎭
π⎛21⎫1π1
=4arcsinçx-⎪=arcsin
⎝55⎭025
⎰3x+6dx
(2019数二)
为什么能这样待定?
⎧P=A
⎧M=2C
(x-1)2
(x2
+x+1)
为什么要这样待定?
⎨Q=B-A⎨
=C+D
3x+6
=A1+B1
+C2x+1+D1
(x-1)2(x2+x+1)
x-1(x-1)2
x2+x+1x2+x+1
3x+6
=Px+Q+
Mx+N
=A(x-1)+B+
C(2x+1)+D
(x-1)2
(x2
+x+1)(x-1)2
x2+x+1
(x-1)2
x2+x+1
分式分解定理
省去一些中间过程
设P(x)
Q(x)
为有理真分式,其中Q(x)=Q1
(x)Q2
(x)且Q1
(x),Q2
(x)互素
则存在唯一一组多项式P1
(x),P2
(x)使得
P(x)
Q(x)
=P1Q1
(x)+
(x)
P2(x)
Q2(x)
其中P1
Q1
(x)P
,
(x)Q2
(x)
(x)
为真分式
⎰3x+6dx
(2019数二)
赋值法
3x+6
=A1
+B1
+C2x+1
+D1
对应系数
(x-1)2(x2+x+1)
x-1
(x-1)2
x2+x+1
x2+x+1
3x+6=(A(x-1)+B)(x2+x+1)+(C(2x+1)+D)(x-1)2
x=1
⇒9=3B
⎧A=-2
x=0x=-1
⇒6=B-A+C+D
⇒3=B-2A-4C+4D0=A+2C
⎪B=3
⎪
⎨C=1
⎪⎩D=0
3x+6
=-21+31
+2x+1
(x-1)2(x2+x+1)
x-1(x-1)2
x2+x+1
⎰3x+6dx
=-2ln
x-1
-31+ln(x2
+x+1)+C
(x-1)2(x2+x+1)
x-1
⎛
lnç1
⎝
⎛
⎫dx(x
⎭
⎫
>0)
⎛
(2009数二)
⎫1
1⎛1⎫
⎰lnç1+
⎪dx
=xlnç1+
⎪-⎰
⋅⋅ç-
⎪⋅xdx
2
⎝⎭⎝
⎭1+
21+x⎝x⎭
x
⎰1⋅1⋅⎛ç-1
⎫⎪⋅xdx
=-1
⎰1dx令=t
1+2
⎝x2⎭
2xç1+
⎝
1+x⎫
⎪
x⎭
1+xx
有理化
=-1
⎰1⋅-2tdt
21
t2-1
(1+t)t
(t2
-1)2
=⎰1dt
(t+1)2(t-1)
⎛
lnç1
⎝
⎫dx(x
⎭
>0)
(2009数二)
t=>1
1
(t+1)2
dt=⎛
(t-1)⎝
1
2(t+1)2
-1+
4(t+1)
1
4(t-1)
⎫dt
⎭
=1
2(t+1)
-1ln(t+1)+
4
1ln(t-1)+C4
1=A1
+
B1
+
C1
赋值法
(t+1)2(t-1)(t+1)2t+1
t-1
对应系数
1=A(t-1)+B(t2-1)+C(t+1)2
t=1
⇒1=4C
C=14
t=-1
⇒1=-2A
A=-12
0=B+CB=-14
a
设⎰0
xe2xdx=
1,则a=
4
(2014数三)
⎰xe2xdx
=(mx+n)e2x+C
xe2x
=(2mx+2n+m)e2x
⎧1
⎧1=2m⎪2
⎨0=2n+m⎨1
⎩⎪n=-
⎩4
⎰xe2xdx
=⎛ç
⎝
1x-1
24
⎫⎪e2x+C
⎭
ç⎛1a-1
⎝24
⎫⎪e2a
⎭
-⎛ç-1⎫⎪
⎝4⎭
a=1
2
设函数f
⎧λe-λx
(x)=⎨
⎩0
x>0,λx≤0
>0,则⎰-∞
xf(x)dx=
(2011数二)
+∞
+∞
xf(x)dx
-∞
λxe-λxdx
0
⎰λxe-λxdx
=(mx+n)e-λx+C
λxe-λx
=(-λmx-λn+m)e-λx
⎧λ=-λm
⎧⎪m=-1
⎨0=-λn+m
⎨n=1
λ
⎰λxe-λxdx
=⎛ç-x-1
⎝λ
⎫⎪e-λx+C
⎭
λxe-λxdx=1
0λ
lim⎰
e-x
sinnxdx=
(2009数二)循环
1
n→∞0
⎰e-x
sinnxdx
=e-x
(asinnx+bcosnx)+C
e-x
sinnx
=e-x
(-asinnx-bcosnx+ancosnx-bnsinnx)
⎧1=-a-bn
⎨0=-b+an
⎧=-
⎨
1
1+n2
n
⎩⎪b=-
⎩
1+n2
⎰e-x
sinnxdx
=e-x
sinnx-ncosnx+C1+n2
1
⎰
e-x
0
sinnxdx
=e-1
sinn-ncosn1+n2
--n1+n2
⎰f(x)g(x)dx
=⎰f
(x)dG(x)
=f(x)G(x)-⎰f
