7.6-bo'lim: Integrals involving quadratics
Shu nuqtaga qadar biz kvadratiklarni o'z ichiga olgan bir nechta integrallarni ko'rdik. A couple of examples are,
\[\int{{\frac{x}{{{x^2} \pm a}}\,dx}} = \frac{1}{2}\ln \left| {{x^2} \pm a} \right| + c\hspace{0.5in}\int{{\frac{1}{{{x^2} + {a^2}}}\,dx = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)+c}}\]Shuningdek, biz \(\sqrt {{b^2}{x^2} - {a^2}} \), \(\sqrt {{a^2} - {b^2}{x^2}} \) va \(\sqrt {{a^2} + {b^2}{x^2}} \) ishtirokidagi integrallarni triglarni almashtirish bilan bajarish mumkinligini ko‘rdik.
E'tibor bering, bu integrallarning barchasida \(x\) atamasi yo'q edi. Ularning barchasi faqat kvadratik haddan va doimiydan iborat.
Umumiy kvadratiklarni o'z ichiga olgan ba'zi integrallarni bajarish juda oson. Masalan, quyidagi integralni tez almashtirish bilan bajarish mumkin.
\[\begin{align*}\int{{\frac{{2x + 3}}{{4{x^2} + 12x - 1}}\,dx}} & = \frac{1}{4}\int{{\frac{1}{u}\,du}}\hspace{0.5in}u = 4{x^2} + 12x - 1\hspace{0.25in}du = 4\left( {2x + 3} \right)dx\\ & = \frac{1}{4}\ln \left| {4{x^2} + 12x - 1} \right| + c\end{align*}\]Kvadratlar bilan ba'zi integrallarni qisman kasrlar bilan bajarish mumkin. Masalan,
\[\int{{\frac{{10x - 6}}{{3{x^2} + 16x + 5}}\,dx}} = \int{{\frac{4}{{x + 5}} - \frac{2}{{3x + 1}}\,dx}} = 4\ln \left| {x + 5} \right| - \frac{2}{3}\ln \left| {3x + 1} \right| + c\]Afsuski, bu usullar ko'p integrallarda ishlamaydi. Oddiy almashtirish faqat ayiruvchi maxraj hosilasining doimiy karrali bo'lsa va qisman kasrlar faqat maxrajni ko'paytiruvchi bo'lsa ishlaydi.
Ushbu bo'limning mavzusi, agar biz shu nuqtaga qadar ko'rib chiqqan usullar ishlamasa, kvadratiklar bilan bog'liq integrallar bilan qanday ishlash kerak.
Orqaga Trigni almashtirish bo'limda biz umumiy kvadratikga ega bo'lgan kvadrat ildizlar bilan qanday ishlashni ko'rdik. Keling, shunga o'xshash boshqasini tezda ko'rib chiqaylik, chunki bunday integralni bajarish g'oyasi ushbu bo'limdagi boshqa integrallar uchun bizga kerak bo'lgan narsadir.
Yechim
Trigni almashtirish bo'limidan eslaylikki, bu erda trigni almashtirishni amalga oshirish uchun birinchi navbatda kvadratdagi kvadratni bajarishimiz kerak edi. Bu beradi,
\[{x^2} + 4x + 5 = {x^2} + 4x + 4 - 4 + 5 = {\left( {x + 2} \right)^2} + 1\]Kvadratni to'ldirgandan so'ng, integral bo'ladi,
\[\int{{\sqrt {{x^2} + 4x + 5} \,dx}} = \int{{\sqrt {{{\left( {x + 2} \right)}^2} + 1} \,dx}}\]Buni amalga oshirgandan so'ng, biz kerakli trig o'rnini aniqlashimiz mumkin. Mana,
\[x + 2 = \tan \theta \hspace{0.5in}x = \tan \theta - 2\hspace{0.5in}dx = {\sec ^2}\theta \,d\theta \] \[\sqrt {{{\left( {x + 2} \right)}^2} + 1} = \sqrt {{{\tan }^2}\theta + 1} = \sqrt {{{\sec }^2}\theta } = \left| {\sec \theta } \right| = \sec \theta \]Eslatib o'tamiz, biz noaniq integralni bajarayotganimiz uchun mutlaq qiymat satrlarini tashlab yuborishimiz mumkin. Ushbu almashtirishdan foydalanib, integral quyidagicha bo'ladi:
\[\begin{align*}\int{{\sqrt {{x^2} + 4x + 5} \,dx}} & = \int{{{{\sec }^3}\theta \,d\theta }}\\ & = \frac{1}{2}\left( {\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|} \right) + c\end{align*}\]Quyidagi to'g'ri burchakli uchburchak bilan integralni tugatishimiz mumkin.
