Solutions
I.4.6. Le changement de variable t = π − x donne
I =
π
0
xf (sin x) dx =
π
0
(π − t)f (sin t) dt = π
π
0
f (sin t) dt − I.
Donc,
π
0
x sin
2n x
sin
2n x + cos 2n x
dx =
π
2
π
0
sin
2n x
sin
2n x + cos 2n x
dx
=
π
2
π/2
0
sin
2n x
sin
2n x + cos 2n x
dx
+
π
2
π
π/2
sin
2n x
sin
2n x + cos 2n x
dx
=
π
2
π/2
0
sin
2n x + cos 2n x
sin
2n x + cos 2n x
dx =
π 2
4
.
I.4.7.
(a) On a
a
−a
f (x) dx =
0
−a
f (−x) dx +
a
0
f (x) dx = 2
a
0
f (x) dx.
(b) On a
a
−a
f (x) dx =
0
−a
−f (−x) dx +
a
0
f (x) dx = 0.
I.4.8. La fonction f étant périodique, on obtient
a+T
a
f (x) dx =
0
a
f (x) dx +
T
0
f (x) dx +
a+T
T
f (x − T ) dx
=
0
a
f (x) dx +
T
0
f (x) dx −
0
a
f (x) dx.
I.4.9. On a
b
a
f (nx) dx =
1
n
nb
na
f (x) dx
=
1
n
na+T
na
f (x) dx + · · · +
nb
na+k(n)T
f (x) dx
,
107