'(x)G(x)dx
G(x)是g(x)的原函数
⎰f(x)g(x)dx
→⎰f
'(x)G(x)dx
消去f(x)
f(x)复杂
f'(x)简洁
简化被积函数
特别地,当f
(x)是多项式时,可以局部实现降次
⎰
+∞lnxdx
1(1+x)2
(2013数一)
⎰f(x)g(x)dx
→⎰f
'(x)G(x)dx
⎰
lnxdx
(1+x)2
lnx⋅1dx
(1+x)2
→1⋅-1dx=
x1+x
⎛ç1-1
⎝1+xx
⎫⎪dx
⎭
=ln
1+x
lnx
+
C=ln
1+x+Cx
+∞ln(1+x)dx
0(1+x)2
(2017数二)
⎰f(x)g(x)dx
→⎰f
'(x)G(x)dx
ln(1+x)dx
(1+x)2
⎰ln(1+x)⋅1dx
→⎰1⋅-1
dx=
1+C
(1+x)2
1+x1+x
1+x
2
⎰
1
1x3
1
exdx
(2006数一)
⎰f(x)g(x)dx
→⎰f
'(x)G(x)dx
令1=
x
t⇒x=1
t
x:
1→2
t:
1→1
2
2
1
⎰1exdx
1
=⎰2
t3et
⎛ç-1
⎫⎪dt
1
=-⎰2
tetdt
1x3
1⎝t2⎭1
⎰t⋅etdt
→⎰1⋅etdt
=et+C
etdt=det
⎰tetdt
=⎰tdet
=tet
⎰1⋅etdt
⎰
arcsinx+lnxdx
x
(2011数三)
⎰f(x)g(x)dx
→⎰f
'(x)G(x)dx
⎰(arcsin
x+lnx)⋅1
dx→⎰⎛ç1⋅1+1
⎫⎪⋅2
xdx
x⎝1-x2xx⎭
⎰⎛ç1+2
⎫⎪dx
=-2
1-x+4
x+C
1dxx
=d(2x)
⎝1-xx⎭
⎰e2x
arctan
ex-1dx
(2018数一)
⎰f(x)g(x)dx
→⎰f
'(x)G(x)dx
→⎰e2x
⋅1⋅edx
x
=1⎰
21+(ex
e2x
dx
-1)22
令
ex-1
ex-1=
t⇒ex
=t2
+1⇒x
=ln(t2
+1)
4ex-1
1(t2+1)22t
=4⎰t
dt
t2+1
=1(t2
2
+1)dt
=1t3
6
+1t+C=1(26
ex-1)3+1
2
ex-1+C
x2arcsinx
dx
0
(2008数二)
⎰f(x)g(x)dx
→⎰f
'(x)G(x)dx
消去根号
x=sint
x=cost
令x=sint
t∈(0
π
,)⇒t2
=arcsinx
x:
0→1
t:
0→π
2
1x2arcsinx
πsin2t⋅tπ
⎰dx
=⎰2⋅costdt
=⎰2
sin2
t⋅tdt
00cost
0
降次⎰sin2
t⋅tdt
→2t-sin2t
⎰
4
⋅1dt
⎰sin2
tdt=
1-cos2tdt=
2
2t-sin2t+C
4
=2t2
+
cos2t+C8
2
⎰0x
cos
xdx
(2010数一)
⎰f(x)g(x)dx
→⎰f
'(x)G(x)dx
令x=
t⇒x
=t2
x:
0→π2
t:
0→π
π2ππ2
⎰0xcosxdx=⎰0tcost⋅2tdt=2⎰0t
costdt
⎰t2
⋅costdt
→⎰2t⋅sintdt
→⎰2⋅(-cost)dt
=-2sint+C
降次
⎰t3
⎰t4
⋅costdt
⋅costdt
⎰tn
⋅costdt
设an
=⎰0x
dx(n
=0,1,2,⋅⋅⋅)
(1)证明:
数列{a
}单调减少,且a
=n-1a
(n=0,1,2,⋅⋅⋅)
(2)求liman
an-a
n-1
=⎰0
(xn
n
-xn-1)
n
dx≤0
n+2
n-2
n→∞a
1
n-1
设an
=⎰0x
dx(n
=0,1,2,⋅⋅⋅)
消去根号
分部积分公式
方法一
(1)证明:
数列{a
}单调减少,且a
=n-1a
(n=0,1,2,⋅⋅⋅)
(2)求liman
nnn+2
n-2
n→∞a
n-1
令x=sinθ
θ∈[0π]
2
x:
0→1
θ:
0→π
2
sinn
θdθ=dF(θ)
1ππ
a=x
0
⎰
π
dx=
2sinnθ
0
1
1-sin2
π
θ⋅cosθdθ=
2sinn0
θcos2
θdθ
=2sinn0
1⎛
θcosθ⋅cosθdθ
π
=n+1
π
2cosθdsinn+1θ
0
π
⎫1+
=çcosθsinn+1θ2-
n+1⎝0
⎰
π
2sinn+10
θ(-sinθ)dθ⎪
⎭
π
=n+1
2sinn0
2θdθ
(n+1)an=
