Trigonometrinių funkcijų integravimas

Straipsnis turėtų prasidėti aiškiu apibrėžimu. Jei galite, apibrėžkite straipsnio dalyką, pagrindinę sąvoką.

Trigonometrinių funkcijų integravimas – integravimo technika, kai trigonometrinės funkcijos yra pakeičiamos kitomis išraiškomis. Trys įprastai naudojamos išraiškos yra apribotas sinusas, apribotas tangentas ir apribotas sekantas.

Pavyzdžiai

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

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

Pavyzdžiai

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

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

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

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

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

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

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

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

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

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

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

Pavyzdžiai

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

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

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

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

Šaltiniai

1. „Trigonometric Substitutions“. sfu.ca. Nuoroda tikrinta 2024-02-03.