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Exercice : Intégrales 3Intégration en mathématiques/Exercices/Intégrales 3 », n'a pu être restituée correctement ci-dessus.
∫
0
π
4
sin
3
x
d
x
{\displaystyle \int _{0}^{\frac {\pi }{4}}\sin 3x\,\mathrm {d} x}
Solution
∫
0
π
4
sin
3
x
d
x
=
−
1
3
[
cos
3
x
]
0
π
4
=
1
+
1
/
2
3
{\displaystyle \int _{0}^{\frac {\pi }{4}}\sin 3x\,\mathrm {d} x={\frac {-1}{3}}\left[\cos 3x\right]_{0}^{\frac {\pi }{4}}={\frac {1+1/{\sqrt {2}}}{3}}}
∫
−
π
4
π
4
tan
2
x
d
x
{\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\tan ^{2}x\,\mathrm {d} x}
Solution
∫
−
π
4
π
4
tan
2
x
d
x
=
[
tan
x
−
x
]
−
π
/
4
π
/
4
=
2
−
π
2
{\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\tan ^{2}x\,\mathrm {d} x=\left[\tan x-x\right]_{-\pi /4}^{\pi /4}=2-{\frac {\pi }{2}}}
∫
−
π
4
π
4
tan
x
cos
x
d
x
{\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}{\frac {\tan x}{\cos x}}\,\mathrm {d} x}
Solution
∫
−
π
4
π
4
tan
x
cos
x
d
x
=
0
{\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}{\frac {\tan x}{\cos x}}\,\mathrm {d} x=0}
∫
0
π
3
cos
2
t
sin
3
t
d
t
{\displaystyle \int _{0}^{\frac {\pi }{3}}\cos 2t\sin ^{3}t\,\mathrm {d} t}
Solution
∫
0
π
3
cos
2
t
sin
3
t
d
t
=
∫
0
π
3
(
2
cos
2
t
−
1
)
(
cos
2
t
−
1
)
(
−
sin
t
)
d
t
=
∫
1
1
/
2
(
2
x
2
−
1
)
(
x
2
−
1
)
d
x
=
∫
1
1
/
2
(
2
x
4
−
3
x
2
+
1
)
d
x
=
[
2
x
5
5
−
x
3
+
x
]
1
1
/
2
=
−
1
80
.
{\displaystyle {\begin{aligned}\int _{0}^{\frac {\pi }{3}}\cos 2t\sin ^{3}t\,\mathrm {d} t&=\int _{0}^{\frac {\pi }{3}}\left(2\cos ^{2}t-1\right)\left(\cos ^{2}t-1\right)(-\sin t)\,\mathrm {d} t\\&=\int _{1}^{1/2}\left(2x^{2}-1\right)\left(x^{2}-1\right)\,\mathrm {d} x\\&=\int _{1}^{1/2}\left(2x^{4}-3x^{2}+1\right)\,\mathrm {d} x\\&=\left[{\frac {2x^{5}}{5}}-x^{3}+x\right]_{1}^{1/2}\\&={\frac {-1}{80}}.\end{aligned}}}
∫
0
π
2
cos
t
cos
2
t
cos
3
t
d
t
{\displaystyle \int _{0}^{\frac {\pi }{2}}\cos t\cos 2t\cos 3t\,\mathrm {d} t}
Solution
2
cos
a
cos
b
=
cos
(
a
+
b
)
+
cos
(
a
−
b
)
{\displaystyle 2\cos a\cos b=\cos(a+b)+\cos(a-b)}
cos
t
cos
2
t
cos
3
t
=
1
2
(
cos
3
t
+
cos
t
)
cos
3
t
=
1
4
(
1
+
cos
6
t
+
cos
4
t
+
cos
2
t
)
{\displaystyle \cos t\cos 2t\cos 3t={\frac {1}{2}}(\cos 3t+\cos t)\cos 3t={\frac {1}{4}}(1+\cos 6t+\cos 4t+\cos 2t)}
1
4
∫
0
π
2
(
1
+
cos
6
t
+
cos
4
t
+
cos
2
t
)
d
t
=
1
4
[
t
+
sin
6
t
6
+
sin
4
t
4
+
sin
2
t
2
]
0
π
2
=
[
t
4
]
0
π
2
=
π
8
{\displaystyle {\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}(1+\cos 6t+\cos 4t+\cos 2t)\,\mathrm {d} t={\frac {1}{4}}\left[t+{\frac {\sin 6t}{6}}+{\frac {\sin 4t}{4}}+{\frac {\sin 2t}{2}}\right]_{0}^{\frac {\pi }{2}}=\left[{\frac {t}{4}}\right]_{0}^{\frac {\pi }{2}}={\frac {\pi }{8}}}
∫
π
4
5
π
4
(
cos
2
t
+
sin
4
t
)
sin
t
d
t
{\displaystyle \int _{\frac {\pi }{4}}^{\frac {5\pi }{4}}\left(\cos ^{2}t+\sin ^{4}t\right)\sin t\,\mathrm {d} t}
Solution
∫
π
4
5
π
4
(
cos
2
t
+
sin
4
t
)
sin
t
d
t
=
−
∫
1
/
2
−
1
/
2
(
x
2
+
(
1
−
x
2
)
2
)
d
x
=
[
x
5
5
−
x
3
3
+
x
]
−
1
/
2
1
/
2
=
2
(
1
20
−
1
6
+
1
)
=
53
60
