С Википедије, слободне енциклопедије
Tehnika u integralnom vrednovanju
Metoda supstitucije je metoda rešavanja integrala u kojoj se deo integrala zamenjuje jednostavnijim simbolom (obično se koristi latinično slovo u), u cilju da bi se dobio integral koji je lakše rešiti.[1]
∫
ln
x
x
d
x
=
?
{\displaystyle \int {\frac {\ln {x}}{x}}dx=?}
u
=
l
n
(
x
)
{\displaystyle u=ln(x)}
d
u
=
1
x
d
x
{\displaystyle du={\frac {1}{x}}dx}
x
d
u
=
d
x
{\displaystyle xdu=dx}
∫
l
n
(
x
)
x
d
x
=
∫
u
x
x
d
u
=
u
2
2
+
C
{\displaystyle \int {\frac {ln(x)}{x}}dx=\int {\frac {u}{x}}xdu={\frac {u^{2}}{2}}+C}
∫
l
n
(
x
)
x
d
x
=
(
ln
(
x
)
)
2
2
+
C
{\displaystyle \int {\frac {ln(x)}{x}}dx={\frac {(\ln(x))^{2}}{2}}+C}
U nekim slučajevima moraće se rešiti za
x
{\displaystyle x}
∫
2
x
5
x
−
20
d
x
=
?
{\displaystyle \int 2x{\sqrt {5x-20}}dx=?}
ω
=
5
x
−
20
{\displaystyle \omega =5x-20}
d
ω
=
5
d
x
{\displaystyle d\omega =5dx}
d
ω
5
=
d
x
{\displaystyle {\frac {d\omega }{5}}=dx}
x
=
ω
+
20
5
{\displaystyle x={\frac {\omega +20}{5}}}
∫
2
x
5
x
−
20
d
x
=
∫
2
(
ω
+
20
5
)
ω
d
ω
5
=
2
25
∫
(
ω
+
20
)
ω
d
ω
=
2
25
∫
(
ω
ω
+
20
ω
)
d
ω
=
2
25
(
∫
ω
ω
d
ω
+
∫
20
ω
d
ω
)
{\displaystyle \int 2x{\sqrt {5x-20}}dx=\int 2\left({\frac {\omega +20}{5}}\right){\sqrt {\omega }}{\frac {d\omega }{5}}={\frac {2}{25}}\int (\omega +20){\sqrt {\omega }}d\omega ={\frac {2}{25}}\int (\omega {\sqrt {\omega }}+20{\sqrt {\omega }})d\omega ={\frac {2}{25}}\left(\int \omega {\sqrt {\omega }}d\omega +\int 20{\sqrt {\omega }}d\omega \right)}
2
25
(
∫
ω
ω
d
ω
+
∫
20
ω
d
ω
)
=
2
25
(
ω
5
2
5
2
+
20
ω
3
2
3
2
)
+
C
{\displaystyle {\frac {2}{25}}\left(\int \omega {\sqrt {\omega }}d\omega +\int 20{\sqrt {\omega }}d\omega \right)={\frac {2}{25}}\left({\frac {\omega ^{\frac {5}{2}}}{\frac {5}{2}}}+20{\frac {\omega ^{\frac {3}{2}}}{\frac {3}{2}}}\right)+C}
∫
2
x
5
x
−
20
d
x
=
2
25
(
(
5
x
−
20
)
5
2
5
2
+
20
(
5
x
−
20
)
3
2
3
2
)
+
C
{\displaystyle \int 2x{\sqrt {5x-20}}dx={\frac {2}{25}}\left({\frac {(5x-20)^{\frac {5}{2}}}{\frac {5}{2}}}+20{\frac {(5x-20)^{\frac {3}{2}}}{\frac {3}{2}}}\right)+C}
Metoda supstitucije se može koristiti za definisanje novih antiderivata.
∫
tan
x
d
x
=
?
{\displaystyle \int \tan {x}dx=?}
∫
tan
x
d
x
=
∫
sin
x
cos
x
d
x
{\displaystyle \int \tan {x}dx=\int {\frac {\sin {x}}{\cos {x}}}dx}
u
=
cos
x
{\displaystyle u=\cos {x}}
d
u
=
−
sin
x
d
x
{\displaystyle du=-\sin {x}dx}
d
u
−
s
i
n
x
=
d
x
{\displaystyle {\frac {du}{-sin{x}}}=dx}
∫
sin
x
cos
x
d
x
=
∫
s
i
n
x
u
d
u
−
sin
x
=
−
∫
1
u
d
u
=
−
ln
u
+
C
{\displaystyle \int {\frac {\sin {x}}{\cos {x}}}dx=\int {\frac {sin{x}}{u}}{\frac {du}{-\sin {x}}}=-\int {\frac {1}{u}}du=-\ln {u}+C}
∫
tan
x
d
x
=
ln
c
o
s
x
+
C
{\displaystyle \int \tan {x}dx=\ln {cos{x}}+C}
Briggs, William; Cochran, Lyle (2011), Calculus /Early Transcendentals (Single Variable изд.), Addison-Wesley, ISBN 978-0-321-66414-3 Ferzola, Anthony P. (1994), „Euler and differentials” , The College Mathematics Journal 25 (2): 102–111, JSTOR 2687130 , doi :10.2307/2687130 Fremlin, D.H. (2010), Measure Theory, Volume 2 , Torres Fremlin, ISBN 978-0-9538129-7-4 Hewitt, Edwin ; Stromberg, Karl (1965), Real and Abstract Analysis , Springer-Verlag, ISBN 978-0-387-04559-7 Katz, V. (1982), „Change of variables in multiple integrals: Euler to Cartan”, Mathematics Magazine 55 (1): 3–11, JSTOR 2689856 , doi :10.2307/2689856 Rudin, Walter (1987), Real and Complex Analysis , McGraw-Hill, ISBN 978-0-07-054234-1 Swokowski, Earl W. (1983), Calculus with analytic geometry (alternate изд.), Prindle, Weber & Schmidt, ISBN 0-87150-341-7 Spivak, Michael (1965), Calculus on Manifolds , Westview Press, ISBN 978-0-8053-9021-6
