Matematika/Trigonometrinių funkcijų integravimas

I. Integralai $\int {\sin }^{m}x{\cos }^{n}xdx,$ kur m, n - sveikieji skaičiai, suvedami į integralą su binominiu diferencialu ir integruojami tik 3 atvejais:

1)n nelyginis;

2)m nelyginis;

3)m+n lyginis.

Jei n nelyginis, taikome keitinį $\sin x=t,$ jei m nelyginis, taikome keitinį $\cos x=t;$ jei $m+n$ lyginis, keičiame $\tan x=t;$ $\sin x=\frac{t}{\sqrt{1+t^2}};\cos x=\frac{1}{\sqrt{1+t^2}};dx=\frac{dt}{1+t^2}.$

Pavyzdžiai

• $\int \frac{dx}{5{\cos }^{2}x+9{\sin }^{2}x}.\tan x=t;\sin x=\frac{t}{\sqrt{1+t^2}};\cos x=\frac{1}{\sqrt{1+t^2}};dx=\frac{dt}{1+t^2}.$

$\int \frac{\frac{dt}{1+t^2}}{5(\frac{1}{\sqrt{1+t^2}}{)}^2+9(\frac{t}{\sqrt{1+t^2}}{)}^2}=\int \frac{dt}{5+9t^2}=\frac{1}{3\sqrt{5}}\arctan \frac{3t}{\sqrt{5}}+C.$

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

$=-\cot x-\frac{3x}{2}-\frac{\sin (2x)}{4}+C.$

• $\int \frac{{\sin }^{2}x}{{\cos }^{6}x}dx.$ Skaičiai m ir n lyginiai, $m=2,$ $n=-6,$ $m+n=-4$ lyginis, todėl taikome keitnį $\tan x=t;$ ${\sin }^{2}x=\frac{t^2}{1+t^2};$ $\cos x=\frac{1}{\sqrt{1+t^2}};$ ${\cos }^{6}x=\frac{1}{(1+t^2{)}^3};$ $dx=\frac{dt}{1+t^2}.$

$\int \frac{{\sin }^{2}x}{{\cos }^{6}x}dx=\int \frac{\frac{t^2}{1+t^2}\cdot \frac{dt}{1+t^2}}{\frac{1}{(1+t^2{)}^3}}=\int \frac{\frac{t^{2}dt}{(1+t^2{)}^2}}{\frac{1}{(1+t^2{)}^3}}=\int \frac{t^2}{(1+t^2{)}^2}\cdot (1+t^2{)}^{3}dt=\int t^2(1+t^2)dt=\frac{t^3}{3}+\frac{t^5}{5}+C=\frac{{\tan }^{3}x}{3}+\frac{{\tan }^{5}x}{5}+C.$

• Apskaičiuosime integralą $\int \frac{\sin x\cos xdx}{{\sin }^{4}x+{\cos }^{4}x}.$

Kadangi pointegralinė funkcija nekeičia reikšmės, kai kartu keičiami $\sin x$ ir $\cos x$ ženklai, tai pagal tam tikras taisykles, pakeitę $t=\tan x,$ gauname

$\int \frac{\sin x\cos xdx}{{\sin }^{4}x+{\cos }^{4}x}=\int \frac{\frac{t}{\sqrt{1+t^2}}\frac{1}{\sqrt{1+t^2}}\frac{dt}{1+t^2}}{\frac{t^4}{(1+t^2{)}^2}+\frac{1}{(1+t^2{)}^2}}=\int \frac{\frac{t}{(1+t^2{)}^2}dt}{\frac{t^4+1}{(1+t^2{)}^2}}=\int \frac{tdt}{t^4+1}=\frac{1}{2}\int \frac{d(t^2)}{(t^2{)}^2+1}=\frac{1}{2}\arctan (t^2)+C=\frac{1}{2}\arctan ({\tan }^{2}x)+C,$

kur ${\sin }^{4}x=\frac{t^4}{(1+t^2{)}^2},{\cos }^{4}x=\frac{1}{(1+t^2{)}^2},dx=\frac{dt}{1+t^2}.$

II.Integralai $\int \sin x\cos xdx$ (be laipnsių) suvedami į racionaliųjų funkcijų integralus keitiniu $\tan \frac{x}{2}=t.$ Tada $\sin x=\frac{2t}{1+t^2};\cos x=\frac{1-t^2}{1+t^2};dx=\frac{2dt}{1+t^2}.$

Pavyzdžiai

• $\int \frac{dx}{3+\sin x+\cos x}.$ $\tan \frac{x}{2}=t;$ $\frac{x}{2}=\arctan t;$ $x=2\arctan t;$ $\sin x=\frac{2t}{1+t^2};$ $\cos x=\frac{1-t^2}{1+t^2};$ $dx=d(2\arctan t)=\frac{2dt}{1+t^2}.$

