Bu birinchi integralda ikkita shart mavjud va ikkalasi ham bir xil almashtirishni talab qiladi. Bu shuni anglatadiki, biz integral uchun maxsus hech narsa qilishimiz shart emas. Bu erda eng keng tarqalgan "xatolar" dan biri bu integralni buzish va har bir qismda alohida almashtirishni amalga oshirishdir. Bu haqiqatan ham xato emas, lekin albatta biz qilishimiz kerak bo'lgan ish hajmini oshiradi. Shunday qilib, integraldagi ikkala atama ham bir xil almashtirishdan foydalanganligi sababli, biz quyidagi almashtirishdan foydalanib, hamma narsani bitta integral sifatida bajaramiz.

\[u = 2t \hspace{0.5in} du = 2dt\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,dt = \frac{1}{2}du\]

U holda integral,

\[\begin{align*}\int{{{{\bf{e}}^{2t}} + \sec \left( {2t} \right)\tan \left( {2t} \right)\,dt}} & = \frac{1}{2}\int{{{{\bf{e}}^u} + \sec \left( u \right)\tan \left( u \right)\,du}}\\ & = \frac{1}{2}\left( {{{\bf{e}}^u} + \sec \left( u \right)} \right) + c\\ & = \frac{1}{2}\left( {{{\bf{e}}^{2t}} + \sec \left( {2t} \right)} \right) + c\end{align*}\]

Ko'pincha almashtirish integralda bir necha marta ishlatilishi mumkin, shuning uchun bu sodir bo'lsa, bundan hayajonlanmang. Shuni ham yodda tutingki, butun integral oldida \(\frac{1}{2}\) bo'lganligi sababli, integralning javobi oldida \(\frac{1}{2}\) ham bo'lishi kerak.

Bu integral avvalgisiga o'xshaydi, lekin birinchi qarashda u o'xshamasligi mumkin. Mana bu muammoni almashtirish,

\[u = \cos \left( t \right)\hspace{0.25in}\,\,\,\,\,du = - \sin \left( t \right)dt\hspace{0.25in} \Rightarrow \hspace{0.25in}\sin \left( t \right)dt = - du\]

Biz almashtirishni muammoga ikki marta kiritamiz (chunki ikkita kosinus mavjud) va faqat hamma narsani ko'paytiruvchi sinus borligi sababli ishlaydi. Oldindagi sinus bo'lmasa, biz bu almashtirishdan foydalana olmaymiz.

Bu holda integral,

\[\begin{align*}\int{{\sin \left( t \right)\left( {4{{\cos }^3}\left( t \right) + 6{{\cos }^2}\left( t \right) - 8} \right)\,dt}} & = - \int{{4{u^3} + 6{u^2}}} - 8\,du\\ & = - \left( {{u^4} + 2{u^3} - 8u} \right) + c\\ & = - \left( {{{\cos }^4}\left( t \right) + 2{{\cos }^3}\left( t \right) - 8\cos \left( t \right)} \right) + c\end{align*}\]

Shunga qaramay, butun integral oldidagi minus belgisi bilan ehtiyot bo'ling.

Ushbu integraldagi har bir atama bir xil almashtirishdan foydalanishi aniq bo'lishi kerak, ammo narsalarni aniqroq qilish uchun narsalarni biroz qayta yozamiz.

\[ \displaystyle \int{{x\cos \left( {{x^2} + 1} \right) + \frac{x}{{{x^2} + 1}}\,dx}} = \int{{x\left( {\cos \left( {{x^2} + 1} \right) + \frac{1}{{{x^2} + 1}}} \right)\,dx}}\]

Har bir atamada \(x\) bo'lganligi sababli, biz differentsial uchun bu kerak bo'ladi, biz uni old qismga kiritish uchun ikkala shartdan ham hisobga oldik. Bu integral endi oldingisiga juda o'xshaydi. Mana almashtirish.

