Čia pateikiami metodai, padedantys integruoti .
Tiesioginis integravimas [ keisti ] Jei
∫
f
(
x
)
d
x
=
F
(
x
)
+
C
,
{\displaystyle \int f(x){\mathsf {d}}x=F(x)+C,\quad }
tai
∫
f
(
u
)
d
u
=
F
(
u
)
+
C
.
{\displaystyle \int f(u){\mathsf {d}}u=F(u)+C.\quad }
Šis metodas pagrįstas pirmos eilės diferencialo formos invariantiškumu .
Pavyzdžiai,
∫
t
3
d
t
=
t
4
4
+
C
{\displaystyle \int t^{3}{\mathsf {d}}t={\frac {t^{4}}{4}}+C}
d
(
x
+
10
)
=
d
x
{\displaystyle {\mathsf {d}}(x+10)={\mathsf {d}}x}
tai:
∫
(
x
+
10
)
3
d
x
=
∫
(
x
+
10
)
3
d
(
x
+
10
)
=
(
x
+
10
)
4
4
+
C
.
{\displaystyle \int (x+10)^{3}{\mathsf {d}}x=\int (x+10)^{3}{\mathsf {d}}(x+10)={\frac {(x+10)^{4}}{4}}+C.}
∫
x
4
d
x
=
x
4
+
1
4
+
1
+
C
=
x
5
5
+
C
.
{\displaystyle \int x^{4}dx={\frac {x^{4+1}}{4+1}}+C={\frac {x^{5}}{5}}+C.}
∫
10
x
d
x
=
10
x
ln
10
+
C
.
{\displaystyle \int 10^{x}dx={\frac {10^{x}}{\ln 10}}+C.}
∫
x
2
+
5
x
−
1
x
d
x
=
∫
(
x
3
/
2
+
5
x
1
/
2
−
x
−
1
/
2
)
d
x
=
∫
x
3
/
2
d
x
+
5
∫
x
1
/
2
d
x
−
∫
x
−
1
/
2
d
x
=
{\displaystyle \int {\frac {x^{2}+5x-1}{\sqrt {x}}}dx=\int (x^{3/2}+5x^{1/2}-x^{-1/2})dx=\int x^{3/2}dx+5\int x^{1/2}dx-\int x^{-1/2}dx=}
=
x
3
/
2
+
1
3
/
2
+
1
+
C
1
+
5
x
1
/
2
+
1
1
/
2
+
1
+
C
2
−
x
−
1
/
2
+
1
−
1
/
2
+
1
+
C
3
=
2
5
x
5
/
2
+
10
3
x
3
/
2
−
2
x
1
/
2
+
C
=
2
x
(
x
2
5
+
5
3
x
−
1
)
{\displaystyle ={\frac {x^{3/2+1}}{3/2+1}}+C_{1}+5{\frac {x^{1/2+1}}{1/2+1}}+C_{2}-{\frac {x^{-1/2+1}}{-1/2+1}}+C_{3}={\frac {2}{5}}x^{5/2}+{\frac {10}{3}}x^{3/2}-2x^{1/2}+C=2{\sqrt {x}}({\frac {x^{2}}{5}}+{\frac {5}{3}}x-1)}
∫
d
x
sin
2
x
⋅
cos
2
x
=
∫
sin
2
x
+
cos
2
x
sin
2
x
⋅
cos
2
x
d
x
=
∫
d
x
cos
2
x
+
∫
d
x
sin
2
x
=
tan
x
−
cot
x
+
C
.
{\displaystyle \int {\frac {dx}{\sin ^{2}x\cdot \cos ^{2}x}}=\int {\frac {\sin ^{2}x+\cos ^{2}x}{\sin ^{2}x\cdot \cos ^{2}x}}dx=\int {\frac {dx}{\cos ^{2}x}}+\int {\frac {dx}{\sin ^{2}x}}=\tan x-\cot x+C.}
∫
tan
2
x
d
x
=
∫
sin
2
x
cos
2
x
d
x
=
∫
1
−
cos
2
x
cos
2
x
d
x
=
∫
d
x
cos
2
x
−
∫
d
x
=
tan
x
−
x
+
C
.
{\displaystyle \int \tan ^{2}xdx=\int {\frac {\sin ^{2}x}{\cos ^{2}x}}dx=\int {\frac {1-\cos ^{2}x}{\cos ^{2}x}}dx=\int {\frac {dx}{\cos ^{2}x}}-\int dx=\tan x-x+C.}
∫
sin
2
x
2
d
x
=
∫
1
−
cos
x
2
d
x
=
1
2
∫
d
x
−
1
2
∫
cos
x
d
x
=
x
2
−
sin
x
2
+
C
.
{\displaystyle \int \sin ^{2}{\frac {x}{2}}dx=\int {\frac {1-\cos x}{2}}dx={\frac {1}{2}}\int dx-{\frac {1}{2}}\int \cos xdx={\frac {x}{2}}-{\frac {\sin x}{2}}+C.}
∫
cos
(
2
x
)
d
x
sin
2
x
⋅
cos
2
x
=
∫
cos
2
x
−
sin
2
x
sin
2
x
⋅
cos
2
x
d
x
=
∫
d
x
sin
2
x
−
∫
d
x
cos
2
x
=
−
cot
x
−
tan
x
+
C
.
