7.2-bo'lim : Trig funksiyalarini o'z ichiga olgan integrallar
5. \( \displaystyle \int{{{{\sec }^6}\left( {3y} \right){{\tan }^2}\left( {3y} \right)\,dy}}\) ni baholang.
Yechim
Bu erda e'tibor berish kerak bo'lgan birinchi narsa shundaki, sekantdagi ko'rsatkich teng va shuning uchun biz ulardan ikkitasini ajratib olishimiz mumkin.
\[ \int{{{{\sec }^6}\left( {3y} \right){{\tan }^2}\left( {3y} \right)\,dy}} = \int{{{{\sec }^4}\left( {3y} \right){{\tan }^2}\left( {3y} \right){{\sec }^2}\left( {3y} \right)\,dy}}\]2-qadam
Endi qolgan sekantlarni tangenslarga aylantirish uchun \({\tan ^2}\theta + 1 = {\sec ^2}\theta \) trig identifikatoridan foydalanishimiz mumkin.
\[\begin{align*}\int{{{{\sec }^6}\left( {3y} \right){{\tan }^2}\left( {3y} \right)\,dy}} & = \int{{{{\left[ {{{\sec }^2}\left( {3y} \right)} \right]}^2}{{\tan }^2}\left( {3y} \right){{\sec }^2}\left( {3y} \right)\,dy}}\\ & = \int{{{{\left[ {{{\tan }^2}\left( {3y} \right) + 1} \right]}^2}{{\tan }^2}\left( {3y} \right){{\sec }^2}\left( {3y} \right)\,dy}}\end{align*}\]3-qadam
Endi biz integralni baholash uchun \(u = \tan \left( {3y} \right)\) almashtirishdan foydalanishimiz mumkin.
\[\begin{align*}\int{{{{\sec }^6}\left( {3y} \right){{\tan }^2}\left( {3y} \right)\,dy}}& = \frac{1}{3}\int{{{{\left[ {{u^2} + 1} \right]}^2}{u^2}\,du}}\\ & = \frac{1}{3}\int{{{u^6} + 2{u^4} + {u^2}\,du}} = \frac{1}{3}\left( {\frac{1}{7}{u^7} + \frac{2}{5}{u^5} + \frac{1}{3}{u^3}} \right) + c\end{align*}\]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!
\[\int{{{{\sec }^6}\left( {3y} \right){{\tan }^2}\left( {3y} \right)\,dy}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{21}}{{\tan }^7}\left( {3y} \right) + \frac{2}{{15}}{{\tan }^5}\left( {3y} \right) + \frac{1}{9}{{\tan }^3}\left( {3y} \right) + c}}\]