Corrigé DST 2

n= 6 a
b
(a+b)6=a6+ 6a5b+ 15a4b2+ 20a3b3+ 15a2b4+ 6ab5+b6.
sin6(t) = eit −e−it
2i6
=1
26i6(eit)6+ 6(eit)5(−e−it) + 15(eit)4(−e−it)2+ 20(eit)3(−e−it)3
+ 15(eit)4(−e−it)2+ 6(eit)5(−e−it)+(eit)6
=−1
64 (e6it +e−6it)−6e4it +e−4it+ 15 e2it +e−2it−20
=−1
64 (2 cos(6t)−12 cos(4t) + 30 cos(2t)−20)
=1
32 (10 −15 cos(2t) + 6 cos(4t)−cos(6t)) .
(x+iy)2= 5−4i x y
x2−y2= 5
2xy =−4
x2+y2=√52+ 42=√41 ⇐⇒ 
2x2=√41 + 5
2y2=√41 −5
2xy =−4
|(x+iy)2|=|15 −8i|
41
x y
s√41 + 5
2+is√41 −5 2
5−4i δ
(1 −2i)−δ
2et (1 −2i) + δ 2.
z=reiθ r z θ
−32 = −25= 25eiπ
z5=−32 ⇐⇒ r5e5iθ = 25eiπ ⇐⇒ r5= 25
5θ=π[2π]=r= 2 θ=π
5[2π 5].
S= 2eiπ
5,2e3iπ 5,2e5iπ 5
|{z}
−2
,2e7iπ
5,2e9iπ 5
−2 3
1
x2−x−6=1 51 x−3−1 x+ 2.
]3,+∞[
x7→ ln 
x−3
x+ 2.
R(2x−1)dx=1
2·R2(2x−1)1/2dx=1 2·2 3(2x−1)3/2
x7→ 1
3(2x−1)3/2
[−1/2,+∞[

x7→ xe−3xR
0
F:R→R
x7→ Rx
0te−3tdt.
x F (x)
F(x) = 1
91−e−3x−3xe−3x.
x7→ ex
2 + e2xR
F:x7→ Zx
0
et
2 + e2tdt
x u =et
C1du=etdt
F(x) = Zx
0
et
2+(et)2dt=Zex
1
1
2 + u2du
=√2
2Zex
1
u
√2
1 + u
√22du
=1
√2arctan u √2ex
1
.
x7→ 1
√2arctan ex √2.
x=π−u
dx=−du
Zπ
0
xf (sin x) dx=Z0
π
(π−u) sin(π−u)
| {z }
=sin(u)
(−1)du
=Zπ
0
(π−x)f(sin x) dx
=πZπ
0
f(sin x) dx−Zπ
0
xf (sin x) dx.
2Zπ
0
xf (sin x) dx=πZπ
0
f(sin x) dx,
2
I
I=Zπ
0
xsin x
2−sin2(x)dx=Zπ
0
xf (sin x) dx,
f:u7→ 1
2−u2[0,1]
I=π
2Zπ
0
f(sin x)dx=π
2Zπ
0
sin x
1 + cos2(x)dx.
u= cos x
du=−sin xdx
I=π
2Z1
−1
1
1 + u2du=π 2[arctan(u)]1
−1== π2
4.
arctan(1) = π
4=−arctan(−1)

∀x∈]0,1] 2 arctan r1−x
x!+ arcsin(2x−1) = π 2.(∗)
f:x7→ 2 arctan q1−x
xg:x7→ arcsin(2x−1)
•arctan Rf(x)
x6= 0 1−x
x
x−∞ 0 1 +∞
1−x+−
x−+
1−x
x−+−
f]0,1]
•arcsin [−1,1] x
−1≤2x−1≤1⇐⇒ 0≤2x≤2⇐⇒ 0≤x≤1.
g[0,1]
u:]0,1[ →R∗
+
x7→ 1−x
x
=1
x−1 et v:R∗
+→R
x7→ √x.
u]0,1[ v
R∗
+v◦u
]0,1[
∀x∈]0,1[ (v◦u)0(x) = u0(x)·v0(u(x)) = −1
x2·1
2q1−x
x
.
2 arctan R
f= 2 arctan ◦(v◦u) ]0,1[ x
f0(x) = (v◦u)0(x)·2
1+(v◦u(x))2
=−1
x2·
2
2q1−x
x·1
1 + 1−x
x
−1
x
2rx
1−x·
x
1−x+x=−1
px(1 −x).
w:]0,1[ →]−1,1[
x7→ 2x−1.
w]0,1[ arcsin
]−1,1[ g=
arcsin ◦w]0,1[ x
g0(x) = w0(x)·arcsin0(2x−1)
=2
p1−(2x−1)2
=2
p(1 −(2x−1)) (1 + (2x−1))
=1
px(1 −x).
f+g
]0,1[ x f0(x)+g0(x) = 0
f+g]0,1[
f(1/2) + g(1/2) = 2 arctan(1) + arcsin(0) = 2 ·π
4+ 0 = π 2.
1
f(1) + g(1) = 2 arctan(0) + arcsin(1) = π
2.

