Bunday holda \(u = 25{x^2} - 4\) almashtirish ishlamaydi (almashtirish uchun kerak bo'lgan suratda bizda \(x\,dx\) mavjud emas) va shuning uchun biz bu integral uchun boshqacha harakat qilishimiz kerak bo'ladi.

Ildizdagi ikkita atamani qandaydir tarzda bitta atamagacha qisqartirsak yaxshi bo'lardi. Quyidagi almashtirish biz uchun buni amalga oshiradi.

\[x = \frac{2}{5}\sec \theta \]

Bu qaerdan kelib chiqqanligi haqida tashvishlanmang. Muammoni ishlayotganimizda, u ishlayotganini va agar bizda muammoda kvadrat ildizning o'xshash turi bo'lsa, biz har doim shunga o'xshash almashtirishdan foydalanishimiz mumkinligini ko'rasiz.

O'zgartirishni amalga oshirishdan oldin, keling, bu ildizdagi ikkita atamani bitta atamaga qisqartirish imkonini beradi degan da'voni tekshirib ko'raylik.

\[\sqrt {25{x^2} - 4} = \sqrt {25\left( {\frac{4}{{25}}} \right){{\sec }^2}\theta - 4} = \sqrt {4\left( {{{\sec }^2}\theta - 1} \right)} = 2\sqrt {{{\sec }^2}\theta - 1} \]

Endi ikkita atamani bitta atamaga qisqartirishimiz kerak bo'lgan narsa munosabatlarni eslash,

\[{\tan ^2}\theta + 1 = {\sec ^2}\theta \hspace{0.25in}\,\,\,\,\, \Rightarrow \hspace{0.5in}{\sec ^2}\theta - 1 = {\tan ^2}\theta \]

Ushbu faktdan foydalanib, kvadrat ildiz bo'ladi,

\[\sqrt {25{x^2} - 4} = 2\sqrt {{{\tan }^2}\theta } = 2\left| {\tan \theta } \right|\]

Shunday qilib, biz nafaqat ikki atamani bitta atamaga qisqartira oldik, biz ham ildizni osongina yo'q qila oldik!

Biroq, mutlaq qiymatlar satrining mavjudligiga e'tibor bering. Bular muhim. Shuni eslang

\[\sqrt {{x^2}} = \left| x \right|\]

Ushbu bosqichda har doim mutlaq qiymat chiziqlari bo'lishi kerak. Agar biz \(\tan \theta \) ning har doim ijobiy yoki har doim salbiy ekanligini bilsak, mutlaq qiymat satrlarini yo'q qilishimiz mumkin:

\[\left| x \right| = \left\{ {\begin{array}{rl}x&{{\mbox{if }}x \ge 0}\\{ - x}&{{\mbox{if }}x < 0}\end{array}} \right.\]

Cheklovlarsiz biz \(\tan \theta \) ning ijobiy yoki salbiy ekanligini aniqlay olmaymiz, ammo integralni bajarish uchun ularni yo'q qilishimiz kerak bo'ladi. Shuning uchun biz noaniq integralni bajarayotganimiz uchun \(\tan \theta \) musbat bo'ladi deb faraz qilamiz va shuning uchun biz mutlaq qiymat satrlarini tashlab qo'yishimiz mumkin. Bu beradi,

\[\sqrt {25{x^2} - 4} = 2\tan \theta \]

Shunday qilib, biz bu almashtirish bilan ildiz ostidagi ikkita atamani bitta atamaga qisqartirishga muvaffaq bo'ldik va bu jarayonda ildizni ham yo'q qildik. Ildizni yo'q qilish - bu almashtirishning yaxshi yon ta'siri, chunki endi muammoni hal qilish biroz osonroq bo'ladi.

Keling, almashtirishni amalga oshiramiz va biz nimaga erishamiz. Almashtirishda biz \(dx\) ni ham almashtirishimiz kerakligini unutmang. Buni almashtirishdan olish juda oson.

\[x = \frac{2}{5}\sec \theta \hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,\,\,dx = \frac{2}{5}\sec \theta \tan \theta \,d\theta \]

Ushbu almashtirishdan foydalanib, integral quyidagicha bo'ladi:

\[\begin{align*}\int{{\frac{{\sqrt {25{x^2} - 4} }}{x}\,dx}} & = \int{{\frac{{2\tan \theta }}{{\frac{2}{5}\sec \theta }}\,\left( {\frac{2}{5}\sec \theta \tan \theta } \right)d\theta }}\\ & = 2\int{{{{\tan }^2}\theta \,d\theta }}\end{align*}\]