\[\tan \theta = \frac{{x + 2}}{1}\hspace{0.5in}\sec \theta = \frac{{\sqrt {{x^2} + 4x + 5} }}{1} = \sqrt {{x^2} + 4x + 5} \]
Shunday qilib, kvadratni to'ldirib, biz umumiy kvadratikga ega bo'lgan integralni oldik va uni ma'lum integrasiya texnikasidan foydalanishga imkon beradigan shaklga aylantira oldik.
Davom etishdan oldin kvadratni to'ldirishni tez ko'rib chiqaylik. Mana biz ishlatadigan kvadrat formulani to'ldirishning umumiy.
\[{x^2} + bx + c = {x^2} + bx + {\left( {\frac{b}{2}} \right)^2} - {\left( {\frac{b}{2}} \right)^2} + c = {\left( {x + \frac{b}{2}} \right)^2} + c - \frac{{{b^2}}}{4}\]Bu har doim umumiy kvadratikni oladi va uni kvadrat va doimiy atama bilan yozadi.
Eslatib o'tamiz, buni amalga oshirish uchun biz \({x^2}\) oldida bitta koeffitsientga ega bo'lishimiz kerak. Agar yo'q bo'lsa, kvadratni to'ldirishdan oldin koeffitsientni hisobga olishimiz kerak. Boshqacha aytganda,
\[a{x^2} + bx + c = a \;\; {\underbrace { \left( {x^2} + \frac{b}{a}x + \frac{c}{a}\right)}_{\scriptstyle{\mbox{complete the }}\atop\scriptstyle{\mbox{square on this!}}}} \]Endi keling, kvadratni to'ldirish biz hozir bajara olmaydigan integrallarni bajarish uchun qanday ishlatilishini ko'rib chiqamiz.
Yechim
OK, bu faktor emas, shuning uchun qisman fraktsiyalar bu ishda ishlamaydi. Xuddi shunday, surat faqat "1" bo'lgani uchun biz \(u = 2{x^2} - 3x + 2\) o'rnini ishlata olmaymiz. Xo'sh, keling, maxrajdagi kvadratni to'ldirsak nima bo'lishini ko'rib chiqamiz.
\[\begin{align*}2{x^2} - 3x + 2 & = 2\left( {{x^2} - \frac{3}{2}x + 1} \right)\\ & = 2\left( {{x^2} - \frac{3}{2}x + \frac{9}{{16}} - \frac{9}{{16}} + 1} \right)\\ & = 2\left( {{{\left( {x - \frac{3}{4}} \right)}^2} + \frac{7}{{16}}} \right)\end{align*}\]Bu bilan integral,
\[\int{{\frac{1}{{2{x^2} - 3x + 2}}\,dx}} = \frac{1}{2}\int{{\frac{1}{{{{\left( {x - \frac{3}{4}} \right)}^2} + \frac{7}{{16}}}}\,dx}}\]Endi bu unchalik katta o'zgarish bo'lib tuyulmasligi mumkin. Biroq, endi biz quyidagi almashtirishdan foydalanishimiz mumkinligiga e'tibor bering.