$\int \frac{dx}{3+\sin x+\cos x}=\int \frac{\frac{2dt}{1+t^2}}{3+\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}=\int \frac{\frac{2dt}{1+t^2}}{\frac{2(t^2+t+2)}{1+t^2}}=\int \frac{dt}{t^2+t+2}=\int \frac{dt}{(t+\frac{1}{2}{)}^2+\frac{7}{4}}=$ $=\frac{2}{\sqrt{7}}\arctan \frac{2t+1}{\sqrt{7}}+C=\frac{2}{\sqrt{7}}\arctan \frac{2\tan \frac{x}{2}+1}{\sqrt{7}}+C.$

• $\int \frac{\cot xdx}{1-\sin x-\cos x}=\int \frac{\cos xdx}{\sin x(1-\sin x-\cos x)}=\int \frac{\frac{1-t^2}{1+t^2}\cdot \frac{2dt}{1+t^2}}{\frac{2t}{1+t^2}(1-\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2})}=$

$=\int \frac{\frac{2(1-t^2)dt}{(1+t^2{)}^2}}{\frac{2t+2t^3-4t^2-2t+2t^3}{(1+t^2{)}^2}}=\int \frac{2(1-t)(1+t)dt}{4t^3-4t^2}=\int \frac{2(1-t)(1+t)dt}{-4t^2(1-t)}=-\frac{1}{2}\int (\frac{1}{t^2}+\frac{1}{t})dt=\frac{1}{2}(\cot \frac{x}{2}-\ln |\tan \frac{x}{2}|)+C,$ kur $\tan \frac{x}{2}=t;\sin x=\frac{2t}{1+t^2};\cos x=\frac{1-t^2}{1+t^2};dx=\frac{2dt}{1+t^2}.$

III. Integralams $\int R(x,\sqrt{a^2-x^2})dx=\int x^{\pm 1}(\sqrt{a^2-x^2}{)}^{\pm 1}dx$ taikomi ketiniai $x=a\sin t,$ $x=a\tan t$ arba $x=\frac{a}{\sin t}.$

Pavyzdžiai

• $\int x\sqrt{9-x^2}dx=\int 3\sin t\sqrt{9-9{\sin }^{2}t}\cdot 3\cos tdt=27\int \sin t\sqrt{1-{\sin }^{2}t}\cdot \cos tdt=$

$=27\int \sin t\cos t\cos tdt=-27\int {\cos }^{2}td(\cos t)=-27\frac{{\cos }^{3}t}{3}+C=-9(1-{\sin }^{2}t)\sqrt{1-{\sin }^{2}t}+C=$ $=-9(1-\frac{x^2}{3^2})\sqrt{1-\frac{x^2}{3^2}}+C=-\frac{1}{3}(9-x^2)\sqrt{9-x^2}+C,$ kur $3\sin t=x;$ $\sin t=\frac{x}{3};$ $dx=3\cos tdt.$

• $\int \frac{dx}{x\sqrt{a^2+x^2}}=\int \frac{adt}{a{\cos }^{2}t\cdot \tan t\sqrt{a^2+a^2{\tan }^{2}t}}=\int \frac{adt}{a^2{\cos }^{2}t\cdot \tan t\sqrt{1+{\tan }^{2}t}}=\int \frac{\cos tdt}{a{\cos }^{2}t\cdot \tan t}=\frac{1}{a}\int \frac{dt}{\sin t}=$

$=\frac{1}{a}\ln |\frac{1-\cos t}{\sin t}|+C=\frac{1}{a}\ln |\frac{1}{\sin t}-\cot t|+C=\frac{1}{a}\ln |\frac{\sqrt{a^2+x^2}-a}{x}|+C,$

kur $x=a\tan t;$ $dx=\frac{a}{{\cos }^{2}t};$ $\tan t=\frac{x}{a};$ $\cot t=\frac{a}{x};$ $\frac{1}{\sin t}=\sqrt{1+{\cot }^{2}t}=\frac{\sqrt{a^2+x^2}}{x};$ $1+{\tan }^{2}t=\frac{1}{{\cos }^{2}t}.$

Integravimas tam tikrų klasių trigonometrinių funkcijų

$\int R(\sin x,\cos x)𝖽x=\int R(\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2})\frac{2𝖽t}{1+t^2}.$

Imamas keitinys $\tan \frac{x}{2}=t,-\pi <x<\pi .$

Funkcijos $\sin (x)$ ir $\cos (x)$ išreiškiamos per tangentą $\tan \frac{x}{2}$ ir taip išreiškiamos per t.

