Integravimo metodai

Čia pateikiami metodai, padedantys integruoti.

Tiesioginis integravimas

Jei

$\int f(x){\mathsf {d}}x=F(x)+C,\quad$

tai

$\int f(u){\mathsf {d}}u=F(u)+C.\quad$

Šis metodas pagrįstas pirmos eilės diferencialo formos invariantiškumu.

Pavyzdžiai,

• kadangi:

$\int t^{3}{\mathsf {d}}t={\frac {t^{4}}{4}}+C$ ir ${\mathsf {d}}(x+10)={\mathsf {d}}x$,

tai:

$\int (x+10)^{3}{\mathsf {d}}x=\int (x+10)^{3}{\mathsf {d}}(x+10)={\frac {(x+10)^{4}}{4}}+C.$

• $\int x^{4}dx={\frac {x^{4+1}}{4+1}}+C={\frac {x^{5}}{5}}+C.$
• $\int 10^{x}dx={\frac {10^{x}}{\ln 10}}+C.$
• $\int {\frac {x^{2}+5x-1}{\sqrt {x}}}dx=\int (x^{3/2}+5x^{1/2}-x^{-1/2})dx=\int x^{3/2}dx+5\int x^{1/2}dx-\int x^{-1/2}dx=$

$={\frac {x^{3/2+1}}{3/2+1}}+C_{1}+5{\frac {x^{1/2+1}}{1/2+1}}+C_{2}-{\frac {x^{-1/2+1}}{-1/2+1}}+C_{3}={\frac {2}{5}}x^{5/2}+{\frac {10}{3}}x^{3/2}-2x^{1/2}+C=2{\sqrt {x}}({\frac {x^{2}}{5}}+{\frac {5}{3}}x-1)$

• $\int {\frac {dx}{\sin ^{2}x\cdot \cos ^{2}x}}=\int {\frac {\sin ^{2}x+\cos ^{2}x}{\sin ^{2}x\cdot \cos ^{2}x}}dx=\int {\frac {dx}{\cos ^{2}x}}+\int {\frac {dx}{\sin ^{2}x}}=\tan x-\cot x+C.$
• $\int \tan ^{2}xdx=\int {\frac {\sin ^{2}x}{\cos ^{2}x}}dx=\int {\frac {1-\cos ^{2}x}{\cos ^{2}x}}dx=\int {\frac {dx}{\cos ^{2}x}}-\int dx=\tan x-x+C.$
• $\int \sin ^{2}{\frac {x}{2}}dx=\int {\frac {1-\cos x}{2}}dx={\frac {1}{2}}\int dx-{\frac {1}{2}}\int \cos xdx={\frac {x}{2}}-{\frac {\sin x}{2}}+C.$
• $\int {\frac {\cos(2x)dx}{\sin ^{2}x\cdot \cos ^{2}x}}=\int {\frac {\cos ^{2}x-\sin ^{2}x}{\sin ^{2}x\cdot \cos ^{2}x}}dx=\int {\frac {dx}{\sin ^{2}x}}-\int {\frac {dx}{\cos ^{2}x}}=-\cot x-\tan x+C.$
• $\int {dx \over x^{2}(4+x^{2})}={1 \over 4}\int {4\;dx \over x^{2}(4+x^{2})}={1 \over 4}\int {4+x^{2}-x^{2} \over x^{2}(4+x^{2})}dx={\frac {1}{4}}\int {dx \over x^{2}}-{\frac {1}{4}}\int {dx \over 4+x^{2}}=$

$=-{\frac {1}{4}}\cdot {1 \over x}-{1 \over 8}\arctan {x \over 2}+C.$

Trigonometrinių funkcijų integravimas taikant dvigubą faktorialą

Panaudojant integravimo dalimis metodą, įrodyta, kad

$\int _{0}^{\pi \over 2}\sin ^{n}x\;dx=\int _{0}^{\pi \over 2}\cos ^{n}x\;dx={(n-1)!! \over n!!}\cdot {\pi \over 2},$ kai n lyginis;

$\int _{0}^{\pi \over 2}\sin ^{n}x\;dx=\int _{0}^{\pi \over 2}\cos ^{n}x\;dx={(n-1)!! \over n!!},$ kai 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: $5!!=1\cdot 3\cdot 5=15,\;6!!=2\cdot 4\cdot 6=48.$

Pavyzdžiai

• $\int _{0}^{\pi }\sin ^{8}{x \over 2}dx=2\int _{0}^{\pi \over 2}\sin ^{8}t\;dt=2\cdot {7!! \over 8!!}\cdot {\pi \over 2}=2\cdot {7\cdot 5\cdot 3 \over 8\cdot 6\cdot 4\cdot 2}\cdot {\pi \over 2}={35\pi \over 128},$ kur ${x \over 2}=t;\;dt={1 \over 2}dx;$ $dx=2dt.$

• $4\int _{0}^{\pi \over 2}(\cos ^{2}x-{2 \over 3}\cos ^{4}x)dx=4({1!! \over 2!!}\cdot {\pi \over 2}-{2 \over 3}\cdot {3!! \over 4!!}\cdot {\pi \over 2})=4({\pi \over 4}-{\pi \over 3}\cdot {3 \over 4\cdot 2})=4({\pi \over 4}-{\pi \over 8})=4\cdot {\pi \over 8}={\pi \over 2}.$

• $\int _{0}^{\pi \over 2}\sin ^{3}x\;dx={(3-1)!! \over 3!!}={2!! \over 3!!}={2 \over 3}.$

• $\int _{0}^{\pi }\sin ^{4}x\;dx=2\int _{0}^{\pi \over 2}\sin ^{4}x\;dx=2\cdot {3!! \over 4!!}\cdot {\pi \over 2}={3 \over 4\cdot 2}\pi ={3 \over 8}\pi .$

• $\int _{-{\pi \over 2}}^{\pi \over 2}\cos ^{4}xdx=2\int _{0}^{\pi \over 2}\cos ^{4}xdx=2\cdot {3!! \over 4!!}\cdot {\pi \over 2}={3 \over 8}\pi .$

Nuorodos

• https://integralaineringa.weebly.com/uploads/6/8/5/1/68515625/neapibreztinis_integralas.pdf