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.

Bu erda e'tibor berish kerak bo'lgan birinchi narsa, tangensdagi ko'rsatkich g'alati va bizda muammolarda sekant bor va shuning uchun biz ularning har biridan birini ajratib olishimiz va qolgan tangenslarni sekantlarga aylantirish uchun \({\tan ^2}\theta + 1 = {\sec ^2}\theta \) trig identifikatoridan foydalanishimiz mumkin.

\[\begin{align*}\int{{{{\tan }^7}\left( z \right){{\sec }^3}\left( z \right)\,dz}} & = \int{{{{\tan }^6}\left( z \right){{\sec }^2}\left( z \right)\,\,\,\tan \left( z \right)\sec \left( z \right)\,dz}}\\ & = \int{{{{\left[ {{{\tan }^2}\left( z \right)} \right]}^3}{{\sec }^2}\left( z \right)\,\,\,\tan \left( z \right)\sec \left( z \right)\,dz}}\\ & = \int{{{{\left[ {{{\sec }^2}\left( z \right) - 1} \right]}^3}{{\sec }^2}\left( z \right)\,\,\,\tan \left( z \right)\sec \left( z \right)\,dz}}\end{align*}\]

2-qadam

Endi biz integralni baholash uchun \(u = \sec \left( z \right)\) almashtirishdan foydalanishimiz mumkin.

\[\begin{align*}\int{{{{\tan }^7}\left( z \right){{\sec }^3}\left( z \right)\,dz}} & = \int{{{{\left[ {{u^2} - 1} \right]}^3}{u^2}\,du}}\\ & = \int{{{u^8} - 3{u^6} + 3{u^4} - {u^2}\,du}} = \frac{1}{9}{u^9} - \frac{3}{7}{u^7} + \frac{3}{5}{u^5} - \frac{1}{3}{u^3} + c\end{align*}\]

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!

\[\int{{{{\tan }^7}\left( z \right){{\sec }^3}\left( z \right)\,dz}} = \frac{1}{9}{\sec ^9}\left( z \right) - \frac{3}{7}{\sec ^7}\left( z \right) + \frac{3}{5}{\sec ^5}\left( z \right) - \frac{1}{3}{\sec ^3}\left( z \right) + c\]

4-qadam

Endi biz qilishimiz kerak bo'lgan narsa chegaralar bilan shug'ullanishdir.

\[\begin{align*}\int_{0}^{{\frac{\pi }{4}}}{{{{\tan }^7}\left( z \right){{\sec }^3}\left( z \right)\,dz}} & = \left. {\left( {\frac{1}{9}{{\sec }^9}\left( z \right) - \frac{3}{7}{{\sec }^7}\left( z \right) + \frac{3}{5}{{\sec }^5}\left( z \right) - \frac{1}{3}{{\sec }^3}\left( z \right)} \right)} \right|_0^{\frac{\pi }{4}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{2}{{315}}\left( {8 + 13\sqrt 2 } \right) = 0.1675}}\end{align*}\]