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En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Intégrales 5Intégration en mathématiques/Exercices/Intégrales 5 », n'a pu être restituée correctement ci-dessus.
Calculer les intégrales suivantes.
Exercice 13-1
∫
0
π
2
x
sin
x
d
x
{\displaystyle \int _{0}^{\frac {\pi }{2}}x\sin x\,\mathrm {d} x}
Solution
[
−
x
cos
x
]
0
π
2
+
∫
0
π
2
cos
=
[
sin
x
−
x
cos
x
]
0
π
2
=
1
{\displaystyle \left[-x\cos x\right]_{0}^{\frac {\pi }{2}}+\int _{0}^{\frac {\pi }{2}}\cos =\left[\sin x-x\cos x\right]_{0}^{\frac {\pi }{2}}=1}
Exercice 13-2
∫
0
π
4
t
2
cos
t
d
t
{\displaystyle \int _{0}^{\frac {\pi }{4}}t^{2}\cos t\,\mathrm {d} t}
Solution
[
t
2
sin
t
]
0
π
4
−
2
∫
0
π
4
t
sin
t
d
t
=
[
t
2
sin
t
+
2
t
cos
t
]
0
π
4
−
2
∫
0
π
4
cos
=
[
t
2
sin
t
+
2
t
cos
t
−
2
sin
t
]
0
π
4
=
1
2
(
π
2
16
+
π
2
−
2
)
{\displaystyle \left[t^{2}\sin t\right]_{0}^{\frac {\pi }{4}}-2\int _{0}^{\frac {\pi }{4}}t\sin t\,\mathrm {d} t=\left[t^{2}\sin t+2t\cos t\right]_{0}^{\frac {\pi }{4}}-2\int _{0}^{\frac {\pi }{4}}\cos =\left[t^{2}\sin t+2t\cos t-2\sin t\right]_{0}^{\frac {\pi }{4}}={\frac {1}{\sqrt {2}}}\left({\frac {\pi ^{2}}{16}}+{\frac {\pi }{2}}-2\right)}
.
Exercice 13-3
∫
0
π
4
t
sin
3
t
d
t
{\displaystyle \int _{0}^{\frac {\pi }{4}}t\sin 3t\,\mathrm {d} t}
Solution
[
−
t
cos
3
t
3
]
0
π
4
+
∫
0
π
4
cos
3
t
3
d
t
=
[
−
t
cos
3
t
3
+
sin
3
t
9
]
0
π
4
=
1
3
2
(
π
4
+
1
3
)
{\displaystyle \left[-{\frac {t\cos 3t}{3}}\right]_{0}^{\frac {\pi }{4}}+\int _{0}^{\frac {\pi }{4}}{\frac {\cos 3t}{3}}\,\mathrm {d} t=\left[-{\frac {t\cos 3t}{3}}+{\frac {\sin 3t}{9}}\right]_{0}^{\frac {\pi }{4}}={\frac {1}{3{\sqrt {2}}}}\left({\frac {\pi }{4}}+{\frac {1}{3}}\right)}
.
Exercice 13-4
∫
0
2
π
t
2
sin
2
t
d
t
{\displaystyle \int _{0}^{2\pi }t^{2}\sin 2t\,\mathrm {d} t}
Solution
[
−
t
2
cos
2
t
2
]
0
2
π
+
∫
0
2
π
t
cos
2
t
d
t
=
[
−
t
2
cos
2
t
2
]
0
2
π
+
[
t
sin
2
t
2
]
0
2
π
−
∫
0
2
π
sin
2
t
2
d
t
=
−
2
π
2
+
0
−
0
=
−
2
π
2
{\displaystyle \left[{\frac {-t^{2}\cos 2t}{2}}\right]_{0}^{2\pi }+\int _{0}^{2\pi }t\cos 2t\,\mathrm {d} t=\left[{\frac {-t^{2}\cos 2t}{2}}\right]_{0}^{2\pi }+\left[{\frac {t\sin 2t}{2}}\right]_{0}^{2\pi }-\int _{0}^{2\pi }{\frac {\sin 2t}{2}}\,\mathrm {d} t=-2\pi ^{2}+0-0=-2\pi ^{2}}
.
