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En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Primitives 3Intégration en mathématiques/Exercices/Primitives 3 », n'a pu être restituée correctement ci-dessus.
f
{\displaystyle f}
F
{\displaystyle F}
f
{\displaystyle f}
f
{\displaystyle f}
F
{\displaystyle F}
f
(
x
)
=
x
tan
x
2
{\displaystyle f(x)=x\tan x^{2}}
Solution
f
{\displaystyle f}
R
\
{
±
π
2
+
k
π
|
k
∈
N
}
{\displaystyle \mathbb {R} \left\backslash \left\{\left.\pm {\sqrt {{\frac {\pi }{2}}+k\pi }}\,\right|\,k\in \mathbb {N} \right\}\right.}
]
k
π
−
π
2
,
k
π
+
π
2
[
{\displaystyle \left]{\sqrt {k\pi -{\frac {\pi }{2}}}},{\sqrt {k\pi +{\frac {\pi }{2}}}}\right[}
]
−
k
π
+
π
2
,
−
k
π
−
π
2
[
{\displaystyle \left]-{\sqrt {k\pi +{\frac {\pi }{2}}}},-{\sqrt {k\pi -{\frac {\pi }{2}}}}\right[}
k
∈
N
∗
{\displaystyle k\in \mathbb {N} ^{*}}
]
−
π
2
,
π
2
[
{\displaystyle \left]-{\sqrt {\frac {\pi }{2}}},{\sqrt {\frac {\pi }{2}}}\right[}
Sur chacun de ces intervalles,
f
=
−
u
′
2
u
{\displaystyle f={\frac {-u'}{2u}}}
u
(
x
)
=
cos
x
2
{\displaystyle u(x)=\cos x^{2}}
f
{\displaystyle f}
F
=
−
ln
|
cos
x
2
|
2
{\displaystyle F={\frac {-\ln \left|\cos x^{2}\right|}{2}}}
f
(
x
)
=
x
cos
x
{\displaystyle f(x)=x\cos x}
f
(
x
)
=
x
sin
x
{\displaystyle f(x)=x\sin x}
f
(
x
)
=
x
cos
(
2
x
)
{\displaystyle f(x)=x\cos(2x)}
f
(
x
)
=
x
sin
(
2
x
)
{\displaystyle f(x)=x\sin(2x)}
f
(
x
)
=
x
2
cos
x
{\displaystyle f(x)=x^{2}\cos x}
Solution
En enchaînant deux intégrations par parties , on montre qu'une primitive de
f
{\displaystyle f}
R
{\displaystyle \mathbb {R} }
F
:
x
↦
x
2
sin
x
+
2
x
cos
x
−
2
sin
x
{\displaystyle F:x\mapsto x^{2}\sin x+2x\cos x-2\sin x}
f
(
x
)
=
x
cos
2
x
{\displaystyle f(x)=x\cos ^{2}x}
Solution
Sachant que
cos
2
x
=
1
+
cos
2
x
2
{\displaystyle \cos ^{2}x={\frac {1+\cos 2x}{2}}}
intégrations par parties montre qu'une primitive de
f
{\displaystyle f}
R
{\displaystyle \mathbb {R} }
F
:
x
↦
x
(
x
2
+
sin
2
x
4
)
−
x
2
4
+
cos
2
x
8
=
x
2
4
+
x
sin
2
x
4
+
cos
2
x
8
{\displaystyle F:x\mapsto x\left({\frac {x}{2}}+{\frac {\sin 2x}{4}}\right)-{\frac {x^{2}}{4}}+{\frac {\cos 2x}{8}}={\frac {x^{2}}{4}}+{\frac {x\sin 2x}{4}}+{\frac {\cos 2x}{8}}}
f
(
x
)
=
(
3
x
2
−
7
x
+
1
)
cos
3
x
{\displaystyle f(x)=(3x^{2}-7x+1)\cos 3x}
Solution
En enchaînant deux intégrations par parties , on montre qu'une primitive de
f
{\displaystyle f}
R
{\displaystyle \mathbb {R} }
F
:
x
↦
(
3
x
2
−
7
x
+
1
)
sin
3
x
3
+
6
x
−
7