\[u = x - \frac{3}{4}\hspace{0.5in}du = dx\]va integral endi,
\[\int{{\frac{1}{{2{x^2} - 3x + 2}}\,dx}} = \frac{1}{2}\int{{\frac{1}{{{u^2} + \frac{7}{{16}}}}\,du}}\]Endi biz bu teskari tangens ekanligini ko'rishimiz mumkin! Shunday qilib, bo'lim boshidan formuladan foydalanib, biz olamiz,
\[\begin{align*}\int{{\frac{1}{{2{x^2} - 3x + 2}}\,dx}} & = \frac{1}{2}\left( {\frac{4}{{\sqrt 7 }}} \right){\tan ^{ - 1}}\left( {\frac{{4u}}{{\sqrt 7 }}} \right) + c\\ & = \frac{2}{{\sqrt 7 }}{\tan ^{ - 1}}\left( {\frac{{4x - 3}}{{\sqrt 7 }}} \right) + c\end{align*}\]Yechim
Bu misol avvalgisidan bir oz farq qiladi. Bu holda bizda hisobda \(x\) bor, biroq hisoblagich hali ham maxrajning hosilasining ko'paytmasi emas va shuning uchun oddiy Hisob I o'rnini bosish ishlamaydi.
Shunday qilib, keling, maxrajdagi kvadratni yana to'ldiramiz va nima olishimizni ko'raylik,
\[{x^2} + 10x + 28 = {x^2} + 10x + 25 - 25 + 28 = {\left( {x + 5} \right)^2} + 3\]Kvadratni to'ldirgandan so'ng, integral bo'ladi,
\[\int{{\frac{{3x - 1}}{{{x^2} + 10x + 28}}\,dx}} = \int{{\frac{{3x - 1}}{{{{\left( {x + 5} \right)}^2} + 3}}\,dx}}\]Shu nuqtada biz avvalgi misolda qilgan almashtirish turidan foydalanishimiz mumkin. Yagona haqiqiy farq shundaki, biz almashtirishni hisoblagichga ham ulaganimizga ishonch hosil qilishimiz kerak.
\[u = x + 5\hspace{0.5in}x = u - 5\hspace{0.5in}dx = du\] \[\begin{align*}\int{{\frac{{3x - 1}}{{{x^2} + 10x + 28}}\,dx}} & = \int{{\frac{{3\left( {u - 5} \right) - 1}}{{{u^2} + 3}}\,du}}\\ & = \int{{\frac{{3u}}{{{u^2} + 3}} - \frac{{16}}{{{u^2} + 3}}\,du}}\\ & = \frac{3}{2}\ln \left| {{u^2} + 3} \right| - \frac{{16}}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{u}{{\sqrt 3 }}} \right) + c\\ & = \frac{3}{2}\ln \left| {{{\left( {x + 5} \right)}^2} + 3} \right| - \frac{{16}}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{x + 5}}{{\sqrt 3 }}} \right) + c\end{align*}\]Shunday qilib, umuman olganda, shakldagi integral bilan ishlashda,
\[\begin{equation}\int{{\frac{{Ax + B}}{{a{x^2} + bx + c}}\,dx}} \label{eq:eq1} \end{equation}\]Bu erda biz maxraj koeffitsientga ega emasligini va hisoblagich maxraj hosilasining doimiy ko'paytmasi emasligini taxmin qilamiz. Bunday hollarda biz maxrajdagi kvadratni to'ldiramiz va keyin hisoblagichning aniq shakliga qarab teskari tangens va/yoki logarifm hosil qiladigan almashtirishni amalga oshiramiz.
Keling, \(\eqref{eq:eq1}\) bilan bir xil umumiy shaklda bo'lgan bir nechta integrallarni ko'rib chiqaylik, bundan tashqari maxraj ham darajaga ko'tariladi. Boshqacha qilib aytganda, integrallarni shaklda ko'rib chiqaylik,
\[\begin{equation}\int{{\frac{{Ax + B}}{{{{\left( {a{x^2} + bx + c} \right)}^n}}}\,dx}} \label{eq:eq2} \end{equation}\]Yechim
Ko'pincha bu integral oldingi ikkitasi bilan bir xil ishlaydi, faqat yo'lda sodir bo'ladigan istisno. Shunday qilib, keling, kvadratni maxrajdagi kvadratni to'ldirishdan boshlaylik.
\[{x^2} - 6x + 11 = {x^2} - 6x + 9 - 9 + 11 = {\left( {x - 3} \right)^2} + 2\]U holda integral,
\[\int{{\frac{x}{{{{\left( {{x^2} - 6x + 11} \right)}^3}}}\,dx}} = \int{{\frac{x}{{{{\left[ {{{\left( {x - 3} \right)}^2} + 2} \right]}^3}}}\,dx}}\]Endi biz avvalgi ikkita misolda shu nuqtada ishlatgan almashtirishdan foydalanamiz.