Ushbu almashtirish orqali biz berilgan integralni trig funksiyalari ishtirok etgan integralga keltira oldik va biz bu muammolarni qanday bajarishni oldingi maqolada ko'rib chiqdik. bo'limi. Keling, integralni tugatamiz.

\[\begin{align*}\int{{\frac{{\sqrt {25{x^2} - 4} }}{x}\,dx}} & = 2\int{{{{\sec }^2}\theta - 1\,d\theta }}\\ & = 2\left( {\tan \theta - \theta } \right) + c\end{align*}\]

Shunday qilib, bizda integral uchun javob bor. Afsuski, \(x\)-da javob kerakli darajada berilmagan. Shunday qilib, biz javobimizni \(x\) shartlarida yozishimiz kerak. Biz buni bir nechta to'g'ri burchakli uchburchak triglari bilan qilishimiz mumkin. Bizning asl almashtirishimizdan bizda,

\[\sec \theta = \frac{{5x}}{2} = \frac{{{\mbox{hypotenuse}}}}{{{\mbox{adjacent}}}}\]

Bu quyidagi to'g'ri burchakli uchburchakni beradi.

Bundan shuni ko'rishimiz mumkinki,

\[\tan \theta = \frac{{\sqrt {25{x^2} - 4} }}{2}\]

Biz \(\theta \) bilan har qanday usullardan birida ishlay olamiz. Bizning almashtirishimizdan shuni ko'rishimiz mumkinki,

\[\theta = {\sec ^{ - 1}}\left( {\frac{{5x}}{2}} \right)\]

Bu \(\theta \) bilan ishlashning mutlaqo maqbul usuli bo'lsa-da, biz oltita mumkin bo'lgan teskari trig funksiyalaridan birini ishlatishimiz mumkin va sinus va kosinus ko'pchilikka tanish bo'lgan ikkita trig funksiyasi bo'lgani uchun biz odatda teskari sinus yoki teskari kosinusdan foydalanamiz. Bu holda biz teskari kosinusdan foydalanamiz.

\[\theta = {\cos ^{ - 1}}\left( {\frac{2}{{5x}}} \right)\]

Shunday qilib, bularning barchasi bilan integral bo'ladi,

\[\begin{align*}\int{{\frac{{\sqrt {25{x^2} - 4} }}{x}\,dx}} & = 2\left( {\frac{{\sqrt {25{x^2} - 4} }}{2} - {{\cos }^{ - 1}}\left( {\frac{2}{{5x}}} \right)} \right) + c\\ & = \sqrt {25{x^2} - 4} - 2{\cos ^{ - 1}}\left( {\frac{2}{{5x}}} \right) + c\end{align*}\]

Endi bizda \(x\) nuqtai nazaridan javob bor.

Bu erdagi cheklovlar almashtirishni o'zgartirmaydi, shuning uchun u bir xil bo'lib qoladi.

\[x = \frac{2}{5}\sec \theta \]

Ushbu almashtirishdan foydalanib, kvadrat ildiz hali ham kamayadi,

\[\sqrt {25{x^2} - 4} = 2\left| {\tan \theta } \right|\]

Biroq, oldingi misoldan farqli o'laroq, biz mutlaq qiymat satrlarini tashlab keta olmaymiz. Bu holda bizda integralda cheklovlar bor va shuning uchun biz mavjud bo'lgan \(\theta \) diapazonini aniqlash uchun chegaralar va almashtirishdan foydalanishimiz mumkin. Buni olganimizdan so'ng biz mutlaq qiymat satrlarini qanday tushirishni aniqlashimiz mumkin.

Mana \(\theta \) chegaralari va shuni yodda tutingki, agar siz trig tenglamalarni sekant bo'yicha yechishda yaxshi bo'lmasangiz, quyida qilganimizdek har doim kosinusga aylantira olasiz.

\[\begin{align*}x & = \frac{2}{5}\,\,\,:\,\,\,\frac{2}{5} = \frac{2}{5}\sec \theta = \frac{2}{5}\frac{1}{{\cos \theta }}\hspace{0.25in} \to \hspace{0.25in}\cos \theta = 1\hspace{0.25in}\,\, \Rightarrow \hspace{0.25in}\,\,\theta = {\cos ^{ - 1}}\left( 1 \right) = 0\\ x & = \frac{4}{5}\,\,\,:\,\,\,\frac{4}{5} = \frac{2}{5}\sec \theta = \frac{2}{5}\frac{1}{{\cos \theta }}\hspace{0.25in} \to \hspace{0.25in}\cos \theta = \frac{1}{2}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\theta = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{3}\end{align*}\]

