1.1. Sodda funksiyalarning aniqmas integrallari jadvali 1. Aniqmas integralning ta’rifi. π(π₯) funksiya (π, π) intervalda aniqlangan bo‘lib, πΉ′(π₯) = π(π₯), π₯ ∈ (π, π), bo‘lsa, u holda πΉ(π₯) funksiya π(π₯) funksiyaning boshlang‘ich funksiyasi deyiladi. Agar πΉ1 (π₯) va πΉ2 (π₯) funksiyalar π(π₯) funksiyaning boshlang‘ich funksiyalari bo‘lsalar, u holda ular bir biridan o‘zgarmasga farq qiladilar. Ya’ni πΉ2 (π₯) = πΉ1 (π₯) + πΆ bo‘ladi. Ta’rif. Agar πΉ(π₯) funksiya π(π₯) funksiyaning boshlang‘ich funksiyasi bo‘lsa, u holda {πΉ(π₯) + πΆ} funksiyalar oilasiga π(π₯) funksiyaning aniqmas integrali deyiladi va quyidagicha belgilanadi. ∫ π(π₯)ππ₯ = πΉ(π₯) + πΆ. 2. Aniqmas integralning asosiy xossalari. a) π[∫ π(π₯)ππ₯] = π(π₯)ππ₯; b) ∫ πΦ(π₯) = Φ(π₯) + πΆ; v) ∫ π΄π(π₯)ππ₯ = π΄ ∫ π(π₯)ππ₯(π΄ = const; π΄ ≠ 0); g) ∫[π(π₯) + π(π₯)] ππ₯ = ∫ π(π₯)ππ₯ + ∫ π(π₯)ππ₯ 3. Sodda funksiyalarning aniqmas integrallari jadvali. I. ∫ π₯ π ππ₯ = II. ∫ ππ₯ π₯ π+1 π+1 + πΆ(n ≠ −1). = ln|π₯| + πΆ(π₯ ≠ 0). ππππ‘ππ₯ + πΆ ππ₯ III. ∫ ={ . 2 1+π₯ −πππππ‘ππ₯ + πΆ ππ₯ 1 1+π₯ IV. ∫ = ln { } + πΆ. 2 1−π₯ 2 1−π₯ ππ₯ arcsinπ₯ + πΆ V. ∫ ={ . √1−π₯ 2 −arccosπ₯ + πΆ ππ₯ VI. ∫ 2 = ln |π₯ + √π₯ 2 ± 1| + πΆ. π₯ √π₯ ±1 π₯ ππ₯ VII. ∫ π ππ₯ = + πΆ(π > 0, π ≠ 1) ; ∫ π π₯ ππ₯ = π π₯ + πΆ. lnπ VIII. ∫ sinπ₯ππ₯ = −cosπ₯ + πΆ XII. ∫ π βπ₯ππ₯ = πβπ₯ + πΆ IX. ∫ cosπ₯ππ₯ = sinπ₯ + πΆ XIII. ∫ πβπ₯ππ₯ = π βπ₯ + πΆ ππ₯ X. ∫ 2 = −ππ‘ππ₯ + πΆ sin π₯ ππ₯ XIV. ∫ π β2 π₯ ππ₯ = −ππ‘β + πΆ XI. ∫ 2 = π‘ππ₯ + πΆ cos π₯ ππ₯ XV. ∫ 2 = π‘βπ₯ + πΆ πβ π₯ Jadval yordamida hisoblashga misollar. 1-misol. Yechish. 3 √π₯−2 √π₯ 2 +1 ππ₯ ∫ 4 √π₯ – integralni hisoblang. 3 3 √π₯ 2 ππ₯ ππ₯ √π₯ − 2√π₯ 2 + 1 √π₯ ∫ ππ₯ = ∫ ππ₯ − 2 ∫ + ∫ = 4 4 4 4 √π₯ √π₯ √π₯ √π₯ 1 5 1 4 12 ⋅ 2 4 = ∫ π₯ 4 ππ₯ − 2 ∫ π₯ 12 ππ₯ + ∫ π₯ −4 ππ₯ = − 4 + 12 − 4 +πΆ 3√π₯ 3 7 √π₯ 7 5√π₯ 5 √1+π₯ 2 +√1−π₯ 2 2-misol. ∫ ππ₯ – integralni hisoblang. √1−π₯ 4 Yechish. √1 − π₯ 2 √1 + π₯ 2 + √1 − π₯ 2 √1 + π₯ 2 ∫ ππ₯ = ∫ ππ₯ + ∫ ππ₯ = √1 − π₯ 4 √(1 + π₯ 2 )(1 − π₯ 2 ) √(1 − π₯ 2 )(1 + π₯ 2 ) 1 1 =∫ ππ₯ + ∫ ππ₯ = arcsinπ₯ + ln |π₯ + √1 + π₯ 2 | + πΆ 2 2 √1 + π₯ √1 − π₯ ππ₯ 3-misol. ∫ – integralni hisoblang. π Yechish. (π₯−π) ππ₯ 1 1 = + πΆ. (π₯ − π)π 1 − π (π₯ − π)π−1 – integralni hisoblang. 2 ∫ 4-misol. ∫ Yechish. ππ₯ √π₯ 2 ±π ππ₯ ∫ =∫ π(π₯/π) √(π₯/π)2 ±1 √π₯ 2 ± π 2 ππ₯ 5-misol. ∫ 2 2 – integralni hisoblang. π +π₯ Yechish. = ln |π₯ + √π₯ 2 ± π2 | + πΆ. 1 aπππ‘π(π₯/π) + πΆ ππ₯ 1 π(π₯/π) π ∫ 2 = ∫ ={ 1 π + π₯ 2 π 1 + (π₯/π)2 − πππππ‘π(π₯/π) + πΆ π 1+cos2 π₯ Mustaqil ishlash uchun misollar ππ₯ 6-misol.∫ ππ₯ 1+cos2π₯ 7-misol.∫ √3+3π₯ 2 8-misol.∫ 9-misol.∫ 2 2 ππ₯ cos π₯sin π₯ (1−π₯)2 π₯ √π₯ ππ₯ π₯ cos2π₯ (1+π₯)2 10-misol.∫ 2sin ππ₯ 2 11-misol.∫ ππ₯ π₯(1+π₯ 2 ) 12-misol.∫ π‘π2 π₯ππ₯ 13-misol.∫ ππ₯ 14-misol.∫ cos2π₯+sin2 π₯ √π₯−π₯ 3 π π₯ +π₯ 2 ππ₯ π₯3 15-misol. ∫(arcsinπ₯ + arccosπ₯) ππ₯