\[u = {x^2} + 1 \hspace{0.5in} du = 2x\,dx\hspace{0.25in}\,\,\,\,\,\,\,\, \Rightarrow \,\hspace{0.25in}\,\hspace{0.25in}x\,dx = \frac{1}{2}du\]

Integral bu,

\[\begin{align*}\int{{x\cos \left( {{x^2} + 1} \right) + \frac{x}{{{x^2} + 1}}\,dx}} & = \frac{1}{2}\int{{\cos \left( u \right) + \frac{1}{u}\,du}}\\ & = \frac{1}{2}\left( {\sin \left( u \right) + \ln \left| u \right|} \right) + c\\ & = \frac{1}{2}\left( {\sin \left( {{x^2} + 1} \right) + \ln \left| {{x^2} + 1} \right|} \right) + c\end{align*}\]

Bu integralda birinchi a'zo hech qanday almashtirishga muhtoj emas, ikkinchi a'zo esa almashtirishga muhtoj. Shunday qilib, buni hal qilish uchun biz integralni quyidagicha ajratishimiz kerak,

\[\int{{{x^2} + {{\bf{e}}^{1 - x}}\,dx}} = \int{{{x^2}\,dx}} + \int{{{{\bf{e}}^{1 - x}}\,dx}}\]

Ikkinchi integralning o'rnini bosish shunday bo'ladi:

\[u = 1 - x\hspace{0.25in}du = - dx\hspace{0.25in}\, \Rightarrow \hspace{0.25in}\,\,dx = - du\]

Integral bu,

\[\begin{align*}\int{{{x^2} + {{\bf{e}}^{1 - x}}\,dx}} & = \int{{{x^2}\,dx}} - \int{{{{\bf{e}}^u}\,du}}\\ & = \frac{1}{3}{x^3} - {{\bf{e}}^u} + c\\ & = \frac{1}{3}{x^3} - {{\bf{e}}^{1 - x}} + c\end{align*}\]

Bunday integral bilan ehtiyot bo'ling. Bu erda eng keng tarqalgan xatolardan biri quyidagi "yorliq" ni bajarishdir.

\[ \int{{{x^2} + {{\bf{e}}^{1 - x}}\,dx}} = - \int{{{x^2}\, + {{\bf{e}}^u}\,du}}\]

Boshqacha qilib aytadigan bo'lsak, ba'zi talabalar integralni buzmasdan ikkinchi davrni almashtirishga harakat qilishadi. Bu borada ikkita muammo bor. Birinchidan, butun integral oldida "-" belgisi mavjud bo'lib, u erda bo'lmasligi kerak. Bu faqat ikkinchi muddatda bo'lishi kerak, chunki bu almashtirishni olish atamasi. Ikkinchidan, ehtimol bundan ham muhimi, integralda \(x\) mavjud va bizda differentsial uchun \(du\) mavjud. Biz bunday o'zgaruvchilarni aralashtira olmaymiz. Biz integrallarni qilganda, integraldagi barcha o'zgaruvchilar differentsialdagi o'zgaruvchiga mos kelishi kerak.

Ushbu integral yuqoridagi 1c-misolga juda o'xshaydi, lekin u boshqacha. Ushbu integralda bizda ikkinchi hadning hisoblagichida \(x\) endi yo'q va bu biz birinchi atama uchun ishlatadigan almashtirish ikkinchi muddat uchun ishlamasligini anglatadi. Aslida, ikkinchi atama almashtirishga mutlaqo muhtoj emas, chunki u shunchaki teskari tangens.

