7.2-bo'lim : Trig funksiyalarini o'z ichiga olgan integrallar
13. \( \displaystyle \int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}}\) ni baholang.
Yechim
Numeratordagi yig'indi tufayli integratsiyani (va keyin integralni) quyidagicha ikkiga bo'lish mantiqiy (umid qilamanki) bo'ladi.
\[ \int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} = \int{{\frac{2}{{{{\cos }^2}\left( z \right)}} + \frac{{7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} = \int{{\frac{2}{{{{\cos }^2}\left( z \right)}}\,dz}} + \int{{\frac{{7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}}\]2-qadam
Endi, birinchi integral bir qarashda qiyin ko'rinadi, lekin biz buni sekantlar nuqtai nazaridan osongina qayta yozishimiz mumkin, bunda u haqiqatan ham oson integralga aylanadi.
Ikkinchi integral uchun, agar u bo'linma o'rniga mahsulot bo'lsa, buni qanday qilishimiz haqida o'ylab ko'ring. Bunday holda, biz shunchaki sinusni ajratib olamiz.
\[ \int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} = \int{{2{{\sec }^2}\left( z \right)\,dz}} + 7\int{{\frac{{{{\sin }^2}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\sin \left( z \right)\,dz}}\]3-qadam
Yuqorida ta'kidlab o'tilganidek, birinchi integral hozir juda oson (buni keyingi bosqichda qilamiz) va ikkinchi integral uchun biz ikkinchi integraldagi qolgan sinuslarni kosinuslarga aylantirish uchun \({\sin ^2}\theta + {\cos ^2}\theta = 1\) trig identifikatoridan foydalanishimiz mumkin.
\[ \int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} = \int{{2{{\sec }^2}\left( z \right)\,dz}} + 7\int{{\frac{{1 - {{\cos }^2}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\sin \left( z \right)\,dz}}\]4-qadam
Endi biz ikkinchi integralni baholash uchun \(u = \cos \left( z \right)\) almashtirishdan foydalanishimiz mumkin. Birinchi integral qo'shimcha ishni talab qilmaydi.
\[\begin{align*}\int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} & = 2\tan \left( z \right) - 7\int{{\frac{{1 - {u^2}}}{{{u^2}}}\,du}}\\ & = 2\tan \left( z \right) - 7\int{{{u^{ - 2}} - 1\,du}} = 2\tan \left( z \right) - 7\left( { - {u^{ - 1}} - u} \right) + c\end{align*}\]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.
5-qadam
\(u\) o'rniga qayta kirishni unutmang!
\[ \int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{2\tan \left( z \right) + 7\frac{1}{{\cos \left( z \right)}} + 7\cos \left( z \right) + c = 2\tan \left( z \right) + 7\sec \left( z \right) + 7\cos \left( z \right) + c}}\]