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

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

Pavyzdžiai

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

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

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

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

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

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

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

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

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

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

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

• $\int x{\sqrt {9-x^{2}}}dx=\int 3\sin t{\sqrt {9-9\sin ^{2}t}}\cdot 3\cos t\;dt=9\int \sin t{\frac {3{\sqrt {9-9\sin ^{2}t}}}{3}}\cdot \cos t\;dt=$

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

• $\int {\frac {dx}{x{\sqrt {a^{2}+x^{2}}}}}=\int {\frac {a\;dt}{a\cos ^{2}t\cdot \tan t{\sqrt {a^{2}\tan ^{2}t}}}}={\frac {1}{a}}\int {\frac {dt}{\sin t}}={\frac {1}{a}}\ln |{\frac {1-\cos t}{\sin t}}|+C=$

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

Integravimas tam tikrų klasių trigonometrinių funkcijų

$\int R(\sin x,\cos x){\mathsf {d}}x=\int R\left({\frac {2t}{1+t^{2}}},\;{\frac {1-t^{2}}{1+t^{2}}}\right){\frac {2{\mathsf {d}}t}{1+t^{2}}}.$

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

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

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

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

Toliau

${\frac {x}{2}}=\arctan t,\;\;x=2\arctan t,\quad {\mathsf {d}}x={\frac {2{\mathsf {d}}t}{1+t^{2}}}.$

Pavyzdžiai

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

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

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

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

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

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

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

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 $\tan x$:

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

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

${\mathsf {d}}x={\mathsf {d}}(\arctan t)={\frac {{\mathsf {d}}t}{1+t^{2}}}.$

Pavyzdžiai

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

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

Patikriname

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

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

${\frac {1}{\sqrt {2}}}{\frac {\mathsf {d}}{{\mathsf {d}}u}}\left(\arctan u\right){\frac {\mathsf {d}}{{\mathsf {d}}x}}\left(\tan x\right)={\frac {1}{\sqrt {2}}}{\frac {1}{1+u^{2}}}{\frac {1}{{\sqrt {2}}\cos ^{2}x}}={\frac {1}{\sqrt {2}}}{\frac {1}{1+({\frac {\tan x}{\sqrt {2}}})^{2}}}{\frac {1}{{\sqrt {2}}\cos ^{2}x}}={\frac {1}{\sqrt {2}}}{\frac {1}{1+{\frac {\tan ^{2}x}{2}}}}{\frac {1}{{\sqrt {2}}\cos ^{2}x}}=$

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

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

Taip pat skaitykite

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