$\sin x=\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{1}=\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}}=\frac{2(\sin \frac{x}{2}\cos \frac{x}{2})/{\cos }^2\frac{x}{2}}{({\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2})/{\cos }^2\frac{x}{2}}=\frac{2\tan \frac{x}{2}}{{\tan }^2\frac{x}{2}+1}=\frac{2t}{1+t^2},$

$\cos x=\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{1}=\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}}=\frac{({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})/{\cos }^2\frac{x}{2}}{({\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2})/{\cos }^2\frac{x}{2}}=\frac{1-{\tan }^2\frac{x}{2}}{{\tan }^2\frac{x}{2}+1}=\frac{1-t^2}{1+t^2}.$

Toliau

$\frac{x}{2}=\arctan t,x=2\arctan t,𝖽x=\frac{2𝖽t}{1+t^2}.$

Pavyzdžiai

• $\int \frac{dx}{\sin x}=\int \frac{\frac{2𝖽t}{1+t^2}}{\frac{2t}{1+t^2}}=\frac{dt}{t}=\ln |t|+C=\ln |\tan \frac{x}{2}|+C.$

• $\int \frac{𝖽x}{1+\sin x}=\int \frac{\frac{2𝖽t}{1+t^2}}{1+\frac{2t}{1+t^2}}=\int \frac{2𝖽t}{(1+t^2)\frac{1+t^2+2t}{1+t^2}}=\int \frac{2𝖽t}{t^2+2t+1}=\int \frac{2𝖽t}{(t+1{)}^2}=$

$=\int \frac{2𝖽(t+1)}{(t+1{)}^2}=-\frac{2}{t+1}+C=-\frac{2}{\tan \frac{x}{2}+1}+C_1.$

• $\int \frac{𝖽x}{9+8\cos x+\sin x}=\int \frac{2𝖽t}{(1+t^2)(9+8\cdot \frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2})}=\int \frac{2𝖽t}{(1+t^2)\cdot \frac{9(1+t^2)+8(1-t^2)+2t}{1+t^2}}=$

$=\int \frac{2𝖽t}{9(1+t^2)+8(1-t^2)+2t}=2\int \frac{𝖽t}{9+9t^2+8-8t^2+2t}=2\int \frac{𝖽t}{t^2+2t+17}=2\int \frac{𝖽t}{(t+1{)}^2+16}=$

$=2\int \frac{𝖽t}{16(\frac{(t+1{)}^2}{16}+1)}=2\int \frac{4𝖽(\frac{t+1}{4})}{16((\frac{t+1}{4}{)}^2+1)}=\frac{1}{2}\int \frac{𝖽(\frac{t+1}{4})}{(\frac{t+1}{4}{)}^2+1}=\frac{1}{2}\arctan \frac{t+1}{4}+C=\frac{1}{2}\arctan \frac{\tan \frac{x}{2}+1}{4}+C,$

kur $𝖽\left(\frac{t+1}{4}\right)=\frac{𝖽t}{4};𝖽t=4𝖽\left(\frac{t+1}{4}\right).$

• $\int \frac{1+\sin x}{\sin x(1+\cos x)}dx=\int \frac{(1+\frac{2t}{1+t^2})\frac{2𝖽t}{1+t^2}}{\frac{2t}{1+t^2}(1+\frac{1-t^2}{1+t^2})}=\int \frac{\frac{1+t^2+2t}{1+t^2}\frac{2𝖽t}{1+t^2}}{\frac{2t}{1+t^2}(\frac{1+t^2}{1+t^2}+\frac{1-t^2}{1+t^2})}=\int \frac{2\cdot \frac{t^2+2t+1}{(1+t^2{)}^2}𝖽t}{\frac{2t}{1+t^2}\cdot \frac{2}{1+t^2}}=\int \frac{2(t^2+2t+1)𝖽t}{4t}=\frac{1}{2}\int (t+2+\frac{1}{t})𝖽t=.$

$=\frac{1}{2}(\frac{t^2}{2}+2t+\ln t)+C=\frac{t^2}{4}+t+\frac{1}{2}\ln t+C=\frac{{\tan }^2\frac{x}{2}}{4}+\tan \frac{x}{2}+\frac{1}{2}\ln \tan \frac{x}{2}+C.$

Integravimas tam tikrų klasių trigonometrinių funkcijų 4)

Jeigu pointegralinė funkcija turi pavidalą $R(\sin (x),\cos (x))$, bet $\sin x$ ir $\cos x$ turi tik lyginius laipsnius, tai daromas keitinys:

$\tan x=t,$

$x=\arctan t,$

nes ${\sin }^{2}x$ ir ${\cos }^{2}x$ išsireiškia racionaliai per $\tan x$:

