TARKIBIDA TRIGONOMETRIK FUNKSIYALAR QATNASHGAN IFODALARNI INTEGRALLASH.

Аннотация

In this article, we learn to integrate expressions involving trigonometric functions in a simplified form. Specifically, we explore convenient methods of working with trigonometric functions by introducing appropriate notations. Additionally, we study how to plot graphs of trigonometric expressions using the Maple software. Keywords: rational function, integral, trigonometric function, odd function, even function. Hamma trigonometrik funksiyalarni integrallash mumkin.Bu ifodani 𝑅(𝑠𝑖𝑛𝑥,𝑐𝑜𝑠𝑥)orqali belgilaymiz.Endi 𝑅(𝑠𝑖𝑛𝑥,𝑐𝑜𝑠𝑥) ko`rinishidagi ifodani integrallaymiz.∫𝑅(𝑠𝑖𝑛𝑥,𝑐𝑜𝑠𝑥)𝑑𝑥 𝑥 Bunday integralni 𝑡𝑔 = 𝑡 belgilash yordamida 𝑡 o`zgaruvchili ratsional 2 funksiyaning integraliga almashtirish mumkin.Integralni bunday almashtirish 𝑥 ratsionallashtirish deyiladi. Ko`rinib turibdiki, 𝑡𝑔 = 𝑡 desak, 2 https://scientific-jl.com 1 1 8 64-son_3-to’plam_Iyun-2025 ISSN:3030- 3 6 1 3 T A D Q I Q O T L A R jahon ilmiy – metodik jurnali 2𝑡𝑔 𝑥 2𝑡 1−𝑡𝑔2𝑥 1−𝑡2 𝑥 𝑠𝑖𝑛𝑥 = 2 = ; 𝑐𝑜𝑠𝑥 = 2 = ; = 𝑎𝑟𝑐𝑡𝑔𝑡 ≫ 𝑥 = 1+𝑡𝑔2𝑥 1+𝑡2 1+𝑡𝑔2𝑥 1+𝑡2 2 2 2 2𝑑𝑡 2𝑎𝑟𝑐𝑡𝑔𝑡 ≫ 𝑑𝑥 = 1+𝑡2 2𝑡 1−𝑡2 2𝑑𝑡 Shuning uchun ∫𝑅(𝑠𝑖𝑛𝑥,𝑐𝑜𝑠𝑥)𝑑𝑥=∫𝑅( ; ) = ∫𝑅 (𝑡)𝑑𝑡 1+𝑡2 1+𝑡2 1+𝑡2 1 bunda 𝑅 (𝑡) − 𝑡 o`zgaruvchili ratsional funksiya.Bunday almashtirish 1 𝑅(𝑠𝑖𝑛𝑥,𝑐𝑜𝑠𝑥) ko`rinishidagi har qanday funksiyani integrallashga imkon beradi, shuning uchun bunday almashtirish ko`pincha ancha murakkab ratsional funksiyaga olib keladi. 10.Agar 𝑅(𝑠𝑖𝑛𝑥,𝑐𝑜𝑠𝑥) ifoda ′′𝑠𝑖𝑛𝑥′′ ga nisbatan toq funksiya,ya`ni 𝑅(−𝑠𝑖𝑛𝑥,𝑐𝑜𝑠𝑥) = −𝑅(𝑠𝑖𝑛𝑥,𝑐𝑜𝑠𝑥) bo`lsa, u holda ′′𝑐𝑎𝑠𝑥 = 𝑡′′ 𝑥 ∈ (0;𝜋) almashtirish bajarilsa (1;1) integral ostidagi ifoda 𝑡 ning ratsional funksiyasiga keltiriladi. 