Endi biz trig tenglamalarni yechishdan bilamizki, aslida biz foydalanishi mumkin bo'lgan cheksiz ko'p javoblar mavjud. Aslida, yuqoridagi ish uchun ko'proq "to'g'ri" javob,

\[\theta = 0 + 2\pi n = 2\pi n\hspace{0.25in}\& \hspace{0.25in} \theta = \frac{\pi }{3} + 2\pi n\hspace{0.25in} n = 0, \pm 1, \pm 2, \pm 3, \ldots \]

Xo'sh, qaysi birini ishlatishimiz kerak? Javob oddiy. Sekant triglarini almashtirish va chegaralarni o'zgartirishda biz doimo \(\theta \) teskari sekant diapazonida deb faraz qilamiz. Yoki,

\[{\mbox{If }}\theta = {\sec ^{ - 1}}\left( x \right)\,\,{\mbox{then}}\,\,0 \le \theta < \frac{\pi }{2}\,\,\text{yoki}\,\,\frac{\pi }{2} < \theta \le \pi \]

E'tibor bering, biz \(\theta = \frac{\pi }{2}\) dan qochishimiz kerak, chunki bu nuqtada sekant mavjud bo'lmaydi. Shuni ham yodda tutingki, \(\theta \) diapazoni sekant bo'yicha berilgan, garchi biz javoblarni olish uchun aslida teskari kosinusdan foydalanganmiz. Bu muammo bo'lmaydi, chunki teskari kosinus \(\theta = \frac{\pi }{2}\) ni berishi mumkin bo'lsa ham, biz uni yuqoridagi ishimizda hech qachon olmaymiz, chunki bu biz sekantning aniqlanmaganligidan boshlashimizni talab qiladi va bu chegaralarni o'zgartirganda sodir bo'lmaydi, chunki bu o'z navbatida chegaralardan birini ham aniqlanmagan bo'lishini talab qiladi!

Shunday qilib, yangi chegaralarni topishda bizga \(\theta \) ning barcha mumkin bo'lgan qiymatlari kerak emas edi, biz chegaralarni o'zgartirganimizda olingan teskari kosinus javoblariga muhtojmiz. Shuning uchun, agar biz \(\frac{2}{5} \le x \le \frac{4}{5}\) oralig'ida bo'lsak, \(\theta \) \(0 \le \theta \le \frac{\pi }{3}\) diapazonida bo'ladi va \(\theta \) diapazonining bu diapazonida tangensi ijobiy bo'ladi va shuning uchun biz mutlaq qiymat satrlarini tashlab qo'yishimiz mumkin.

Keling, almashtirishni qilaylik. E'tibor bering, ish oldingi misol bilan bir xil va shuning uchun uning ko'p qismi qoldirilgan. Biz oxirgi integralni tanlaymiz va keyin almashtirishni qilamiz.

\[\begin{align*}\int_{{\,\frac{2}{5}}}^{{\,\frac{4}{5}}}{{\frac{{\sqrt {25{x^2} - 4} }}{x}\,dx}} & = 2\int_{{\,0}}^{{\,\frac{\pi }{3}}}{{{{\sec }^2}\theta - 1\,d\theta }}\\ & = 2\left. {\left( {\tan \theta - \theta } \right)} \right|_0^{{\pi }/{3}\;}\\ & = 2\sqrt 3 - \frac{{2\pi }}{3}\end{align*}\]

E'tibor bering, cheklovlar tufayli biz muammoni hal qilish uchun to'g'ri burchakli uchburchakka murojaat qilishimiz shart emas edi.

Shunga qaramay, almashtirish va kvadrat ildiz birinchi ikkita misol bilan bir xil.

\[x = \frac{2}{5}\sec \theta \hspace{0.5in}\sqrt {25{x^2} - 4} = 2\left| {\tan \theta } \right|\]

Keling, ushbu muammo uchun \(\theta \) chegaralarini ko'rib chiqaylik.

\[\begin{align*}x & = - \frac{2}{5}\,\,\,:\,\,\, - \frac{2}{5} = \frac{2}{5}\sec \theta = \frac{2}{5}\frac{1}{{\cos \theta }}\hspace{0.25in} \to \hspace{0.25in}\cos \theta = - 1\hspace{0.25in}\,\, \Rightarrow \hspace{0.25in}\,\,\theta = {\cos ^{ - 1}}\left( { - 1} \right) = \pi \\ x & = - \frac{4}{5}\,\,\,:\,\,\, - \frac{4}{5} = \frac{2}{5}\sec \theta = \frac{2}{5}\frac{1}{{\cos \theta }}\hspace{0.25in} \to \hspace{0.25in}\cos \theta = - \frac{1}{2}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\theta = {\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = \frac{{2\pi }}{3}\end{align*}\]

