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I n = π
2
0 sinn xdx
I 0 I 1
(I n)
I n I n−2
(n + 1) I nI n+1
limn→+∞
I n limn→+∞
I n
I n+1lim
n→+∞√
nI n
I 2n I 2n+1 π
I 0 = π
2
0 dx = π2 , I 1 =
π2
0 sin xdx = [− cos x]π
2
0 = 1 I 2 = π
2
0 sin2 xdx
u = π2 − x I 2 =
π
2−π2
π
2−0 sin
2 π2 − u
(−du) = π20 cos2 udu
π2
0 cos2 udu +
π2
0 sin2 xdx =
π2
0
cos2 x + sin2 x
dx =
π2
0 dx = π2 = 2I 2
I 2
I 0 = π
2
I 1 = 1 I 2 = π
4
(I n)
x ∈ 0,π2 0 sin x 1 sinn x
x ∈ 0,π2 0 sinn+1 x sinn x
0,π2
, 0
π2
0 sinn+1 xdx
π2
0 sinn xdx
0 I n+1 I n
(I n)
I n = π
2
0
sinn xdx = π
2
0
sinn−1 x sin xdx
I n = π
2
0 sin x sinn−1 xdx =
− cos x sinn−1 xπ20
=0
n1
+
π2
0cos x (n− 1)cos x sinn−2 xdx
n2
I n = (n− 1) π
2
0
1− sin2 x sinn−2 xdx = (n− 1) π20
sinn−2 xdx− π20 sinn xdx = (n− 1)
(I n−2 − I n)
∀n 2 I n = n−1n
I n−2
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n 2
I n =
n−1n
I n−2I n+1 =
nn+1I n−1
I nI n+1 = n−1n
nn+1I n−2I n−1 ⇔ (n + 1)
I nI n+1 = (n− 1) I n−2I n−1
(2n + 1) I 2nI 2n+1 = (2n− 1)
I 2n−2I 2n−1 = . . . =
I 0I 1 = π2(2n)
I 2n−1I 2n = (2n− 2) I 2n−3I 2n−2 =
. . . = 2I 1I 2 = π2
∀n 0 (n + 1) I nI n+1 = π2
limn→+∞
I n = l, limn→+∞
I nI n+1 = l2 = lim
n→+∞π
2(n+1) = 0 limn→+∞I n = 0
(I n) ∀n 0 I n >
0 un =
I nI n+1
(I n) un 1
unun+1 =
I nI n+1I n+1I n+2
= I nI n+2
= n+2n+1 limn→+∞
unun+1 = limn→+∞
n+2n+1 = 1
un
un+2=
I n
I n+1
I n+3
I n+2=
I n
I n+1
n+1n+2I n+1
nn+1I n
= (n + 1)2
n (n + 2) =
n2 + 2n + 1
n2 + 2n 1
(u2n) (u2n+1) 1
limn→+∞
u2n = a 1 limn→+∞
u2n+1 = b 1 ∀n 1 un 1
limn→+∞
unun+1 = 1 = ab a 1 b 1
a = b = 1
un = I nI n+1
→n→+∞
1
∀n 0 (n + 1) I nI n+1 = π2
(n + 1) I 2n+1 = π2
I n+1I n →n→+∞π2 limn→+∞
√ nI n =
π2
I 2n = 2n−1
2n I 2n−2 = 2n−1
2n2n−32n−2I 2n−4 =
2n−12n
2n−32n−2
2n−52n−4I 2n−6 =
2n−12n
2n−32n−2 . . .
3412I 0
I 2n = (2n− 1)× (2n− 3) . . . 5× 3
(2n)× (2n− 2) . . . 4× 2 I 0 =2n!
(2n)×(2n−2)...4×2[2n] [ 2 (n− 1)] . . . [2× 2][2× 1]
π2
I 2n =
(2n)!