\[u = x - 3\hspace{0.5in}x = u + 3\hspace{0.5in}dx = du\] \[\begin{align*}\int{{\frac{x}{{{{\left( {{x^2} - 6x + 11} \right)}^3}}}\,dx}} & = \int{{\frac{{u + 3}}{{{{\left( {{u^2} + 2} \right)}^3}}}\,du}}\\ & = \int{{\frac{u}{{{{\left( {{u^2} + 2} \right)}^3}}}\,du}} + \int{{\frac{3}{{{{\left( {{u^2} + 2} \right)}^3}}}\,du}}\end{align*}\]Endi bu erda farqlar paydo bo'la boshlaydi. Birinchi integral \(v = {u^2} + 2\) almashtirish bilan amalga oshirilishi mumkin va unchalik qiyin emas. Biroq, ikkinchi integralni birinchi integralda qo'llanilgan almashtirish bilan bajarib bo'lmaydi va u teskari tangens emas.
Ma'lum bo'lishicha, trigni almashtirish ikkinchi integralda yaxshi ishlaydi va u biz qilgandek bo'ladi. qachon muammoda kvadrat ildizlarimiz bor edi.
\[u = \sqrt 2 \tan \theta \hspace{0.5in}du = \sqrt 2 {\sec ^2}\theta \,d\theta \]Ushbu ikkita almashtirish bilan integrallar,
\[\begin{align*}\int{{\frac{x}{{{{\left( {{x^2} - 6x + 11} \right)}^3}}}\,dx}} & = \frac{1}{2}\int{{\frac{1}{{{v^3}}}\,dv}} + \int{{\frac{3}{{{{\left( {2{{\tan }^2}\theta + 2} \right)}^3}}}\,\left( {\sqrt 2 {{\sec }^2}\theta } \right)d\theta }}\\ & = - \frac{1}{4}\frac{1}{{{v^2}}} + \int{{\frac{{3\sqrt 2 {{\sec }^2}\theta }}{{8{{\left( {{{\tan }^2}\theta + 1} \right)}^3}}}\,d\theta }}\\ & = - \frac{1}{4}\frac{1}{{{{\left( {{u^2} + 2} \right)}^2}}} + \frac{{3\sqrt 2 }}{8}\int{{\frac{{{{\sec }^2}\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^3}}}\,d\theta }}\\ & = - \frac{1}{4}\frac{1}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{{3\sqrt 2 }}{8}\int{{\frac{1}{{{{\sec }^4}\theta }}\,d\theta }}\\ & = - \frac{1}{4}\frac{1}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{{3\sqrt 2 }}{8}\int{{{{\cos }^4}\theta \,d\theta }}\end{align*}\]OK, hozir bizda qolgan integral uchun ikkita variant bor. Biz o'rgangan g'oyalarimizdan foydalanishimiz mumkin bo'limi trig integrallarini o'z ichiga olgan integrallar haqida yoki biz quyidagi formuladan foydalanishimiz mumkin.
\[\int{{{{\cos }^m}\theta \,d\theta }} = \frac{1}{m}\sin \theta {\cos ^{m - 1}}\theta + \frac{{m - 1}}{m}\int{{{{\cos }^{m - 2}}\theta \,d\theta }}\]Keling, integralni bajarish uchun ushbu formuladan foydalanamiz.
\[\begin{align*}\int{{{{\cos }^4}\theta \,d\theta }} & = \frac{1}{4}\sin \theta {\cos ^3}\theta + \frac{3}{4}\int{{{{\cos }^2}\theta \,d\theta }}\\ & = \frac{1}{4}\sin \theta {\cos ^3}\theta + \frac{3}{4}\left( {\frac{1}{2}\sin \theta \cos \theta + \frac{1}{2}\int{{{{\cos }^0}\theta \,d\theta }}} \right)\hspace{0.25in}{\cos ^0}\theta = 1!\\ & = \frac{1}{4}\sin \theta {\cos ^3}\theta + \frac{3}{8}\sin \theta \cos \theta + \frac{3}{8}\theta \end{align*}\]Keyinchalik, buni \(x\)-ga qaytarish uchun quyidagi to'g'ri burchakli uchburchakdan foydalanamiz.