Esda tutingki, chegaralarni konvertatsiya qilishda biz teskari sekant/kosinus natijalaridan foydalanamiz. Shunday qilib, \(x\) diapazoni uchun bizda \(\frac{{2\pi }}{3} \le \theta \le \pi \) bor va bu diapazonda \(\theta \) tangensi manfiy va shuning uchun bu holda biz mutlaq qiymat satrlarini tashlab qo'yishimiz mumkin, lekin buning uchun minus belgisini qo'shishimiz kerak bo'ladi. Boshqacha aytganda,

\[\sqrt {25{x^2} - 4} = - 2\tan \theta \]

Demak, bu integratsiya jarayonida qiladigan yagona o‘zgarish bu integral oldiga minus belgisi qo‘yishdir. U holda integral,

\[\begin{align*}\int_{{\, - \,\frac{4}{5}}}^{{\, - \,\frac{2}{5}}}{{\frac{{\sqrt {25{x^2} - 4} }}{x}\,dx}} & = - 2\int_{{\,\frac{{2\pi }}{3}}}^{{\,\pi }}{{{{\sec }^2}\theta - 1\,d\theta }}\\ & = - 2\left. {\left( {\tan \theta - \theta } \right)} \right|_{{{2\pi }}/{3}\;}^\pi \\ & = \frac{{2\pi }}{3} - 2\sqrt 3 \end{align*}\]

Endi, bu muammoning ildizi ostidagi atamalar avvalgilari bilan (deyarli) bir xil ko'rinadi, shuning uchun keling, bir xil turdagi almashtirishni sinab ko'raylik va u bu erda ham ishlaydimi yoki yo'qligini bilib olaylik.

\[x = 3\sec \theta \]

Ushbu almashtirishdan foydalanib, kvadrat ildiz quyidagicha bo'ladi:

\[\sqrt {9 - {x^2}} = \sqrt {9 - 9{{\sec }^2}\theta } = 3\sqrt {1 - {{\sec }^2}\theta } = 3\sqrt { - {{\tan }^2}\theta } \]

Shunday qilib, bu almashtirishdan foydalanib, biz kvadrat ildiz ostida salbiy miqdorga ega bo'lamiz (tangens kvadrati har doim ijobiy bo'ladi) va bu muammo bo'ladi. Ushbu almashtirishdan foydalanish murakkab qiymatlarni beradi va biz buni xohlamaymiz. Shunday qilib, almashtirish uchun sekantdan foydalanish ishlamaydi.

Biroq, quyidagi almashtirish (va differentsial) ishlaydi.

\[x = 3\sin \theta \hspace{0.5in}\hspace{0.25in}dx = 3\cos \theta \,d\theta \]

Ushbu almashtirish bilan kvadrat ildiz,

\[\sqrt {9 - {x^2}} = 3\sqrt {1 - {{\sin }^2}\theta } = 3\sqrt {{{\cos }^2}\theta } = 3\left| {\cos \theta } \right| = 3\cos \theta \]

Biz mutlaq qiymat satrlarini tashlab qo'yishga muvaffaq bo'ldik, chunki biz noaniq integral qilyapmiz va shuning uchun biz hamma narsani ijobiy deb hisoblaymiz.

Integral endi,

\[\begin{align*}\int{{\frac{1}{{{x^4}\sqrt {9 - {x^2}} }}\,dx}} & = \int{{\frac{1}{{81{{\sin }^4}\theta \left( {3\cos \theta } \right)}}\,3\cos \theta \,d\theta }}\\ & = \frac{1}{{81}}\int{{\frac{1}{{{{\sin }^4}\theta }}\,d\theta }}\\ & = \frac{1}{{81}}\int{{{{\csc }^4}\theta \,d\theta }}\end{align*}\]

Oldingida biz sekantdagi ko'rsatkichi juft bo'lgan integrallar bilan qanday ishlashni ko'rib chiqdik va kosekantlar sekantlarga juda o'xshash bo'lganligi sababli, biz bunga o'xshash narsani qilishimiz kerak.