3
cos
3
x
3
−
2
sin
3
x
9
=
(
9
x
2
−
21
x
+
1
)
sin
3
x
+
(
6
x
−
7
)
cos
3
x
9
{\displaystyle F:x\mapsto (3x^{2}-7x+1){\frac {\sin 3x}{3}}+{\frac {6x-7}{3}}{\frac {\cos 3x}{3}}-2{\frac {\sin 3x}{9}}={\frac {(9x^{2}-21x+1)\sin 3x+(6x-7)\cos 3x}{9}}}
f
(
x
)
=
x
cos
2
x
{\displaystyle f(x)={\frac {x}{\cos ^{2}x}}}
f
(
x
)
=
9
x
sin
2
x
cos
x
{\displaystyle f(x)=9x\sin ^{2}x\cos x}
f
(
x
)
=
2
x
(
sin
4
x
−
3
sin
2
x
cos
2
x
)
{\displaystyle f(x)=2x\left(\sin ^{4}x-3\sin ^{2}x\cos ^{2}x\right)}
Solution
f
(
x
)
=
x
(
cos
4
x
−
cos
2
x
)
{\displaystyle f(x)=x\left(\cos 4x-\cos 2x\right)}
f
{\displaystyle f}
R
{\displaystyle \mathbb {R} }
F
:
x
↦
x
(
sin
4
x
4
−
sin
2
x
2
)
+
cos
4
x
16
−
cos
2
x
4
{\displaystyle F:x\mapsto x\left({\frac {\sin 4x}{4}}-{\frac {\sin 2x}{2}}\right)+{\frac {\cos 4x}{16}}-{\frac {\cos 2x}{4}}}
f
(
x
)
=
(
3
x
2
−
7
x
+
1
)
[
cos
(
3
x
−
1
)
+
2
sin
(
3
x
+
1
)
]
{\displaystyle f(x)=(3x^{2}-7x+1)\left[\cos(3x-1)+2\sin(3x+1)\right]}
Solution
En enchaînant deux intégrations par parties , on montre qu'une primitive de
f
{\displaystyle f}
R
{\displaystyle \mathbb {R} }
F
:
x
↦
9
x
2
−
21
x
+
1
9
[
sin
(
3
x
−
1
)
−
2
cos
(
3
x
+
1
)
]
+
6
x
−
7
9
[
cos
(
3
x
−
1
)
+
2
sin
(
3
x
+
1
)
]
{\displaystyle F:x\mapsto {\frac {9x^{2}-21x+1}{9}}\left[\sin(3x-1)-2\cos(3x+1)\right]+{\frac {6x-7}{9}}\left[\cos(3x-1)+2\sin(3x+1)\right]}
![{\displaystyle \left]{\sqrt {k\pi -{\frac {\pi }{2}}}},{\sqrt {k\pi +{\frac {\pi }{2}}}}\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/555e3fe54379864fdf21b802ce0e44a17b2c3bda)
![{\displaystyle \left]-{\sqrt {k\pi +{\frac {\pi }{2}}}},-{\sqrt {k\pi -{\frac {\pi }{2}}}}\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3adc62fe14c3b1f32a04b0cdf6b174e1d2d67977)
![{\displaystyle \left]-{\sqrt {\frac {\pi }{2}}},{\sqrt {\frac {\pi }{2}}}\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42540ea0f4becca54931a17960f1b6af5fe99f13)
![{\displaystyle \left]k\pi -{\frac {\pi }{2}},k\pi +{\frac {\pi }{2}}\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed28e882190d9885e60372b85b0896744bfd5fb2)
![{\displaystyle f(x)=(3x^{2}-7x+1)\left[\cos(3x-1)+2\sin(3x+1)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d251b899b320e336501863adde1cc9c5219ff21)
![{\displaystyle F:x\mapsto {\frac {9x^{2}-21x+1}{9}}\left[\sin(3x-1)-2\cos(3x+1)\right]+{\frac {6x-7}{9}}\left[\cos(3x-1)+2\sin(3x+1)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36b0d9634e264662fc9f0f1457bfe3b97257f05f)
