7.2-bo'lim : Trig funksiyalarini o'z ichiga olgan integrallar
14. \( \displaystyle \int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}}\) ni baholang.
Yechim
Ushbu muammoni hal qilishni boshlash uchun, avvalo, integrandni faqat sinuslar va kosinuslarga qisqartirishimiz mumkinligini bilib olishimiz kerak. Buni sinus nuqtai nazaridan kosekant ta'rifini eslash orqali amalga oshirish juda oson.
\[\begin{align*}\int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}} & = \int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]\frac{1}{{{{\sin }^4}\left( {3x} \right)}}\,dx}}\\ & = \int{{9\sin \left( {3x} \right) - 2\frac{{{{\cos }^3}\left( {3x} \right)}}{{{{\sin }^4}\left( {3x} \right)}}\,dx}}\end{align*}\]2-qadam
Birinchi integral qo'shimcha ishlamasdan bajarish uchun etarlicha sodda.
Ikkinchi integral uchun, agar u bo'linma o'rniga mahsulot bo'lsa, buni qanday qilishimiz haqida o'ylab ko'ring. Bunday holda biz shunchaki kosinusni ajratib olamiz.
\[ \int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}} = \int{{9\sin \left( {3x} \right) - 2\frac{{{{\cos }^2}\left( {3x} \right)}}{{{{\sin }^4}\left( {3x} \right)}}\cos \left( {3x} \right)\,dx}}\]3-qadam
Ikkinchi integral uchun biz qolgan kosinuslarni sinuslarga aylantirish uchun \({\sin ^2}\theta + {\cos ^2}\theta = 1\) trig identifikatoridan foydalanishimiz mumkin.
\[ \int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}} = \int{{9\sin \left( {3x} \right)\,dx}} - 2\int{{\frac{{1 - {{\sin }^2}\left( {3x} \right)}}{{{{\sin }^4}\left( {3x} \right)}}\cos \left( {3x} \right)\,dx}}\]4-qadam
Endi biz ikkinchi integralni baholash uchun \(u = \sin \left( {3x} \right)\) almashtirishdan foydalanishimiz mumkin. Birinchi integral qo'shimcha ishni talab qilmaydi.
\[\begin{align*}\int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}} & = \int{{9\sin \left( {3x} \right)\,dx}} - \frac{2}{3}\int{{\frac{{1 - {u^2}}}{{{u^4}}}\,du}}\\ & = \int{{9\sin \left( {3x} \right)\,dx}} - \frac{2}{3}\int{{{u^{ - 4}} - {u^{ - 2}}\,du}}\\ & = - 3\cos \left( {3x} \right) - \frac{2}{3}\left( { - \frac{1}{3}{u^{ - 3}} + {u^{ - 1}}} \right) + c\end{align*}\]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!
\[\begin{align*}\int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}} & = - 3\cos \left( {3x} \right) + \frac{2}{9}\frac{1}{{{{\sin }^3}\left( {3x} \right)}} - \frac{2}{3}\frac{1}{{\sin \left( {3x} \right)}} + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 3\cos \left( {3x} \right) + \frac{2}{9}{{\csc }^3}\left( {3x} \right) - \frac{2}{3}\csc \left( {3x} \right) + c}}\end{align*}\]