Mana integral.

\[\begin{align*}\int{{\frac{1}{{{x^4}\sqrt {9 - {x^2}} }}\,dx}} & = \frac{1}{{81}}\int{{{{\csc }^2}\theta \,{{\csc }^2}\theta \,d\theta }}\\ & = \frac{1}{{81}}\int{{\left( {{{\cot }^2}\theta + 1} \right)\,{{\csc }^2}\theta \,d\theta }}\hspace{0.5in}u = \cot \theta \\ & = - \frac{1}{{81}}\int{{{u^2} + 1\,du}}\\ & = - \frac{1}{{81}}\left( {\frac{1}{3}{{\cot }^3}\theta + \cot \theta } \right) + c\end{align*}\]

Endi biz to'g'ri burchakli uchburchak yordamida \(x\)-ga qaytishimiz kerak. Mana bu muammo uchun to'g'ri uchburchak va bu muammo uchun trig funktsiyalari.

\[\sin \theta = \frac{x}{3}\hspace{0.5in}\cot \theta = \frac{{\sqrt {9 - {x^2}} }}{x}\]

U holda integral,

\[\begin{align*}\int{{\frac{1}{{{x^4}\sqrt {9 - {x^2}} }}\,dx}} & = - \frac{1}{{81}}\left( {\frac{1}{3}{{\left( {\frac{{\sqrt {9 - {x^2}} }}{x}} \right)}^3} + \frac{{\sqrt {9 - {x^2}} }}{x}} \right) + c\\ & = - \frac{{{{\left( {9 - {x^2}} \right)}^{\frac{3}{2}}}}}{{243{x^3}}} - \frac{{\sqrt {9 - {x^2}} }}{{81x}} + c\end{align*}\]

Birinchidan, aniq yozilmagan bo'lsa ham, bu muammoda haqiqatan ham kvadrat ildiz borligiga e'tibor bering. Ildizni ko'rish uchun narsalarni biroz qayta yozamiz.

\[{\left( {36{x^2} + 1} \right)^{\frac{3}{2}}} = {\left( {{{\left( {36{x^2} + 1} \right)}^{\frac{1}{2}}}} \right)^3} = {\left( {\sqrt {36{x^2} + 1} } \right)^3}\]

Ildiz ostidagi bu atamalar biz oldingi misollarda ko'rgan shaklda emas. Bu erda biz ushbu ildiz uchun almashtirishdan foydalanamiz.

\[x = \frac{1}{6}\tan \theta \hspace{0.5in}\hspace{0.25in}dx = \frac{1}{6}{\sec ^2}\theta \,d\theta \]

Ushbu almashtirish bilan maxraj bo'ladi,

\[{\left( {\sqrt {36{x^2} + 1} } \right)^3} = {\left( {\sqrt {{{\tan }^2}\theta + 1} } \right)^3} = {\left( {\sqrt {{{\sec }^2}\theta } } \right)^3} = {\left| {\sec \theta } \right|^3}\]

Endi bizda cheklovlar borligi sababli ularni \(\theta \) ga aylantirishimiz kerak, shunda biz mutlaq qiymat satrlarini qanday tushirishni aniqlashimiz mumkin.

\[\begin{align*}x & = 0\,\,\,:\,\,\,0 = \frac{1}{6}\tan \theta \hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}\,\,\theta = {\tan ^{ - 1}}\left( 0 \right) = 0\\ x &= \frac{1}{6}\,\,\,:\,\,\,\frac{1}{6} = \frac{1}{6}\tan \theta \hspace{0.25in}\, \Rightarrow \hspace{0.25in}\,\,\theta = {\tan ^{ - 1}}\left( 1 \right) = \frac{\pi }{4}\end{align*}\]

Oldingi ikkita holatda bo'lgani kabi, bu erda chegaralarni o'zgartirishda biz teskari tangens natijalaridan foydalanamiz yoki,

\[{\mbox{If }}\theta = {\tan ^{ - 1}}\left( x \right)\,\,{\mbox{then}}\,\, - \frac{\pi }{2} < \theta < \frac{\pi }{2}\]

Shunday qilib, \(\theta \) ning ushbu diapazonida sekant musbat va shuning uchun biz mutlaq qiymat satrlarini tashlab yuborishimiz mumkin.

Mana integral,

\[\begin{align*}\int_{{\,0}}^{{\,\frac{1}{6}}}{{\frac{{{x^5}}}{{{{\left( {36{x^2} + 1} \right)}^{\frac{3}{2}}}}}\,dx}} & = \int_{{\,0}}^{{\,\frac{\pi }{4}}}{{\frac{{\frac{1}{{7776}}{{\tan }^5}\theta }}{{{{\sec }^3}\theta }}\,\left( {\frac{1}{6}{{\sec }^2}\theta } \right)d\theta }}\\ & = \frac{1}{{46656}}\int_{{\,0}}^{{\,\frac{\pi }{4}}}{{\frac{{{{\tan }^5}\theta }}{{\sec \theta }}d\theta }}\end{align*}\]