arccos + arcsin
cos a= sin π
2−ax∈[−1,1]
cos (arccos(x)) = x= sin π
2−arccos(x).
π
2−arccos(x)xsin
arccos(x)∈[0, π]π
2−arccos(x)∈−π 2,π 2
π
2−arccos(x)∈−π 2,π 2x
−π
2,π 2arcsin(x)
arcsin(x) = π
2−arccos(x).
u∈[0,π
2[
fcos2(u)+gcos2(u)
= 2 arctan s1−cos2(u)
sin2(u)!+ arcsin 2 cos2(u)−1
= 2 arctan (|tan(u)|) + arcsin (cos(2u)) .
u[0,π
2[ tan(u)
arcsin (cos(2u)) = π
2−arccos (cos(2u)) = π 2−2u.
2u∈[0, π]
fcos2(u)+gcos2(u)= 2 arctan (tan(u)) + π
2−2u
= 2u+π
2−2u=π 2
f+g]0,1]
x x = cos(u) [0,π
2[
θ]0, π[
Cn=
n
X
k=0
cosk(θ) cos(kθ), Sn=
n
X
k=0
cosk(θ) sin(kθ) et An=Cn+iSn.
1−cos(θ)eiθ = 1 −eiθ +e−iθ
2eiθ
= 1 −e2iθ + 1
2=1−e2iθ 2
=eiθ e−iθ −eiθ
2
=eiθ −
2isin θ
2= sin(θ)eiθ−iπ
2.
θ∈]0, π[ sin θ
π
2−θ
An=Cn+iSn=
n
X
k=0
cosk(θ) (cos(kθ) + isin(kθ))
| {z }
eikθ
=
n
X
k=0 cos(θ)eiθk.
|cos(θ)eiθ|=
|cos(θ)| 6= 1 θ∈]0, π[
An=1−cos(θ)eiθn+1
1−cos(θ)eiθ =1−cos(θ)n+1ei(n+1)θ
sin(θ)ei(θ−π
2)
=ei(π
2−θ)−cos(θ)n+1ei(π 2+nθ)
sin θ.

Cn= Re (An) =
Re ei(π
2−θ)−cos(θ)n+1Re ei(π 2+nθ)
sin θ
=cos(π
2−θ)−cos(θ)n+1 cos(π 2+nθ)
sin θ
=sin(θ) + cos(θ)n+1 sin(nθ)
sin θ
= 1 + cos(θ)n+1 sin(nθ)
sin θ.
Sn= Im (An) = cos(θ)−cos(θ)n+1 cos(nθ)
sin θ.
n
X
ω∈Un
ω=
n−1
X
k=0
e2ikπ
n=
n−1
X
k=0 e2iπ
nk=
1−e2iπ
nn
1−e2iπ
n
=1−1
1−e2iπ
n
= 0.
•zC
(z+ 1)n=znz
(0 + 1)n= 0n1 = 0
(z+ 1)n
zn= 1 z+ 1 zn
= 1.
z+1
zn
k0n−1
z+ 1
z=e2ikπ
n.
z+ 1 = ze2ikπ
nz1−e2ikπ n=−1.
e2ikπ
n= 1 0 = −1k6= 0
z=−1
1−e2ikπ
n
=−1
eikπ
n−2isin kπ
n=−ie−ikπ
n
2 sin kπ
n.
n−1z
−ie−
ikπ
n
2 sin(kπ
n)k∈J1, n−1K
•
k∈J1, n−1Kz=−ie−
ikπ
n
2 sin(kπ
n)
z=−1
1−e2ikπ
n
z+ 1 = ze2ikπ
nz+ 1
zn
=e2iknπ
n= 1,
(z+ 1)n=znz
•n−1−ie−
ikπ
n
2 sin(kπ
n)
k∈J1, n−1K