2sinn+20
θdθ
⎰
π
(n-1)an-2=
2sinn0
θdθ
π
(n-1)a
n-2
-(n+1)an=
2(sinn0
θ-sinn+2
θ)dθ=
2sinn0
θcos2
θdθ=an
设an
=⎰0x
dx(n
=0,1,2,⋅⋅⋅)
方法二
(1)证明:
数列{a
}单调减少,且a
=n-1a
(n=0,1,2,⋅⋅⋅)
(2)求liman
nnn+2
ππ
n-2
π
n→∞a
n-1
an=
2sinn0
θcos2
θdθ=
2sinn0
θcosθ⋅cosθdθ=
2sinn0
θcosθdsinθ
=sinn+1
π
θcosθ2-
π
2sinθ(nsinn-10
θcos2
θ-sinn+1
θ)dθ
π
=-n2sinn0
θcos2
π
θdθ+2sinn+20
θdθ
=-nan
π
+2sinn+20
θdθ
π
(n+1)an=
π
2sinn+2
0
π
θdθ
(n-1)a
n-2
-(n+1)an=
2(sinn
0
θ-sinn+2
θ)dθ=
2sinn
0
θcos2
θdθ=an
设an
=⎰0x
dx(n
=0,1,2,⋅⋅⋅)
方法三
(1)证明:
数列{a
}单调减少,且a
=n-1a
(n=0,1,2,⋅⋅⋅)
(2)求liman
nnn+2
πππ
n-2
n→∞a
n-1
an=
2sinn0
θcos2
θdθ=
2sinn0
θdθ-2
0
sinn+2
θdθ
华里士公式
ππ
=In
In+2
In=
2sinnxdx=
0
2cosnxdx
0
=In
n+1In+2n
In=
n-1In
n-2
a=1I
a=1I
⎧(n-1)!
⋅π
n是正偶数
nn+2n
n-2nn-2
I=⎪n!
2
n⎨(n-1)!
anI
nn-1
n-1
⎪n是正奇数
n=
an-2
n+2
⋅n=
In-2
n+2⋅n
=n+2
⎩n!
设an
=⎰0x
dx(n
=0,1,2,⋅⋅⋅)
方法四
(1)证明:
数列{a
}单调减少,且a
=n-1a
(n=0,1,2,⋅⋅⋅)
(2)求liman
nnn+2
n-2
n→∞a
n-1
an=⎰x
dx=⎰
xn+1
d
xn+11
=1-x2
-⎰xn+1
-
2x
dx
00n+1
n+1
00n+12
1-x2
=1⎰1
xn+2
dx
形式上接近
n+1
=1
0
1xn+2
()xn+2
n+1⎰0
dx
1-x2
n+1
an=⎰0
dx
1-x2
1(xn
-xn+2)=n
(n-1)an-2
-(n+1)an
=⎰0
1-x2
dx⎰0x
dx=an
设an
=⎰0x
dx(n
=0,1,2,⋅⋅⋅)
方法五
(1)证明:
数列{a
}单调减少,且a
=n-1a
(n=0,1,2,⋅⋅⋅)
(2)求liman
n
1n1
n-1
nn+2
n-2
11
n-1
n→∞a
3
2
n-1
an=
⎰0x
1⎛
dx=⎰0x⋅x
311
dx=-3⎰0x
3
d(1-x)2
⎫
=-çxn-1
3⎝
(1-x2)2
-⎰0
0
(1-x2)2
⋅(n-1)xn-2dx⎪
3
⎭
=n-1⎰1
xn-2
(1-x2)2dx
=n-1⎰1
xn-2
(1-x2)
dx形式上接近
30
=n-1⎛ç1
xn-2
30
dx-xn
dx⎫⎪=
n-1(a
-a)
3⎝⎰0
⎰0⎭3
n-2n
dx=d1⎛çx
2⎝
+
arcsinx⎫⎪
⎭
设an
=⎰0x
dx(n
=0,1,2,⋅⋅⋅)
方法五
(1)证明:
数列{a
}单调减少,且a
=n-1a
(n=0,1,2,⋅⋅⋅)
(2)求liman
an-a
n-1
=⎰0
n
1
(xn
-xn-1)
n
dx≤0
n+2
n-2
n→∞a
n-1
an⋅
an-1=
an=
n-1
liman=a分析
an-1
an-2
an-2
n+2
n→∞an-1
a2=1⇒a=1
n=
n+3
n-1=
n+2
an+1≤
an-1
an≤
an-2
an≤1an-1
0≤e-x
sinx
≤e-x
e-xdx收敛⇒
0
e-x
0
sinx
dx收敛
比较审敛原理
e-x
sinx
dx=
lim⎰
e-x
sinx
dx=
nπ
lim
e-x
sinxdx
归结原理
nπ-x
n-1
=
(k+1)π
e-x
sinx
dx=
n-1
(-1)k
(k+1)π
e-x
sinxdx
n-1