Bu nuqtadan harakat qilishning bir necha yo'li mavjud. Odatda tangensdagi g'alati ko'rsatkich bilan biz ulardan birini ajratib olib, sekantlarga aylantiramiz. Biroq, bu bizda hisoblagichda mavjud bo'lmagan sekantga ega bo'lishini talab qiladi. Shuning uchun, buni qilishning eng yaxshi usuli integrandni sinuslar va kosinuslarga aylantirish kabi ko'rinadi.

\[\begin{align*}\int_{{\,0}}^{{\,\frac{1}{6}}}{{\frac{{{x^5}}}{{{{\left( {36{x^2} + 1} \right)}^{\frac{3}{2}}}}}\,dx}} & = \frac{1}{{46656}}\int_{{\,0}}^{{\,\frac{\pi }{4}}}{{\frac{{{{\sin }^5}\theta }}{{{{\cos }^4}\theta }}d\theta }}\\ & = \frac{1}{{46656}}\int_{{\,0}}^{{\,\frac{\pi }{4}}}{{\frac{{{{\left( {1 - {{\cos }^2}\theta } \right)}^2}}}{{{{\cos }^4}\theta }}\sin \theta d\theta }}\end{align*}\]

Endi biz \(u = \cos \theta \) almashtirishdan foydalanishimiz mumkin va biz chegaralarni ham o'zgartirishimiz mumkin.

\[\begin{align*}\theta & = 0 & \hspace{0.75in} & u = \cos 0 = 1\\ \theta & = \frac{\pi }{4}& \hspace{0.75in} & u = \cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2}\end{align*}\]

U holda integral,

\[\begin{align*}\int_{{\,0}}^{{\,\frac{1}{6}}}{{\frac{{{x^5}}}{{{{\left( {36{x^2} + 1} \right)}^{\frac{3}{2}}}}}\,dx}} & = - \frac{1}{{46656}}\int_{{\,1}}^{{\,\frac{{\sqrt 2 }}{2}}}{{{u^{ - 4}} - 2{u^{ - 2}} + 1\,du}}\\ & = - \frac{1}{{46656}}\left. {\left( { - \frac{1}{{3{u^3}}} + \frac{2}{u} + u} \right)} \right|_1^{\frac{{\sqrt 2 }}{2}}\\ & = \frac{1}{{17496}} - \frac{{11\sqrt 2 }}{{279936}}\end{align*}\]

Bunday holda, ildiz ostidagi miqdor biz yuqorida ko'rib chiqqan holatlarning hech biriga to'g'ri kelmaydi va aslida biz oldingi misollarda ko'rgan shakllarning hech birida emas. Ammo shuni yodda tutingki, agar kvadratni kvadrat bo'yicha to'ldirsak, uni yuqoridagi integrallarga o'xshash qilishimiz mumkin.

Kvadratni to'ldirish uchun \({x^2}\) oldida bitta koeffitsient talab qilinishini unutmang. Buni olganimizdan so'ng, biz \(x\) koeffitsientining yarmini olamiz, kvadratga olamiz va keyin uni miqdorga qo'shamiz va ayiramiz. Mana bu muammoning kvadratini to'ldirish.

\[2\left( {{x^2} - 2x - \frac{7}{2}} \right) = 2\left( {{x^2} - 2x + 1 - 1 - \frac{7}{2}} \right) = 2\left( {{{\left( {x - 1} \right)}^2} - \frac{9}{2}} \right) = 2{\left( {x - 1} \right)^2} - 9\]

Shunday qilib, ildiz bo'ladi,

\[\sqrt {2{x^2} - 4x - 7} = \sqrt {2{{\left( {x - 1} \right)}^2} - 9} \]

Endi bu (juda) noaniq ko'rinadi \({\sec ^2}\theta - 1\) (i.e. bir narsa kvadrat minus raqam) bundan mustasno, bizda kvadrat terminda murakkabroq narsa bor. To'g'ri, biz hali ham sekant almashtirishni amalga oshirishimiz mumkin va u deyarli xuddi shunday ishlaydi.

\[x - 1 = \frac{3}{{\sqrt 2 }}\sec \theta \hspace{0.25in}x = 1 + \frac{3}{{\sqrt 2 }}\sec \theta \hspace{0.25in}dx = \frac{3}{{\sqrt 2 }}\sec \theta \tan \theta \,d\theta \]

Ushbu almashtirishdan foydalanib, ildiz quyidagiga kamayadi:

\[\sqrt {2{x^2} - 4x - 7} = \sqrt {2{{\left( {x - 1} \right)}^2} - 9} = \sqrt {9{{\sec }^2}\theta - 9} = 3\sqrt {{{\tan }^2}\theta } = 3\left| {\tan \theta } \right| = 3\tan \theta \]