Birinchi muddat uchun almashtirish, keyin,

\[u = {x^2} + 1 \hspace{0.5in} du = 2x\,dx\hspace{0.25in}\,\,\,\,\,\,\,\, \Rightarrow \,\hspace{0.25in}\,\hspace{0.25in}x\,dx = \frac{1}{2}du\]

Endi integralni bajaramiz. O'zgartirishni ishlatishdan oldin uni ikki shartga bo'lishni unutmang.

\[\begin{align*}\int{{x\cos \left( {{x^2} + 1} \right) + \frac{1}{{{x^2} + 1}}\,dx}} & = \int{{x\cos \left( {{x^2} + 1} \right)\,dx}} + \int{{\frac{1}{{{x^2} + 1}}\,dx}}\\ & = \frac{1}{2}\int{{\cos \left( u \right)du}} + \int{{\frac{1}{{{x^2} + 1}}\,dx}}\\ & = \frac{1}{2}\sin \left( u \right) + {\tan ^{ - 1}}\left( x \right) + c\\ & = \frac{1}{2}\sin \left( {{x^2} + 1} \right) + {\tan ^{ - 1}}\left( x \right) + c\end{align*}\]

Ushbu integralda, biz hali qilgan har qanday integraldan farqli o'laroq, ikkita atama mavjud va ularning har biri boshqa almashtirishni talab qiladi. Demak, bu integralni bajarish uchun avvalo integralni quyidagicha ajratishimiz kerak:

\[\int{{{{\bf{e}}^{ - z}} + {{\sec }^2}\left( {\frac{z}{{10}}} \right)\,dz}} = \int{{{{\bf{e}}^{ - z}}dz}} + \int{{{{\sec }^2}\left( {\frac{z}{{10}}} \right)\,dz}}\]

Here are the substitutions for each integral.

\[\begin{align*}u & = - z \hspace{0.5in} du = - dz\hspace{0.25in} \Rightarrow \hspace{0.25in}\,dz = - du\\ v & = \frac{z}{{10}} \hspace{0.5in} dv = \frac{1}{{10}}dz\hspace{0.25in} \Rightarrow \hspace{0.25in}dz = 10dv\end{align*}\]

E'tibor bering, biz \(u\) va \(v\)-ni qayta ulashga kirganimizda chalkashmaslik uchun har bir almashtirish uchun turli harflardan foydalanganmiz.

Mana integral.

\[\begin{align*}\int{{{{\bf{e}}^{ - z}} + {{\sec }^2}\left( {\frac{z}{{10}}} \right)\,dz}} & = - \int{{{{\bf{e}}^u}du}} + 10\int{{{{\sec }^2}\left( v \right)\,dv}}\\ & = - {{\bf{e}}^u} + 10\tan \left( v \right) + c\\ & = - {{\bf{e}}^{ - z}} + 10\tan \left( {\frac{z}{{10}}} \right) + c\end{align*}\]

Oxirgi muammoda bo'lgani kabi, bu integral ikkita alohida almashtirishni talab qiladi. Let’s first break up the integral.

\[\begin{align*}\int{{\sin w\sqrt {1 - 2\cos w} + \frac{1}{{7w + 2}}\,dw}} & = \int{{\sin w{{\left( {1 - 2\cos w} \right)}^{\frac{1}{2}}}\,dw}} + \int{{\frac{1}{{7w + 2}}\,dw}}\\ & \end{align*}\]

Here are the substitutions for this integral.

\[\begin{align*}u & = 1 - 2\cos \left( w \right)\hspace{0.25in}du = 2\sin \left( w \right)dw\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\sin \left( w \right)dw = \frac{1}{2}du\\ v & = 7w + 2 \hspace{0.5in} dv = 7dw \hspace{0.5in} \Rightarrow \hspace{0.25in}dw = \frac{1}{7}dv\end{align*}\]

U holda integral,

\[\begin{align*}\int{{\sin w\sqrt {1 - 2\cos w} + \frac{1}{{7w + 2}}\,dw}} & = \frac{1}{2}\int{{{u^{\frac{1}{2}}}\,du}} + \frac{1}{7}\int{{\frac{1}{v}\,dv}}\\ & = \frac{1}{2}\left( {\frac{2}{3}} \right){u^{\frac{3}{2}}} + \frac{1}{7}\ln \left| v \right| + c\\ & = \frac{1}{3}{\left( {1 - 2\cos w} \right)^{\frac{3}{2}}} + \frac{1}{7}\ln \left| {7w + 2} \right| + c\end{align*}\]

The last problem in this set can be tricky. Agar hisoblagichda faqat \(x\) bo'lsa, tabiiy logarifmani olish uchun tez almashtirishni amalga oshirishimiz mumkin edi. Xuddi shunday, agar suratda \(x\) bo'lmasa, biz tez almashtirishdan keyin teskari tangensga ega bo'lardik.

