7.1-bo'lim : Qismlar bo'yicha integratsiya
7. \( \displaystyle \int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}}\) ni baholang.
Yechim
Bu erda birinchi qadam \(u\) va \(dv\) ni tanlashdir va bu holda biz ularni qanday tanlaganimizga ehtiyot bo'lishimiz kerak.
Agar biz ko'plab misollar/amaliy masalalar modeliga amal qilsak, \(u\) ni \({t^7}\) va \(dv\) \(\sin \left( {2{t^4}} \right)\) bo'lsin.
Biroq, bu ba'zi haqiqiy muammolarga olib keladi. \(v\) ni hisoblash uchun biz sinusni birlashtirishimiz kerak va argumentdagi \({t^4}\) tufayli bu mumkin emas. Sinusni integrallash uchun biz quyida ko'rsatilgandek almashtirish uchun integranda \({t^3}\) ga ega bo'lishimiz kerak,
\[\int{{{t^3}\sin \left( {2{t^4}} \right)\,dt}} = \frac{1}{8}\int{{\sin \left( w \right)\,dw}} = - \frac{1}{8}\cos \left( {2{t^4}} \right) + c\hspace{0.25in}w = 2{t^4}\]Endi bu muammo bo'lib tuyulishi mumkin, lekin aslida bu alohida integral uchun muammo emas. E'tibor bering, bizda integralda 7 ta \(t\) mavjud va biz ularni quyidagicha bo'lolmasligimiz uchun hech qanday sabab yo'q,
\[ \int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = \int{{{t^4}\,{t^3}\sin \left( {2{t^4}} \right)\,dt}}\]Buni qilgandan so'ng, biz \(u\) va \(dv\) ni quyidagicha tanlashimiz mumkin,
\[u = {t^4}\hspace{0.5in}dv = {t^3}\sin \left( {2{t^4}} \right)\,dt\]2-qadam
Keyinchalik, \(du\) (\(u\) farqlash orqali) va \(v\) (\(dv\) ni integratsiyalash orqali) hisoblashimiz kerak.
\[\begin{align*}u & = {t^4} & \hspace{0.5in} & \to & \hspace{0.25in}du & = 4{t^3}dt\\ dv & = \,{t^3}\sin \left( {2{t^4}} \right)\,dt & \hspace{0.5in} & \to & \hspace{0.25in}v & = - \frac{1}{8}\cos \left( {2{t^4}} \right)\end{align*}\]3-qadam
\(u\), \(du\), \(v\) va \(dv\) ni qismlar bo‘yicha integratsiyaga ulash formulasini beradi:
\[ \int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = - \frac{1}{8}{t^4}\cos \left( {2{t^4}} \right) + \frac{1}{2}\int{{{t^3}\cos \left( {2{t^4}} \right)\,dt}}\]4-qadam
Shu o‘rinda, yangi integral biz \(v\) ni topishda foydalangan bir xil Hisob I o‘rnini talab qilishiga e’tibor bering. Shunday qilib, biz qilishimiz kerak bo'lgan narsa - yangi integralni baholash va biz tugatamiz.
\[ \int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{8}{t^4}\cos \left( {2{t^4}} \right) + \frac{1}{{16}}\sin \left( {2{t^4}} \right) + c}}\]Ushbu muammolar uchun naqshlarga shunchalik yopishib qolmangki, siz ularni muayyan turdagi muammolar qanday ishlashini "qoidalarga" aylantirasiz. Oson ko'rinadigan naqshlarning ko'pchiligi ham osonlikcha buziladi (bu muammo ko'rsatilgandek).
Biz (o'qituvchi sifatida) ko'plab "oson" muammolarni hal qilishga moyil bo'lganimiz sababli, dastlab ular osongina ko'rish mumkin bo'lgan naqshlarga mos keladi. Bu o'quvchilarni naqshlar har doim ishlaydi degan fikrga olib keladi va keyin naqsh ishlamayotgan joyga duch kelganida, ular muammoga duch kelishadi. Shunday ekan, ehtiyot bo'ling!
Shuni ham yodda tutingki, biz naqshlar mavjud emas va ularni tanib olish foydali emas deb aytmayapmiz. Siz shunchaki ehtiyot bo'lishingiz va tushunishingiz kerakki, ba'zida muammolar paydo bo'lishi mumkin, u siz tan olgan naqshga o'xshaydi, lekin aslida naqshga to'liq mos kelmaydi va muammoni hal qilish uchun boshqa yondashuv kerak bo'ladi.
Muqobil yechimni ko'rsatish
E'tibor bering, bu muammoning muqobil echimi bor. Birinchi qadam sifatida biz \(w = 2{t^4}\) almashtirishdan quyidagi tarzda foydalanishimiz mumkin.
\[w = 2{t^4}\hspace{0.5in} \to \hspace{0.5in}dw = 8{t^3}dt \hspace{0.5in} \& \hspace{0.5in} {t^4} = \frac{1}{2}w\] \[ \int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = \int{{{t^4}\,{t^3}\sin \left( {2{t^4}} \right)\,dt}} = \int{{\left( {\frac{1}{2}w} \right)\left( {\frac{1}{8}} \right)\sin \left( w \right)\,dw}} = \int{{\frac{1}{{16}}w\sin \left( w \right)\,dw}}\]Bu erda ko'rib turganimizdek, biz qismlar bo'yicha integratsiyadan qochmaymiz, lekin bu safar buni ko'rish biroz osonroq. Mana, bu muammo uchun ishning qolgan qismi.
\[\begin{align*}u & = \frac{1}{{16}}w & \hspace{0.5in} & \to & \hspace{0.25in}du & = \frac{1}{{16}}dw\\ dv & = \,\sin \left( w \right)\,dw & \hspace{0.5in} & \to & \hspace{0.25in}v & = - \cos \left( w \right)\end{align*}\] \[\int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = - \frac{1}{{16}}w\cos \left( w \right) + \frac{1}{{16}}\int{{\cos \left( w \right)\,dw}} = - \frac{1}{{16}}w\cos \left( w \right) + \frac{1}{{16}}\sin \left( w \right) + c\]Yakuniy qadam sifatida biz faqat \(w\) ga almashtirishimiz kerak.
\[\int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = - \frac{1}{8}{t^4}\cos \left( {2{t^4}} \right) + \frac{1}{{16}}\sin \left( {2{t^4}} \right) + c\]