\[\tan \theta = \frac{u}{{\sqrt 2 }} = \frac{{x - 3}}{{\sqrt 2 }}\hspace{0.5in}\sin \theta = \frac{{x - 3}}{{\sqrt {{{\left( {x - 3} \right)}^2} + 2} }}\hspace{0.5in}\cos \theta = \frac{{\sqrt 2 }}{{\sqrt {{{\left( {x - 3} \right)}^2} + 2} }}\]
U holda kosinus integrali,
\[\begin{align*}\int{{{{\cos }^4}\theta \,d\theta }} & = \frac{1}{4}\frac{{2\sqrt 2 \left( {x - 3} \right)}}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{3}{8}\frac{{\sqrt 2 \left( {x - 3} \right)}}{{{{\left( {x - 3} \right)}^2} + 2}} + \frac{3}{8}{\tan ^{ - 1}}\left( {\frac{{x - 3}}{{\sqrt 2 }}} \right)\\ & = \frac{{\sqrt 2 }}{2}\frac{{x - 3}}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{{3\sqrt 2 }}{8}\frac{{x - 3}}{{{{\left( {x - 3} \right)}^2} + 2}} + \frac{3}{8}{\tan ^{ - 1}}\left( {\frac{{x - 3}}{{\sqrt 2 }}} \right)\end{align*}\]Aytgancha, asl integral:
\[\begin{align*}\int{{\frac{x}{{{{\left( {{x^2} - 6x + 11} \right)}^3}}}\,dx}} & = - \frac{1}{4}\frac{1}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \\ & \hspace{1.0in}\frac{{3\sqrt 2 }}{8}\left( {\frac{{\sqrt 2 }}{2}\frac{{x - 3}}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{{3\sqrt 2 }}{8}\frac{{x - 3}}{{{{\left( {x - 3} \right)}^2} + 2}} + \frac{3}{8}{{\tan }^{ - 1}}\left( {\frac{{x - 3}}{{\sqrt 2 }}} \right)} \right)\\ & \\ & = \frac{1}{8}\frac{{3x - 11}}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{9}{{32}}\frac{{x - 3}}{{{{\left( {x - 3} \right)}^2} + 2}} + \frac{{9\sqrt 2 }}{{64}}{\tan ^{ - 1}}\left( {\frac{{x - 3}}{{\sqrt 2 }}} \right) + c\end{align*}\]Bu uzoq va chalkash javob, lekin u erda.
Yechim
Boshqa masalalarda bo'lgani kabi, biz birinchi navbatda maxrajdagi kvadratni to'ldiramiz.
\[4 - 2x - {x^2} = - \left( {{x^2} + 2x - 4} \right) = - \left( {{x^2} + 2x + 1 - 1 - 4} \right) = - \left( {{{\left( {x + 1} \right)}^2} - 5} \right) = 5 - {\left( {x + 1} \right)^2}\]Integral bu,
\[\int{{\frac{{x - 3}}{{{{\left( {4 - 2x - {x^2}} \right)}^2}}}\,dx}} = \int{{\frac{{x - 3}}{{{{\left[ {5 - {{\left( {x + 1} \right)}^2}} \right]}^2}}}\,dx}}\]Keling, almashtirishni qilaylik.
