Integravimo metodai

Čia pateikiami metodai, padedantys integruoti.

Tiesioginis integravimas

Jei

$\int f(x)𝖽x=F(x)+C,$

tai

$\int f(u)𝖽u=F(u)+C.$

Šis metodas pagrįstas pirmos eilės diferencialo formos invariantiškumu.

Pavyzdžiai,

• kadangi:

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

tai:

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

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

$=\frac{x^{3/2+1}}{3/2+1}+C_1+5\frac{x^{1/2+1}}{1/2+1}+C_2-\frac{x^{-1/2+1}}{-1/2+1}+C_3=\frac{2}{5}{x}^{5/2}+\frac{10}{3}{x}^{3/2}-2x^{1/2}+C=2\sqrt{x}(\frac{x^2}{5}+\frac{5}{3}x-1)$

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

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

Trigonometrinių funkcijų integravimas taikant dvigubą faktorialą

Panaudojant integravimo dalimis metodą, įrodyta, kad

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n}xdx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{n}xdx=\frac{(n-1)!!}{n!!}\cdot \frac{\pi }{2},$ kai n lyginis;

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n}xdx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{n}xdx=\frac{(n-1)!!}{n!!},$ 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

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

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

• $\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{3}xdx=\frac{(3-1)!!}{3!!}=\frac{2!!}{3!!}=\frac{2}{3}.$

• $\underset{0}{\overset{\pi }{\int }}{\sin }^{4}xdx=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{4}xdx=2\cdot \frac{3!!}{4!!}\cdot \frac{\pi }{2}=\frac{3}{4\cdot 2}\pi =\frac{3}{8}\pi .$

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

Nuorodos

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