7.2-bo'lim : Trig funksiyalarini o'z ichiga olgan integrallar
12. \( \displaystyle \int{{\frac{{{{\sec }^4}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}\,dt}}\) ni baholang.
Yechim
Agar bu sekantlar va tangenslarning hosilasi bo'lsa, biz buni qanday qilishni bilamiz. Xuddi shu g'oyalar bu erda ishlaydi, faqat suratga e'tibor qaratishimiz kerak. Biz maxrajdan (umuman) hech narsani olib tashlay olmaymiz va u xuddi shunday ishlashini kutamiz. Biz faqat numeratordan narsalarni olib tashlashimiz mumkin.
E'tibor bering, sekantdagi ko'rsatkich teng va shuning uchun biz ulardan ikkitasini ajratib olishimiz mumkin.
\[ \int{{\frac{{{{\sec }^4}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}\,dt}} = \int{{\frac{{{{\sec }^2}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}{{\sec }^2}\left( {2t} \right)\,dt}}\]2-qadam
Endi qolgan sekantlarni tangenslarga aylantirish uchun \({\tan ^2}\theta + 1 = {\sec ^2}\theta \) trig identifikatoridan foydalanishimiz mumkin.
\[ \int{{\frac{{{{\sec }^4}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}\,dt}} = \int{{\frac{{{{\tan }^2}\left( {2t} \right) + 1}}{{{{\tan }^9}\left( {2t} \right)}}{{\sec }^2}\left( {2t} \right)\,dt}}\]3-qadam
Endi biz integralni baholash uchun \(u = \tan \left( {2t} \right)\) almashtirishdan foydalanishimiz mumkin.
\[\int{{\frac{{{{\sec }^4}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}\,dt}} = \frac{1}{2}\int{{\frac{{{u^2} + 1}}{{{u^9}}}\,du}} = \frac{1}{2}\int{{{u^{ - 7}} + {u^{ - 9}}\,du}} = \frac{1}{2}\left[ { - \frac{1}{6}{u^{ - 6}} - \frac{1}{8}{u^{ - 8}}} \right] + c\]E'tibor bering, biz bu erda haqiqiy almashtirish ishlarini qilmaymiz. Shu nuqtada, agar kerak bo'lsa, tafsilotlarni to'ldirish uchun almashtirishni yaxshi eslaysiz deb taxmin qilinadi. Agar siz almashtirishda zanglagan bo'lsangiz, ehtimol hisob-kitoblar bo'yicha muammolarga qaytishingiz va almashtirishlar bo'yicha mashq qilishingiz kerak.
4-qadam
\(u\) o'rniga qayta kirishni unutmang!
\[\int{{\frac{{{{\sec }^4}\left( {2t} \right)}}{{{{\tan }^9}\left( {2t} \right)}}\,dt}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{{12}}\frac{1}{{{{\tan }^6}\left( {2t} \right)}} - \frac{1}{{16}}\frac{1}{{{{\tan }^8}\left( {2t} \right)}} + c = - \frac{1}{{12}}{{\cot }^6}\left( {2t} \right) - \frac{1}{{16}}{{\cot }^8}\left( {2t} \right) + c}}\]