\[u = x + 1\hspace{0.5in}x = u - 1\hspace{0.5in}dx = du\]va integral endi,
\[\begin{align*}\int{{\frac{{x - 3}}{{{{\left( {4 - 2x - {x^2}} \right)}^2}}}\,dx}} & = \int{{\frac{{u - 4}}{{{{\left( {5 - {u^2}} \right)}^2}}}\,du}}\\ & = \int{{\frac{u}{{{{\left( {5 - {u^2}} \right)}^2}}}\,du}} - \int{{\frac{4}{{{{\left( {5 - {u^2}} \right)}^2}}}\,du}}\end{align*}\]Birinchi integralda biz almashtirishdan foydalanamiz
\[v = 5 - {u^2}\]va ikkinchi integralda biz quyidagi trig almashtirishdan foydalanamiz
\[u = \sqrt 5 \sin \theta \hspace{0.5in}du = \sqrt 5 \cos \theta \,d\theta \]Ushbu almashtirishlardan foydalanib, integral shunday bo'ladi:
\[\begin{align*}\int{{\frac{{x - 3}}{{{{\left( {4 - 2x - {x^2}} \right)}^2}}}\,dx}} & = - \frac{1}{2}\int{{\frac{1}{{{v^2}}}\,dv}} - \int{{\frac{4}{{{{\left( {5 - 5{{\sin }^2}\theta } \right)}^2}}}\left( {\sqrt 5 \cos \theta } \right)\,d\theta }}\\ & = \frac{1}{2}\frac{1}{v} - \frac{{4\sqrt 5 }}{{25}}\int{{\frac{{\cos \theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^2}}}\,d\theta }}\\ & = \frac{1}{2}\frac{1}{v} - \frac{{4\sqrt 5 }}{{25}}\int{{\frac{{\cos \theta }}{{{{\cos }^4}\theta }}\,d\theta }}\\ & = \frac{1}{2}\frac{1}{v} - \frac{{4\sqrt 5 }}{{25}}\int{{{{\sec }^3}\theta \,d\theta }}\\ & = \frac{1}{2}\frac{1}{v} - \frac{{2\sqrt 5 }}{{25}}\left( {\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|} \right) + c\end{align*}\]Ushbu integralni tugatish uchun bizga quyidagi to'g'ri burchakli uchburchak kerak bo'ladi.
\[\sin \theta = \frac{u}{{\sqrt 5 }} = \frac{{x + 1}}{{\sqrt 5 }}\hspace{0.5in}\sec \theta = \frac{{\sqrt 5 }}{{\sqrt {5 - {{\left( {x + 1} \right)}^2}} }}\hspace{0.5in}\tan \theta = \frac{{x + 1}}{{\sqrt {5 - {{\left( {x + 1} \right)}^2}} }}\]
Shunday qilib, \(x\) ning integraliga qaytsak,
\[\begin{align*}\int{{\frac{{x - 3}}{{{{\left( {4 - 2x - {x^2}} \right)}^2}}}\,dx}} & = \frac{1}{2}\frac{1}{{5 - {u^2}}} - \frac{{2\sqrt 5 }}{{25}}\left( {\frac{{\sqrt 5 \left( {x + 1} \right)}}{{5 - {{\left( {x + 1} \right)}^2}}} + \ln \left| {\frac{{\sqrt 5 }}{{\sqrt {5 - {{\left( {x + 1} \right)}^2}} }} + \frac{{x + 1}}{{\sqrt {5 - {{\left( {x + 1} \right)}^2}} }}} \right|} \right) + c\\ & = \frac{1}{{10}}\frac{{1 - 4x}}{{5 - {{\left( {x + 1} \right)}^2}}} - \frac{{2\sqrt 5 }}{{25}}\ln \left| {\frac{{x + 1 + \sqrt 5 }}{{\sqrt {5 - {{\left( {x + 1} \right)}^2}} }}} \right| + c\end{align*}\]Ko'pincha biz oldingi misolda ishlatgan trigni almashtirishdan foydalanganda quyidagi formula kerak bo'ladi.
\[\int{{{{\sec }^m}\theta \,d\theta }} = \frac{1}{{m - 1}}\tan \theta {\sec ^{m - 2}}\theta + \frac{{m - 2}}{{m - 1}}\int{{{{\sec }^{m - 2}}\theta \,d\theta }}\]E'tibor bering, biz bu erda ishlatgan ikkita trigni almashtirish (sinus va tangens) kerak bo'ladi. Biz ishlatgan uchinchi trigni almashtirish (sekant) bu erda kerak bo'lmaydi. Sekant almashtirishdan foydalanishi mumkin bo'lgan har qanday kvadratikni kvadratdan minus belgisini koeffitsientlash orqali sinus almashtirishga aylantirish mumkin. E'tibor bering, biz buni bunday muammolar uchun qila olamiz, chunki bizda ildiz yo'q va shuning uchun minus belgisi integratsiyadan butunlay chiqarib tashlanishi mumkin, lekin biz Trig almashtirish bo'limidagi muammolarning ildizlari bilan buni qila olmadik.