{\displaystyle \int {\frac {\cos(2x)dx}{\sin ^{2}x\cdot \cos ^{2}x}}=\int {\frac {\cos ^{2}x-\sin ^{2}x}{\sin ^{2}x\cdot \cos ^{2}x}}dx=\int {\frac {dx}{\sin ^{2}x}}-\int {\frac {dx}{\cos ^{2}x}}=-\cot x-\tan x+C.}
∫
d
x
x
2
(
4
+
x
2
)
=
1
4
∫
4
d
x
x
2
(
4
+
x
2
)
=
1
4
∫
4
+
x
2
−
x
2
x
2
(
4
+
x
2
)
d
x
=
1
4
∫
d
x
x
2
−
1
4
∫
d
x
4
+
x
2
=
{\displaystyle \int {dx \over x^{2}(4+x^{2})}={1 \over 4}\int {4\;dx \over x^{2}(4+x^{2})}={1 \over 4}\int {4+x^{2}-x^{2} \over x^{2}(4+x^{2})}dx={\frac {1}{4}}\int {dx \over x^{2}}-{\frac {1}{4}}\int {dx \over 4+x^{2}}=}
=
−
1
4
⋅
1
x
−
1
8
arctan
x
2
+
C
.
{\displaystyle =-{\frac {1}{4}}\cdot {1 \over x}-{1 \over 8}\arctan {x \over 2}+C.}
[ keisti ] Panaudojant integravimo dalimis metodą, įrodyta, kad
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
(
n
−
1
)
!
!
n
!
!
⋅
π
2
,
{\displaystyle \int _{0}^{\pi \over 2}\sin ^{n}x\;dx=\int _{0}^{\pi \over 2}\cos ^{n}x\;dx={(n-1)!! \over n!!}\cdot {\pi \over 2},}
n lyginis;
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
(
n
−
1
)
!
!
n
!
!
,
{\displaystyle \int _{0}^{\pi \over 2}\sin ^{n}x\;dx=\int _{0}^{\pi \over 2}\cos ^{n}x\;dx={(n-1)!! \over n!!},}
n nelyginis.Du šauktukai (n!!) yra dvigubas faktorialas. Šiuo simboliu pažymėsime vien tik lyginių skaičių iki n sandaugą, jei n - lyginis, ir vien tik nelyginių skaičių sandaugą, jei n nelyginis. Pavyzdžiui:
5
!
!
=
1
⋅
3
⋅
5
=
15
,
6
!
!
=
2
⋅
4
⋅
6
=
48.
{\displaystyle 5!!=1\cdot 3\cdot 5=15,\;6!!=2\cdot 4\cdot 6=48.}
Pavyzdžiai
∫
0
π
sin
8
x
2
d
x
=
2
∫
0
π
2
sin
8
t
d
t
=
2
⋅
7
!
!
8
!
!
⋅
π
2
=
2
⋅
7
⋅
5
⋅
3
8
⋅
6
⋅
4
⋅
2
⋅
π
2
=
35
π
128
,
{\displaystyle \int _{0}^{\pi }\sin ^{8}{x \over 2}dx=2\int _{0}^{\pi \over 2}\sin ^{8}t\;dt=2\cdot {7!! \over 8!!}\cdot {\pi \over 2}=2\cdot {7\cdot 5\cdot 3 \over 8\cdot 6\cdot 4\cdot 2}\cdot {\pi \over 2}={35\pi \over 128},}
x
2
=
t
;
d
t
=
1
2
d
x
;
{\displaystyle {x \over 2}=t;\;dt={1 \over 2}dx;}
d
x
=
2
d
t
.
{\displaystyle dx=2dt.}
4
∫
0
π
2
(
cos
2
x
−
2
3
cos
4
x
)
d
x
=
4
(
1
!
!
2
!
!
⋅
π
2
−
2
3
⋅
3
!
!
4
!
!
⋅
π
2
)
=
4
(
π
4
−
π
3
⋅
3
4
⋅
2
)
=
4
(
π
4
−
π
8
)
=
4
⋅
π
8
=
π
2
.
{\displaystyle 4\int _{0}^{\pi \over 2}(\cos ^{2}x-{2 \over 3}\cos ^{4}x)dx=4({1!! \over 2!!}\cdot {\pi \over 2}-{2 \over 3}\cdot {3!! \over 4!!}\cdot {\pi \over 2})=4({\pi \over 4}-{\pi \over 3}\cdot {3 \over 4\cdot 2})=4({\pi \over 4}-{\pi \over 8})=4\cdot {\pi \over 8}={\pi \over 2}.}
∫
0
π
2
sin
3
x
d
x
=
(
3
−
1
)
!
!
3
!
!
=
2
!
!
3
!
!
=
2
3
.
{\displaystyle \int _{0}^{\pi \over 2}\sin ^{3}x\;dx={(3-1)!! \over 3!!}={2!! \over 3!!}={2 \over 3}.}
∫
0
π
sin
4
x
d
x
=
2
∫
0
π
2
sin
4
x
d
x
=
2
⋅
3
!
!
4
!
!
⋅
π
2
=
3
4
⋅
2
π
=
3
8
π
.
{\displaystyle \int _{0}^{\pi }\sin ^{4}x\;dx=2\int _{0}^{\pi \over 2}\sin ^{4}x\;dx=2\cdot {3!! \over 4!!}\cdot {\pi \over 2}={3 \over 4\cdot 2}\pi ={3 \over 8}\pi .}
∫
−
π
2
π
2
cos
4
x
d
x
=
2
∫
0
π
2
cos
4
x
d
x
=
2
⋅
3
!
!
4
!
!
⋅
π
2
=
3
8
π
.
{\displaystyle \int _{-{\pi \over 2}}^{\pi \over 2}\cos ^{4}xdx=2\int _{0}^{\pi \over 2}\cos ^{4}xdx=2\cdot {3!! \over 4!!}\cdot {\pi \over 2}={3 \over 8}\pi .}
