Integrali di funzioni goniometriche
1.
∫ sen
2
x ⋅ dx
Dalla formula di duplicazione del coseno si ha: cos 2 x = cos 2 x − sen 2 x
= 1 − sen 2 x − sen 2 x = 1 − 2 sen 2 x . Pertanto dalla formula cos 2 x = 1 − 2 sen 2 x
=
si ottiene:
1
⋅ (1 − cos 2 x ) . Sostituendo si ha:
2
1
1
2
∫ sen x ⋅ dx = ∫ 2 ⋅ (1 − cos 2 x ) ⋅ dx = 2 ⋅ ∫1 ⋅ dx − ∫ cos 2 x ⋅ dx =
1 ⎡
1 ⎡
1 ⎡
1
1
1
⎤
⎤
⎤
⋅ ⎢ x − sen 2 x ⎥ + c =
⋅ ⎢ x − ⋅ 2 sen x ⋅ cos x ⎥ + c =
⋅ ⎢ ∫ 1 ⋅ dx − ⋅ ∫ 2 ⋅ cos 2 x ⋅ dx ⎥ =
2 ⎣
2 ⎣
2 ⎣
2
2
2
⎦
⎦
⎦
1
=
⋅ x − sen x ⋅ cos x + c
2
2 sen 2 x = 1 − cos 2 x ; sen 2 x =
[
[
2.
∫ cos
2
]
]
x ⋅ dx
(
Dalla formula di duplicazione del coseno si ha: cos 2 x = cos 2 x − sen 2 x
= cos 2 x − 1 − cos 2 x
)
=
= − 1 + 2 cos 2 x . Pertanto dalla formula cos 2 x = − 1 + 2 cos 2 x
si
1
⋅ (1 + cos 2 x ) . Sostituendo si ha:
2
1
1
2
∫ cos x ⋅ dx = ∫ 2 ⋅ (1 + cos 2 x ) ⋅ dx = 2 ⋅ ∫ 1 ⋅ dx + ∫ cos 2 x ⋅ dx =
1 ⎡
1
1 ⎡
1
1 ⎡
1
⎤
⎤
⎤
⋅ ⎢ x + sen 2 x ⎥ + c =
⋅ ⎢ x + ⋅ 2 sen x ⋅ cos x ⎥ + c =
⋅ ⎢ ∫ 1 ⋅ dx + ⋅ ∫ 2 ⋅ cos 2 x ⋅ dx ⎥ =
2 ⎣
2
2 ⎣
2
2 ⎣
2
⎦
⎦
⎦
1
⋅ x + sen x ⋅ cos x + c
=
2
ottiene: 2 cos 2 x = 1 + cos 2 x ; cos 2 x =
[
[
3.
1
∫ sen 2 x ⋅ cos 2 x dx
sen 2 x + cos 2 x
∫ sen 2 x ⋅ cos 2 x dx =
]
]
Ricordando che: sen 2 x + cos 2 x = 1
si ha:
sen 2 x
cos 2 x
dx
+
∫ sen 2 x ⋅ cos 2 x
∫ sen 2 x ⋅ cos 2 x dx =
1
1
∫ cos 2 x dx + ∫ sen 2 x dx
= tg x − cot g x + c .
Matematica
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1
=
4.
1
∫ senx dx
Ricordando che: sen 2 x = 2 sen x ⋅ cos x
si ha: sen x = 2 ⋅ sen
x
x
⋅ cos
2
2
1
dx Applicando la relazione fondamentale sen 2 x + cos 2 x = 1 si ha:
x
x
2 ⋅ sen ⋅ cos
2
2
x
x
x
x
sen 2 + cos 2
sen 2
cos 2
2
2 dx =
2
2
= ∫
dx + ∫
dx =
∫
x
x
x
x
x
x
2 ⋅ sen ⋅ cos
2 ⋅ sen ⋅ cos
2 ⋅ sen ⋅ cos
2
2
2
2
2
2
x
x
1
x
1
x
− sen
sen
cos
cos
1
1
1
1
2 dx + ⋅
2 dx =
2 dx + ⋅ 2 ⋅ 2
2 dx =
=
⋅
⋅ (− 2 ) ⋅ ∫ 2
∫
x
x
2 ∫ cos x
2 ∫ sen x
2
2
cos
sen
2
2
2
2
=
∫
=
− log cos
x
x
+ log sen + c
2
2
1
5.
∫ cos x dx
=
∫
⎛
⎝
x+
Matematica
tg
sen x 2
+c
cos x 2
Ricordando che: cos x = sen ⎜ x +
1
dx
π⎞
⎛
sen ⎜ x + ⎟
2⎠
⎝
= log
= log
2
= log
tg
x
+c
2
π⎞
⎟ si ha:
2⎠
Applicando l’integrale dimostrato precedentemente si ha:
π
2 + c = log
⎛ x π⎞
tg ⎜ + ⎟ + c
⎝2 4⎠
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