${\cos }^{2}x=\frac{1}{2}(1+\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x})=\frac{1}{2}\cdot \frac{1+{\tan }^{2}x+1-{\tan }^{2}x}{1+{\tan }^{2}x}=\frac{1}{2}\cdot \frac{2}{1+{\tan }^{2}x}=\frac{1}{1+{\tan }^{2}x}=\frac{1}{1+t^2},$ arba

${\cos }^{2}x=\frac{{\cos }^{2}x}{{\cos }^{2}x+{\sin }^{2}x}=\frac{1}{\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}}=\frac{1}{1+{\tan }^{2}x}=\frac{1}{1+t^2};$

${\sin }^{2}x=\frac{1}{2}(1-\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x})=\frac{1}{2}\cdot \frac{1+{\tan }^{2}x-(1-{\tan }^{2}x)}{1+{\tan }^{2}x}=\frac{1}{2}\cdot \frac{2{\tan }^{2}x}{1+{\tan }^{2}x}=\frac{{\tan }^{2}x}{1+{\tan }^{2}x}=\frac{t^2}{1+t^2},$ arba

${\sin }^{2}x=\frac{{\sin }^{2}x}{{\cos }^{2}x+{\sin }^{2}x}=\frac{\frac{{\sin }^{2}x}{{\cos }^{2}x}}{\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}}=\frac{{\tan }^{2}x}{1+{\tan }^{2}x}=\frac{t^2}{1+t^2};$

$𝖽x=𝖽(\arctan t)=\frac{𝖽t}{1+t^2}.$

Pavyzdžiai

• $\int \frac{𝖽x}{2-{\sin }^{2}x}=\int \frac{𝖽t}{(2-\frac{t^2}{1+t^2})(1+t^2)}=\int \frac{𝖽t}{\left(\frac{2(1+t^2)-t^2}{1+t^2}\right)(1+t^2)}=\int \frac{𝖽t}{\left(\frac{2+2t^2-t^2}{1+t^2}\right)(1+t^2)}=$

$=\int \frac{𝖽t}{2+t^2}=\int \frac{1}{\sqrt{2}}\arctan \frac{t}{\sqrt{2}}+C=\frac{1}{\sqrt{2}}\arctan \left(\frac{\tan x}{\sqrt{2}}\right)+C.$

Patikriname

$\frac{𝖽}{𝖽x}(\frac{1}{\sqrt{2}}\arctan \left(\frac{\tan x}{\sqrt{2}}\right)+C)=\frac{1}{\sqrt{2}}\frac{𝖽}{𝖽x}(\arctan \left(\frac{\tan x}{\sqrt{2}}\right)).$

Turime, kad $\frac{𝖽}{𝖽u}\arctan u=\frac{1}{1+u^2},$ kur $u=\frac{\tan x}{\sqrt{2}}$ ir $\frac{𝖽}{𝖽x}(\frac{\tan x}{\sqrt{2}})=\frac{1}{\sqrt{2}{\cos }^{2}x}.$ Todėl

$\frac{1}{\sqrt{2}}\frac{𝖽}{𝖽u}(\arctan u)\frac{𝖽}{𝖽x}(\tan x)=\frac{1}{\sqrt{2}}\frac{1}{1+u^2}\frac{1}{\sqrt{2}{\cos }^{2}x}=\frac{1}{\sqrt{2}}\frac{1}{1+(\frac{\tan x}{\sqrt{2}}{)}^2}\frac{1}{\sqrt{2}{\cos }^{2}x}=\frac{1}{\sqrt{2}}\frac{1}{1+\frac{{\tan }^{2}x}{2}}\frac{1}{\sqrt{2}{\cos }^{2}x}=$

$=\frac{1}{2}\frac{2}{2+{\tan }^{2}x}\frac{1}{{\cos }^{2}x}=\frac{1}{2}\frac{2}{2+\frac{{\sin }^{2}x}{{\cos }^{2}x}}\frac{1}{{\cos }^{2}x}=\frac{1}{2}\frac{2}{\frac{2{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}}\frac{1}{{\cos }^{2}x}=\frac{1}{2}\frac{2{\cos }^{2}x}{2{\cos }^{2}x+{\sin }^{2}x}\frac{1}{{\cos }^{2}x}=\frac{1}{2}\frac{2}{2{\cos }^{2}x+{\sin }^{2}x}=$

$=\frac{1}{2(1-{\sin }^{2}x)+{\sin }^{2}x}=\frac{1}{2-2{\sin }^{2}x+{\sin }^{2}x}=\frac{1}{2-{\sin }^{2}x}.$

Taip pat skaitykite

• Integravimas keičiant kintamąjį
• Integravimas dalimis
• Racionaliųjų funkcijų integravimas
• Integralų lentelė
• Integravimo metodai