Integration by Substitution
Indefinite integral
Let
u
=
g
(
x
)
and
F
′
(
x
)
=
f
(
x
)
∫
f
(
g
(
x
)
)
g
′
(
x
)
d
x
=
∫
f
(
u
)
d
u
=
F
(
u
)
+
𝒸
=
F
(
g
(
x
)
)
+
𝒸
Riemann integral
Let
u
=
g
(
x
)
and
F
′
(
x
)
=
f
(
x
)
∫
𝒶
𝒷
f
(
g
(
x
)
)
g
′
(
x
)
d
x
=
∫
g
(
𝒶
)
g
(
𝒷
)
f
(
u
)
d
u
Symmetries
if
f
(
-
x
)
=
f
(
x
)
then
∫
-
𝒶
𝒶
f
(
x
)
d
x
=
2
⁢
∫
0
𝒶
f
(
x
)
d
x
if
f
(
-
x
)
=
-
f
(
x
)
then
∫
-
𝒶
𝒶
f
(
x
)
d
x
=
0
Examples
∫
6
⁢
x
⁢
(
3
⁢
x
2
+
4
)
4
d
x
∫
ℯ
x
⁢
ℯ
x
+
1
d
x
∫
3
x
-
10
d
x
∫
1
9
-
4
⁢
x
2
d
x
∫
-1
2
x
2
⁢
(
2
⁢
x
3
+
1
)
5
d
x
∫
1
ℯ
ln
x
x
d
x
∫
1
2
ℯ
1
x
x
2
d
x
∫
0
π
2
sin
x
1
+
cos
x
d
x
∫
-2
2
x
4
+
3
⁢
x
2
-
7
d
x
∫
0
2
x
4
+
3
⁢
x
2
-
7
d
x
∫
-2
2
x
5
-
8
⁢
sin
x
d
x
Further examples
Indefinite substitution
∫
z
z
2
-
5
d
z
∫
sin
t
cos
3
t
d
t
∫
ℯ
-
x
d
x
Indefinite substitution
∫
3
⁢
x
2
⁢
ℯ
2
⁢
x
3
d
x
∫
2
⁢
x
3
+
3
⁢
x
x
4
+
3
⁢
x
2
d
x
∫
1
1
+
4
⁢
x
2
d
x
Definite substitution
∫
0
π
2
cos
2
x
d
x
∫
0
1
x
⁢
ℯ
4
⁢
x
2
+
3
d
x
∫
1
2
ℯ
-
x
+
1
d
x
Definite substitution
∫
0
1
1
1
-
x
2
d
x
∫
0
3
1
1
+
x
2
d
x