E'tibor bering, biz mutlaq qiymat satrlarini tashlab yuborishimiz mumkin, chunki biz noaniq integral qilamiz. Mana integral.

\[\begin{align*}\int{{\frac{x}{{\sqrt {2{x^2} - 4x - 7} }}\,dx}} & = \int{{\frac{{1 + \frac{3}{{\sqrt 2 }}\sec \theta }}{{3\tan \theta }}\left( {\frac{3}{{\sqrt 2 }}\sec \theta \tan \theta } \right)\,d\theta }}\\ & = \int{{\frac{1}{{\sqrt 2 }}\sec \theta + \frac{3}{2}{{\sec }^2}\theta \,d\theta }}\\ & = \frac{1}{{\sqrt 2 }}\ln \left| {\sec \theta + \tan \theta } \right| + \frac{3}{2}\tan \theta + c\end{align*}\]

Va bu muammo uchun to'g'ri uchburchak.

\[\sec \theta = \frac{{\sqrt 2 \left( {x - 1} \right)}}{3}\hspace{0.5in}\tan \theta = \frac{{\sqrt {2{x^2} - 4x - 7} }}{3}\]

U holda integral,

\[\int{{\frac{x}{{\sqrt {2{x^2} - 4x - 7} }}\,dx}} = \frac{1}{{\sqrt 2 }}\ln \left| {\frac{{\sqrt 2 \left( {x - 1} \right)}}{3} + \frac{{\sqrt {2{x^2} - 4x - 7} }}{3}} \right| + \frac{{\sqrt {2{x^2} - 4x - 7} }}{2} + c\]

Bu ushbu bo'limdagi boshqa muammolarga o'xshamaydi. Biroq shunday. Buni ko'rish uchun birinchi navbatda shuni ta'kidlashimiz kerak,

\[{{\bf{e}}^{2x}} = {\left( {{{\bf{e}}^x}} \right)^2}\]

Buni payqagach, biz quyidagi standart Calculus I almashtirishdan foydalanishimiz mumkin.

\[u = {{\bf{e}}^x}\hspace{0.5in}du = {{\bf{e}}^x}\,dx\]

Biroq, biz differentsial ishda biroz ehtiyot bo'lishimiz kerak. Bizda ildiz oldida faqat \({{\bf{e}}^x}\) yo'q. Buning o'rniga bizda \({{\bf{e}}^{4x}}\) bor. Shunday qilib, biz ulardan birini farqlash uchun olib tashlashimiz va qolganlarida almashtirishdan foydalanishimiz kerak. Mana almashtirish ishi.

\[\begin{align*}\int{{{{\bf{e}}^{4x}}\sqrt {1 + {{\bf{e}}^{2x}}} \,dx}} & = \int{{{{\bf{e}}^{3x}}{{\bf{e}}^x}\sqrt {1 + {{\bf{e}}^{2x}}} \,dx}}\\ & = \int{{{{\left( {{{\bf{e}}^x}} \right)}^3}\sqrt {1 + {{\left( {{{\bf{e}}^x}} \right)}^2}} \,{{\bf{e}}^x}dx}} = \int{{{u^3}\sqrt {1 + {u^2}} \,du}}\end{align*}\]

Bu endi juda aniq trigni almashtirish (umid qilamanki). Ildiz ostidagi miqdor deyarli \(1 + {\tan ^2}\theta \) ga o'xshaydi va shuning uchun biz tangens almashtirishdan foydalanishimiz mumkin. Mana o'sha ish.

\[u = \tan \theta \hspace{0.25in}du = {\sec ^2}\theta \,d\theta \hspace{0.5in}\sqrt {1 + {u^2}} = \sqrt {1 + {{\tan }^2}\theta } = \sqrt {{{\sec }^2}\theta } = \left| {\sec \theta } \right|\]

Biz noaniq integralni bajarayotganimiz uchun sekant musbat deb faraz qilishimiz va mutlaq qiymat satrlarini tushirishimiz mumkin. Ushbu almashtirishni integralga qo'llash,

\[\int{{{{\bf{e}}^{4x}}\sqrt {1 + {{\bf{e}}^{2x}}} \,dx}} = \int{{{{\tan }^3}\theta \left( {\sec \theta } \right)\,\left( {{{\sec }^2}\theta } \right)d\theta }} = \int{{\tan^3 \theta \sec^3\theta d\theta }}\]

Biz bu integralni birozdan keyin tugatamiz. Bunga kirishdan oldin, yuqoridagi almashtirishlarni amalga oshirishning "tezroq" (ammo aniq bo'lmasa ham) usuli bor. Keling, avval buni yoritib olaylik, keyin qaytib kelib, integral bilan ishlashni tugatamiz.