𝑐𝑜𝑠𝑥 = 𝑡 𝑥 = 𝑎𝑟𝑐𝑐𝑜𝑠𝑡 1-misol.∫𝑠𝑖𝑛3𝑥 ∙ 𝑐𝑜𝑠3𝑥𝑑𝑥 = | 𝑑𝑡 | 𝑠𝑖𝑛3𝑥 = √(1 − 𝑐𝑜𝑠2𝑥)3 𝑑𝑥 = − √1−𝑡2 √(1−𝑡2)3𝑡2(−1)𝑑𝑡 𝑡5 𝑡3 1 1 = ∫ = ∫(1 − 𝑡2)𝑡2𝑑𝑡 = − = 𝑐𝑜𝑠5𝑥 − 𝑐𝑜𝑠3𝑥 + 𝐶 √1−𝑡2 5 3 5 3 > f:=(((sin(x))^3)*((cos(x))^3)); f := sin(x)3cos(x)3 > Int (f,x); sin(x)3cos(x)3dx > Int(f,x)=int (f,x); > F:plot((((sin(x))^3)*((cos(x))^3)), x=-Pi..Pi, y=-Pi..Pi); 1 1 sin(x)3cos(x)3dx sin(x)2cos(x)4 cos(x)4  6 12 https://scientific-jl.com 1 1 9 64-son_3-to’plam_Iyun-2025 ISSN:3030- 3 6 1 3 T A D Q I Q O T L A R jahon ilmiy – metodik jurnali Demak, bu integralning chizmasi yuqoridagi ko`rinishda ekan. 20. Agar 𝑅(−𝑠𝑖𝑛𝑥,−𝑐𝑜𝑠𝑥) = 𝑅(𝑠𝑖𝑛𝑥,𝑐𝑜𝑠𝑥) bo`lsa, u holda 𝑡 = 𝑡𝑔𝑥 𝜋 𝜋 𝑥 ∈ ( ;− ) almashtirishdan biri bajariladi. 2 2 𝑥 1−𝑡2 2 1−𝑡2 2-misol.∫ 3𝑠𝑖𝑛𝑥−2𝑐𝑜𝑠𝑥 𝑑𝑥 = | 𝑡𝑔 2 =𝑡 𝑥=2𝑎𝑟𝑐𝑡𝑔𝑡 𝑐𝑜𝑠𝑥= 1+𝑡2| =∫ 2(3 1+𝑡2 −2 1+𝑡2 ) 𝑑𝑡 = 1+𝑐𝑜𝑠𝑥 𝑑𝑥= 1+ 2 𝑡2 𝑠𝑖𝑛𝑥= 1+ 2𝑡 𝑡2 (1+ 1 1 − + 𝑡 𝑡 2 2 )(1+𝑡2) 𝑡2+1−2+8𝑡 𝑑𝑡 𝑑(1+𝑡2) 2∫ 𝑑𝑡 = 2∫𝑑𝑡 − 4∫ + 6∫ = 2𝑡 − 4𝑎𝑟𝑐𝑡𝑔𝑡 + 1+𝑡2 1+𝑡2 1+𝑡2 2 𝑥 𝑥 +3𝑙𝑛|1 + 𝑡2| + 𝐶=2tg( ) + 3ln(𝑡𝑔( ) + 1) − 2𝑥 + 𝐶; 2 2 Endi bu misolni yechimini va grafigini maple dasturida ko`ramiz: > f:=((3*sin(x)-2*cos(x))/(1+cos(x))); 3sin(x)2cos(x) f := 1cos(x) > Int(f,x); 3sin(x)2cos(x)  dx  1cos(x) > int(f,x);  x     x  2   2tan 3ln tan > Int(f,x)=int(f,x);   3sin(x)2cos(x)  x     x  2      1cos(x) dx2tan    2   3ln   tan    2   1   2x > plot(((3*sin(x)-2*cos(x))/(1+cos(x))), x=-4..4, y=-4..4); https://scientific-jl.com 1 2 0 64-son_3-to’plam_Iyun-2025 ISSN:3030- 3 6 1 3 T A D Q I Q O T L A R jahon ilmiy – metodik jurnali Demak, bu integralning chizmasi yuqoridagi ko`rinishda ekan. 30. ∫𝑠𝑖𝑛𝛼 ∙ 𝑥𝑐𝑜𝑠𝛽𝑥𝑑𝑥 , ∫𝑠𝑖𝑛𝛼 ∙ 𝑥𝑠𝑖𝑛𝛽𝑥𝑑𝑥, ∫𝑐𝑜𝑠𝛼𝑥 ∙ 𝑐𝑜𝑠𝛽𝑥𝑑𝑥, ko`rinishidagi integrallarni hisoblash. Foydalanuvchi formulalar: 1 1) 𝑠𝑖𝑛𝛼𝑥 ∙ 𝑐𝑜𝑠𝛽𝑥 = [sin(𝛼 + 𝛽)𝑥 + sin(𝛼 − 𝛽)𝑥] ; 2 1 2) 𝑠𝑖𝑛𝛼𝑥 ∙ 𝑠𝑖𝑛𝛽𝑥 = [cos(𝛼 − 𝛽)𝑥 − cos(𝛼 + 𝛽)𝑥]; 2 1 3) 𝑐𝑜𝑠𝛼𝑥 ∙ 𝑐𝑜𝑠𝛽𝑥 = [cos(𝛼 + 𝛽)𝑥 + cos(𝛼 − 𝛽)𝑥]; 2 1 1 3-misol.