7.2-bo'lim : Trig funksiyalarini o'z ichiga olgan integrallar
1. \( \displaystyle \int{{{{\sin }^3}\left( {\frac{2}{3}x} \right){{\cos }^4}\left( {\frac{2}{3}x} \right)\,dx}}\) ni baholang.
Yechim
Bu erda e'tibor berish kerak bo'lgan birinchi narsa - sinusdagi ko'rsatkich g'alati va shuning uchun biz ulardan birini ajratib olishimiz mumkin.
\[ \int{{{{\sin }^3}\left( {\frac{2}{3}x} \right){{\cos }^4}\left( {\frac{2}{3}x} \right)\,dx}} = \int{{{{\sin }^2}\left( {\frac{2}{3}x} \right){{\cos }^4}\left( {\frac{2}{3}x} \right)\sin \left( {\frac{2}{3}x} \right)\,dx}}\]2-qadam
Endi qolgan sinuslarni kosinuslarga aylantirish uchun \({\sin ^2}\theta + {\cos ^2}\theta = 1\) trig identifikatoridan foydalanishimiz mumkin.
\[\int{{{{\sin }^3}\left( {\frac{2}{3}x} \right){{\cos }^4}\left( {\frac{2}{3}x} \right)\,dx}} = \int{{\left( {1 - {{\cos }^2}\left( {\frac{2}{3}x} \right)} \right){{\cos }^4}\left( {\frac{2}{3}x} \right)\sin \left( {\frac{2}{3}x} \right)\,dx}}\]3-qadam
Endi biz integralni baholash uchun \(u = \cos \left( {\frac{2}{3}x} \right)\) almashtirishdan foydalanishimiz mumkin.
\[\begin{align*}\int{{{{\sin }^3}\left( {\frac{2}{3}x} \right){{\cos }^4}\left( {\frac{2}{3}x} \right)\,dx}} & = - \frac{3}{2}\int{{\left( {1 - {u^2}} \right){u^4}\,du}}\\ & = - \frac{3}{2}\int{{{u^4} - {u^6}\,du}} = - \frac{3}{2}\left( {\frac{1}{5}{u^5} - \frac{1}{7}{u^7}} \right) + c\end{align*}\]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.
4-qadam
\(u\) o'rniga qayta kirishni unutmang!
\[\int{{{{\sin }^3}\left( {\frac{2}{3}x} \right){{\cos }^4}\left( {\frac{2}{3}x} \right)\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{{14}}{{\cos }^7}\left( {\frac{2}{3}x} \right) - \frac{3}{{10}}{{\cos }^5}\left( {\frac{2}{3}x} \right) + c}}\]