Exercice 13-5
∫
0
π
3
x
cos
2
x
d
x
{\displaystyle \int _{0}^{\frac {\pi }{3}}{\frac {x}{\cos ^{2}x}}\,\mathrm {d} x}
Solution
[
x
tan
x
]
0
π
3
−
∫
0
π
3
tan
=
[
x
tan
x
+
ln
cos
x
]
0
π
/
3
=
π
3
3
−
ln
2
{\displaystyle \left[x\tan x\right]_{0}^{\frac {\pi }{3}}-\int _{0}^{\frac {\pi }{3}}\tan =\left[x\tan x+\ln \cos x\right]_{0}^{\pi /3}={\frac {\pi {\sqrt {3}}}{3}}-\ln 2}
.
Exercice 13-6
∫
π
6
π
3
(
tan
t
+
1
tan
t
)
d
t
{\displaystyle \int _{\frac {\pi }{6}}^{\frac {\pi }{3}}\left(\tan t+{\frac {1}{\tan t}}\right)\,\mathrm {d} t}
Solution
[
ln
tan
]
π
/
6
π
/
3
=
ln
3
{\displaystyle \left[\ln \tan \right]_{\pi /6}^{\pi /3}=\ln 3}
.
Exercice 13-7
∫
0
π
2
(
x
cos
2
x
+
sin
4
x
)
d
x
{\displaystyle \int _{0}^{\frac {\pi }{2}}\left(x\cos ^{2}x+\sin ^{4}x\right)\,\mathrm {d} x}
Solution
∫
0
π
2
(
x
1
+
cos
2
x
2
+
3
8
−
cos
2
x
2
+
cos
4
x
8
)
d
x
=
{\displaystyle \int _{0}^{\frac {\pi }{2}}\left(x{\frac {1+\cos 2x}{2}}+{\frac {3}{8}}-{\frac {\cos 2x}{2}}+{\frac {\cos 4x}{8}}\right)\,\mathrm {d} x=}
[
x
(
x
2
+
sin
2
x
4
)
]
0
π
2
−
∫
0
π
2
(
x
2
+
sin
2
x
4
)
d
x
+
[
3
x
8
−
sin
2
x
4
+
sin
4
x
32
]
0
π
2
=
{\displaystyle \left[x\left({\frac {x}{2}}+{\frac {\sin 2x}{4}}\right)\right]_{0}^{\frac {\pi }{2}}-\int _{0}^{\frac {\pi }{2}}\left({\frac {x}{2}}+{\frac {\sin 2x}{4}}\right)\,\mathrm {d} x+\left[{\frac {3x}{8}}-{\frac {\sin 2x}{4}}+{\frac {\sin 4x}{32}}\right]_{0}^{\frac {\pi }{2}}=}
[
x
2
4
+
x
sin
2
x
4
+
cos
2
x
8
+
3
x
8
−
sin
2
x
4
+
sin
4
x
32
]
0
π
2
=
π
2
16
−
1
4
+
3
π
16
{\displaystyle \left[{\frac {x^{2}}{4}}+{\frac {x\sin 2x}{4}}+{\frac {\cos 2x}{8}}+{\frac {3x}{8}}-{\frac {\sin 2x}{4}}+{\frac {\sin 4x}{32}}\right]_{0}^{\frac {\pi }{2}}={\frac {\pi ^{2}}{16}}-{\frac {1}{4}}+{\frac {3\pi }{16}}}
.
Exercice 13-8
∫
0
π
2
(
2
cos
2
t
sin
t
+
4
sin
2
t
)
d
t
{\displaystyle \int _{0}^{\frac {\pi }{2}}\left(2\cos ^{2}t\sin t+4\sin ^{2}t\right)\,\mathrm {d} t}
Solution
∫
1
0
−
2
x
2
d
x
+
2
∫
0
π
2
(
1
−
cos
2
t
)
d
t
=
2
3
+
π
{\displaystyle \int _{1}^{0}-2x^{2}\,\mathrm {d} x+2\int _{0}^{\frac {\pi }{2}}\left(1-\cos 2t\right)\,\mathrm {d} t={\frac {2}{3}}+\pi }
.