7.2-bo'lim : Trig funksiyalarini o'z ichiga olgan integrallar
4. \( \displaystyle \int_{\pi }^{{2\pi }}{{{{\cos }^3}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\,dw}}\) ni baholang.
Yechim
Limitlar bilan shug'ullanish uchun bizda ikkita variant bor. Biz integralni bajarayotganda chegaralar bilan shug'ullanishimiz yoki noaniq integralni baholashimiz va oxirgi bosqichda chegaralar haqida g'amxo'rlik qilishimiz mumkin. Biz ushbu muammoning chegaralarini hal qilishning oxirgi usulidan foydalanamiz.
Bunday holda, ikkala ko'rsatkich ham toq ekanligiga e'tibor bering. Bu shuni anglatadiki, biz kosinusni ajratib, qolganini sinuslarga aylantirishimiz yoki sinusni ajratib, qolganini kosinuslarga aylantirishimiz mumkin. Ikkalasi ham mutlaqo qabul qilinadigan echimlar. Biroq, kosinusdagi ko'rsatkich kichikroq va shuning uchun agar biz kosinusni ajratib olsak va qolgan kosinuslarni sinuslarga aylantirsak, konvertatsiya ishi kamroq bo'ladi.
Mana o'sha ish.
\[\begin{align*}\int{{{{\cos }^3}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\,dw}} & = \int{{{{\cos }^2}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\cos \left( {\frac{1}{2}w} \right)\,dw}}\\ & = \int{{\left( {1 - {{\sin }^2}\left( {\frac{1}{2}w} \right)} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\cos \left( {\frac{1}{2}w} \right)\,dw}}\end{align*}\]2-qadam
Endi biz integralni baholash uchun \(u = \sin \left( {\frac{1}{2}w} \right)\) almashtirishdan foydalanishimiz mumkin.
\[\begin{align*}\int{{{{\cos }^3}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\,dw}} & = 2\int{{\left( {1 - {u^2}} \right){u^5}\,du}}\\ & = 2\int{{{u^5} - {u^7}\,du}} = 2\left( {\frac{1}{6}{u^6} - \frac{1}{8}{u^8}} \right) + c\end{align*}\]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.
3-qadam
\(u\) o'rniga qayta kirishni unutmang!
\[\int{{{{\cos }^3}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\,dw}} = \frac{1}{3}{\sin ^6}\left( {\frac{1}{2}w} \right) - \frac{1}{4}{\sin ^8}\left( {\frac{1}{2}w} \right) + c\]4-qadam
Endi biz qilishimiz kerak bo'lgan narsa chegaralar bilan shug'ullanishdir.
\[ \int_{\pi }^{{2\pi }}{{{{\cos }^3}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\,dw}} = \left. {\left( {\frac{1}{3}{{\sin }^6}\left( {\frac{1}{2}w} \right) - \frac{1}{4}{{\sin }^8}\left( {\frac{1}{2}w} \right)} \right)} \right|_\pi ^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{{12}}}}\]Muqobil yechimni ko'rsatish
Yuqorida aytib o'tganimizdek, agar xohlasak, biz sinusni osongina ajratib olishimiz va qolganini kosinuslarga aylantirishimiz mumkin edi. Biz bu ishni bu erga qo'ymaymiz, lekin bu erda siz buni shunday qilganingizda olishingiz kerak bo'lgan noaniq integraldir.
\[\int{{{{\cos }^3}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\,dw}} = - \frac{1}{2}{\cos ^4}\left( {\frac{1}{2}w} \right) + \frac{2}{3}{\cos ^6}\left( {\frac{1}{2}w} \right) - \frac{1}{4}{\cos ^8}\left( {\frac{1}{2}w} \right) + c\]Shuni ham yodda tutingki, biz noaniq integralni bajarishda qaysi yondashuvdan foydalanmasak ham, aniq integralning qiymati bir xil bo'ladi.