{\displaystyle \int _{\frac {\pi }{4}}^{\frac {5\pi }{4}}\left(\cos ^{2}t+\sin ^{4}t\right)\sin t\,\mathrm {d} t=-\int _{1/{\sqrt {2}}}^{-1/{\sqrt {2}}}\left(x^{2}+(1-x^{2})^{2}\right)\,\mathrm {d} x=\left[{\frac {x^{5}}{5}}-{\frac {x^{3}}{3}}+x\right]_{-1/{\sqrt {2}}}^{1/{\sqrt {2}}}={\sqrt {2}}\left({\frac {1}{20}}-{\frac {1}{6}}+1\right)={\frac {53}{60}}}
∫
0
π
2
cos
t
5
−
3
sin
t
d
t
{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\cos t}{5-3\sin t}}\,\mathrm {d} t}
Solution
∫
0
π
2
cos
t
5
−
3
sin
t
d
t
=
∫
0
1
d
x
5
−
3
x
=
−
1
3
[
ln
(
5
−
3
x
)
]
0
1
=
ln
(
5
/
2
)
3
{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\cos t}{5-3\sin t}}\,\mathrm {d} t=\int _{0}^{1}{\frac {\mathrm {d} x}{5-3x}}={\frac {-1}{3}}\left[\ln(5-3x)\right]_{0}^{1}={\frac {\ln(5/2)}{3}}}
∫
π
6
π
3
2
sin
t
1
+
cos
2
t
2
d
t
{\displaystyle \int _{\frac {\pi }{6}}^{\frac {\pi }{3}}{\frac {2\sin t}{1+\cos ^{2}{\frac {t}{2}}}}\,\mathrm {d} t}
Solution
∫
π
6
π
3
2
sin
t
1
+
cos
2
t
2
d
t
=
−
4
∫
7
/
4
3
/
4
d
x
x
=
4
ln
(
7
/
3
)
{\displaystyle \int _{\frac {\pi }{6}}^{\frac {\pi }{3}}{\frac {2\sin t}{1+\cos ^{2}{\frac {t}{2}}}}\,\mathrm {d} t=-4\int _{7/4}^{3/4}{\frac {\mathrm {d} x}{x}}=4\ln(7/3)}
![{\displaystyle \int _{0}^{\frac {\pi }{4}}\sin 3x\,\mathrm {d} x={\frac {-1}{3}}\left[\cos 3x\right]_{0}^{\frac {\pi }{4}}={\frac {1+1/{\sqrt {2}}}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52615a6fc8d43ed3157daf2dcf8bd3fbfb89517a)
![{\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\tan ^{2}x\,\mathrm {d} x=\left[\tan x-x\right]_{-\pi /4}^{\pi /4}=2-{\frac {\pi }{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a32e5ee8c0bc95b7aaa988661c5554be9d2df18e)
![{\displaystyle {\begin{aligned}\int _{0}^{\frac {\pi }{3}}\cos 2t\sin ^{3}t\,\mathrm {d} t&=\int _{0}^{\frac {\pi }{3}}\left(2\cos ^{2}t-1\right)\left(\cos ^{2}t-1\right)(-\sin t)\,\mathrm {d} t\\&=\int _{1}^{1/2}\left(2x^{2}-1\right)\left(x^{2}-1\right)\,\mathrm {d} x\\&=\int _{1}^{1/2}\left(2x^{4}-3x^{2}+1\right)\,\mathrm {d} x\\&=\left[{\frac {2x^{5}}{5}}-x^{3}+x\right]_{1}^{1/2}\\&={\frac {-1}{80}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72f3c4cbb8cb388056847eb21763f9f1604249ab)
![{\displaystyle {\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}(1+\cos 6t+\cos 4t+\cos 2t)\,\mathrm {d} t={\frac {1}{4}}\left[t+{\frac {\sin 6t}{6}}+{\frac {\sin 4t}{4}}+{\frac {\sin 2t}{2}}\right]_{0}^{\frac {\pi }{2}}=\left[{\frac {t}{4}}\right]_{0}^{\frac {\pi }{2}}={\frac {\pi }{8}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84de86122621eb57ca482340542b0f2cb16317d0)
![{\displaystyle \int _{\frac {\pi }{4}}^{\frac {5\pi }{4}}\left(\cos ^{2}t+\sin ^{4}t\right)\sin t\,\mathrm {d} t=-\int _{1/{\sqrt {2}}}^{-1/{\sqrt {2}}}\left(x^{2}+(1-x^{2})^{2}\right)\,\mathrm {d} x=\left[{\frac {x^{5}}{5}}-{\frac {x^{3}}{3}}+x\right]_{-1/{\sqrt {2}}}^{1/{\sqrt {2}}}={\sqrt {2}}\left({\frac {1}{20}}-{\frac {1}{6}}+1\right)={\frac {53}{60}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e26674bd79292728f45d7a437c64346a4bc2b66)
![{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\cos t}{5-3\sin t}}\,\mathrm {d} t=\int _{0}^{1}{\frac {\mathrm {d} x}{5-3x}}={\frac {-1}{3}}\left[\ln(5-3x)\right]_{0}^{1}={\frac {\ln(5/2)}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a8cacef5b6f15e991fb84e0cfc8b37adb12f430)