∫𝑐𝑜𝑠4𝑥 ∙ 𝑐𝑜𝑠𝑥𝑑𝑥 = (𝑐𝑜𝑠5𝑥 − 𝑐𝑜𝑠3𝑥)𝑑𝑥 = ∫𝑐𝑜𝑠5𝑥𝑑𝑥 + 2 2 1 1 1 ∫𝑐𝑜𝑠3𝑥𝑑𝑥 = 𝑠𝑖𝑛5𝑥 + 𝑠𝑖𝑛3𝑥 + 𝐶. 2 10 6 > f:=((cos(4*x)*cos(x))); f := cos(4x)cos(x) > Int(f,x); cos(4x)cos(x)dx > int(f,x); 1 1 sin(3x) sin(5x) 6 10 > Int(f,x)=int(f,x); 1 1 cos(4x)cos(x)dx sin(3x) sin(5x)  6 10 > plot(((cos(4*x)*cos(x))), x=-4..4, y=-4..4); https://scientific-jl.com 1 2 1 64-son_3-to’plam_Iyun-2025 ISSN:3030- 3 6 1 3 T A D Q I Q O T L A R jahon ilmiy – metodik jurnali Demak, bu integralning chizmasi yuqoridagi ko`rinishda ekan. 40. ∫𝑠𝑖𝑛𝑚𝑥 ∙𝑐𝑜𝑠𝑛𝑥𝑑𝑥 (𝑛,𝑚 ∈ 𝑧) ko`rinishidagi integrallarni hisoblash. 𝐼. 𝑛,𝑚 lar manfiy bo`lmagan (𝑛,𝑚 ∈ 𝑧) 𝑛 > 0,𝑚 > 0 juft son bo`lgan holat.Bu holatda darajani pasaytirish, ya`ni 1 1 𝑠𝑖𝑛2𝑥 = (1 − 𝑐𝑜𝑠𝑥) 𝑐𝑜𝑠2𝑥 = (1 + 𝑐𝑜𝑠2𝑥) 2 2 Formula qo`llaniladi. 1 1 1−𝑐𝑜𝑠2𝑥 1−𝑐𝑜𝑠4𝑥 4-misol.∫𝑠𝑖𝑛4𝑥 ∙ 𝑐𝑜𝑠2𝑥𝑑𝑥 = ∫𝑠𝑖𝑛2𝑥 ∙ 𝑠𝑖𝑛22𝑥𝑑𝑥 = ∫ ∙ 𝑑𝑥 = 4 4 2 2 1 1 1 1 ∫(1 − 𝑐𝑜𝑠4𝑥 − 𝑐𝑜𝑠2𝑥 + 𝑐𝑜𝑠2𝑥 ∙ 𝑐𝑜𝑠4𝑥)𝑑𝑥 = 𝑥 − 𝑠𝑖𝑛4𝑥 − 𝑠𝑖𝑛2𝑥 + 16 16 64 32 1 1 1 1 1 ∫(𝑐𝑜𝑠2𝑥 + 𝑐𝑜𝑠6𝑥)𝑑𝑥 = 𝑥 − 𝑠𝑖𝑛2𝑥 − 𝑠𝑖𝑛4𝑥 + 𝑠𝑖𝑛6𝑥 + 𝐶. 16 16 64 64 182 𝐼𝐼.𝑛,𝑚 larning ikkalasi ham butun manfiy 𝑛,𝑚 ∈ 𝑧 va juft yoki toq bo`lganda 𝑡𝑔𝑥 = 1 1 𝑡 yoki 𝑐𝑡𝑔𝑥 = 𝑡 olinib 1 + 𝑡𝑔2𝑥 = 1 + 𝑐𝑡𝑔2𝑥 = formuladan 𝑐𝑜𝑠2𝑥 𝑠𝑖𝑛2𝑥 foydalaniladi. 1 1 𝑑𝑥 1 5-misol.∫ 𝑑𝑥 = ∫ = ∫ = −∫ ∙ 𝑠𝑖𝑛3𝑥∙𝑐𝑜𝑠6𝑥 𝑠𝑖𝑛8𝑥∙ 𝑐𝑜𝑠5𝑥 𝑠𝑖𝑛2𝑥∙𝑠𝑖𝑛6𝑥∙ 𝑐𝑜𝑠5𝑥 𝑐𝑡𝑔5𝑥 𝑠𝑖𝑛5𝑥 𝑠𝑖𝑛5𝑥 (1 + +𝑐𝑡𝑔2𝑥)3𝑑(𝑐𝑡𝑔𝑥) = | 𝑐𝑡𝑔𝑥=𝑡 | = 1 𝑡−4 + 3 𝑡−2 − 3ln|𝑡| − 𝑡2 + 𝐶 = 1 ∙ 𝑑𝑐𝑡𝑔𝑥=𝑑𝑡 4 2 2 4 1 3 1 𝑐𝑡𝑔2𝑥 + ∙ − 3𝑙𝑛|𝑐𝑡𝑔𝑥| + + 𝐶. 𝑐𝑡𝑔4𝑥 2 𝑐𝑡𝑔2𝑥 2 > f:=(1/((sin(x))^3)*((cos(x)