22n (n!)2π
2
(n + 1) I nI n+1 = π2
I 2n+1 =
22n (n!)2
(2n + 1)!
limn→+∞
√ nI n =
π2 limn→+∞
(2n + 1) I 22n+1 = π2
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limn→+∞
2 (2n + 1)
22n (n!)2
(2n + 1)!
2
= π
1√ 1− x + √ 1 + x dx
1√ 1− x + √ 1 + x =
√ 1− x−√ 1 + x√
1− x +√ 1 + x √ 1− x + √ 1
+ x =√
1− x−√ 1 + x(1− x)− (1 + x) =
√ 1− x−√ 1 + x
−2x
1√ 1− x +√ 1 + x dx =
√ 1− x−2x dx
=F (x)+
√ 1 + x
2x dx
=G(x) F (x)
u =√
1− x x = 1− u2 dx = −2udu
F (x) = √ 1− x
−2x dx = u−2 (1− u2) (−2udu)
F (x) = u2
1− u2 du = u2 − 1 + 1
1− u2 du = −du + du
1− u2
F (x) =
−u + 12 11 + u +
1
1− u du = −u + 12 ln
|1 + u
| − 12 ln
|1
−u
|+ C
F (x) = −√ 1− x + 12 ln1 +
√ 1− x− 12 ln 1−√ 1−
x + C
G (x)
v = −x dx = −dv
G (x) = √ 1 + x
2x dx =
√ 1− v−2v (−dv) = −F (v) =
−F (−x)
1√ 1− x + √ 1 + x dx =
−√ 1− x + 12 ln 1+√ 1−x
1−√ 1−x + √ 1 + x − 12 ln
1+√ 1+x1−√ 1+x + C
f : [a,b] → R∗+ g : [a,b] →
R a < b
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h (x) = ba |f (t) + xg
(t)| dt
∃z > 0 ∀x ∈ [−z,z] h (x) h
(0)
b
a g (t) dt = 0
f : [a,b] → R+
h.
f [a,b]
N = inf x∈[a,b]
f (x) > 0
g [a,b] , M =
supx∈[a,b]
|g (x)|+ 1
x → ba
f (t) + xg (t) dt
= ba
f (t) dt + x ba
g (t) dt
0 ba
g (t) dt = 0
h (0) = ba |f (t)| dt =
b
a f (t) dt |b− a|N > 0,
I = [−y,y] ⊂
[−z,z] x ∀x ∈ I t →
f (t) + xg (t)
f (t) + xg (t) f (t)− |xg (t)|
N − |xg (t)| N − yM
y min N M ,z =⇒ ∀x ∈ [−y,y]
∀t ∈ [a,b] f (t) + xg (t) 0 x ∈
[−y,y] h (x) = b
a |f (t) + xg (t)| dt =
b
a [f (t) + xg (t)] dt =
ba
f (t) dt + x ba
g (t) dt
h (x) = ba
f (t) dt + x ba
g (t) dt ba
f (t) dt = h (0) ⇐⇒ x
ba
g (t) dt 0
x → x ba
g (t) dt
[−y,y]
x = −y y [−y,y] ,
h x = 0,
b
a g (t) dt = 0
R+
R+
N = inf x∈[a,b]
f (x) > 0
f : [a,b] → R+ f
f
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[a,b] = [−1,1] f (x) = x g (x) =
1
h (x) =
b
a |f (t) + xg (t)| dt =
1−1 |t + x| dt
x ∈ [−1,1] h (x) =
−x−1 |t + x| dt + 1−x
|t + x| dth (x) = − −x−1 (t + x)
dt + 1−x (t + x) dth (x) =
− (t+x)22
−x−1
+(t+x)2
2
1−x
= (−1 + x)2
2 +
(1 + x)2
2 = 12
(x− 1)2 + (x + 1)2
h (x) = 12
x2 − 2x + 1 + x2 + 2x + 1 = x2 +
1
h
∀x ∈ [−1,1] h (x) h (0)
1−1 g (t) dt = 2 = 0
R+.