Hisoblash Idagi \(u\) va trig almashtirish bir xil \(u\) ekanligini payqashimiz mumkin va shuning uchun ularni quyidagi almashtirishga birlashtira olamiz.

\[{{\bf{e}}^x} = \tan \theta \]

Keyin biz differentsialni hisoblashimiz mumkin. Shuni yodda tutingki, biz faqat ikkala tomonni farqlashimiz va keyin \(dx\) yoki \(d\theta \)-ni tegishli tomonga yopishtirishimizdir. Buni qilish,

\[{{\bf{e}}^x}\,dx = {\sec ^2}\theta \,d\theta \]

Ushbu almashtirish bilan kvadrat ildiz bo'ladi,

\[\sqrt {1 + {{\bf{e}}^{2x}}} = \sqrt {1 + {{\left( {{{\bf{e}}^x}} \right)}^2}} = \sqrt {1 + {{\tan }^2}\theta } = \sqrt {{{\sec }^2}\theta } = \left| {\sec \theta } \right| = \sec \theta \]

Shunga qaramay, biz mutlaq qiymat satrlarini tashlab qo'yishimiz mumkin, chunki biz noaniq integral qilyapmiz. Keyin integral bo'ladi,

\[\begin{align*}\int{{{{\bf{e}}^{4x}}\sqrt {1 + {{\bf{e}}^{2x}}} \,dx}} & = \int{{{{\bf{e}}^{3x}}{{\bf{e}}^x}\sqrt {1 + {{\bf{e}}^{2x}}} \,dx}}\\ & = \int{{{{\left( {{{\bf{e}}^x}} \right)}^3}\sqrt {1 + {{\bf{e}}^{2x}}} \,\left( {{{\bf{e}}^x}} \right)dx}}\\ & = \int{{{{\tan }^3}\theta \left( {\sec \theta } \right)}}\left( {{{\sec }^2}\theta } \right)\,d\theta = \int{{{{\tan }^3}\theta \,{{\sec }^3}\theta d\theta }}\end{align*}\]

Shunday qilib, kamroq ish bilan bir xil integral. Biroq, bu ikkita almashtirishni bitta almashtirishga birlashtira olishingizni talab qiladi. Ushbu turdagi muammolarni qanday hal qilish sizga bog'liq, lekin agar siz bitta almashtirishdan o'zingizni qulay his qilmasangiz (va buni qilmasangiz, hech qanday yomon narsa yo'q!), ikkita alohida almashtirishni bajaring. Yagona almashtirish usuli faqat buni amalga oshirish mumkinligini ko'rsatish uchun berilgan, shunda ikkala turdagi almashtirishda ham haqiqatan ham qulay bo'lganlar ishni biroz tezroq bajarishlari mumkin.

Endi integral ishni tugatamiz.

\[\begin{align*}\int{{{{\bf{e}}^{4x}}\sqrt {1 + {{\bf{e}}^{2x}}} \,dx}} & = \int{{{{\tan }^3}\theta \,{{\sec }^3}\theta d\theta }}\\ & = \int{{\left( {{{\sec }^2}\theta - 1} \right)}}{\sec ^2}\theta \,\sec \theta \tan \theta \,d\theta \hspace{0.5in}v = \sec \theta \\ & = \int{{{v^4} - {v^2}\,dv}}\\ & = \frac{1}{5}{\sec ^5}\theta - \frac{1}{3}{\sec ^3}\theta + c\end{align*}\]

Mana bu integral uchun to'g'ri burchakli uchburchak.

\[\tan \theta = \frac{{{{\bf{e}}^x}}}{1}\hspace{0.5in}\sec \theta = \frac{{\sqrt {1 + {{\bf{e}}^{2x}}} }}{1} = \sqrt {1 + {{\bf{e}}^{2x}}} \]

U holda integral,

\[\int{{{{\bf{e}}^{4x}}\sqrt {1 + {{\bf{e}}^{2x}}} \,dx}} = \frac{1}{5}{\left( {1 + {{\bf{e}}^{2x}}} \right)^{\frac{5}{2}}} - \frac{1}{3}{\left( {1 + {{\bf{e}}^{2x}}} \right)^{\frac{3}{2}}} + c\]

Bu chalkash muammo edi, lekin biz keyingi bo'limlarda ba'zida bunday integral turlarini ko'ramiz, shuning uchun kamida bittasini ko'rganingizga ishonch hosil qilishimiz kerak edi.