Ushbu integralni biz ishlashimiz mumkin bo'lgan shaklga aylantirish uchun avval uni quyidagicha parchalashimiz kerak bo'ladi.

\[\begin{align*}\int{{\frac{{10x + 3}}{{{x^2} + 16}}\,dx}} & = \int{{\frac{{10x}}{{{x^2} + 16}}\,dx}} + \int{{\frac{3}{{{x^2} + 16}}\,dx}}\\ & = \int{{\frac{{10x}}{{{x^2} + 16}}\,dx}} + \frac{1}{{16}}\int{{\frac{3}{{\frac{{{x^2}}}{{16}} + 1}}\,dx}}\end{align*}\]

Endi bizda ikkita integral bor, ularning har biri boshqa almashtirishni talab qiladi. Yuqoridagi integrallarning har biri uchun almashtirishlar:

\[\begin{align*}u & = {x^2} + 16 \hspace{0.5in} du = 2xdx\hspace{0.25in}\,\,\,\,\,\, \Rightarrow \hspace{0.5in} xdx = \frac{1}{2}du\\ v & = \frac{x}{4} \hspace{0.5in} \hspace{0.25in}dv = \frac{1}{4}dx\hspace{0.25in}\,\,\,\,\,\, \Rightarrow \hspace{0.5in} dx = 4dv\end{align*}\]

U holda integral,

\[\begin{align*}\int{{\frac{{10x + 3}}{{{x^2} + 16}}\,dx}} & = 5\int{{\frac{1}{u}\,du + \frac{3}{4}\int{{\frac{1}{{{v^2} + 1}}\,dv}}}}\\ & = 5\ln \left| u \right| + \frac{3}{4}{\tan ^{ - 1}}\left( v \right) + c\\ & = 5\ln \left| {{x^2} + 16} \right| + \frac{3}{4}{\tan ^{ - 1}}\left( {\frac{x}{4}} \right) + c\end{align*}\]

Bu muammo haqida birinchi savol, ehtimol, nima uchun bu erda? O'zgartirish qoidasi bilan bog'liq muammolar odatda bittadan ko'proq funktsiyani talab qiladi. Bu muammoning kaliti bu erda haqiqatan ham ikkita funktsiya mavjudligini tushunishdir. Biz qilishimiz kerak bo'lgan narsa tangens ta'rifini eslab qolish va biz integralni quyidagicha yozishimiz mumkin:

\[\int{{\tan x\,dx}} = \int{{\frac{{\sin x}}{{\cos x}}\,dx}}\]

Shu tarzda yozilgan bo'lsak, quyidagi almashtirish biz uchun ishlashini ko'rishimiz mumkin,

\[u = \cos x\hspace{0.25in}du = - \sin x\,dx \hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,\,\sin x\,dx = - du\]

U holda integral,

\[\begin{align*}\int{{\tan x\,dx}} & = - \int{{\frac{1}{u}\,du}}\\ & = - \ln \left| u \right| + c\\ & = - \ln \left| {\cos x} \right| + c\end{align*}\]

Endi, bu juda mos javob bo'lsa-da, agar ehtiyot bo'lmasak, oldingi minus belgisi muammolarga olib kelishi mumkin. So, let’s rewrite things a little. Eslatib o'tamiz, a mulk logarifmlardan oldingi minus belgisini kosinusdagi ko'rsatkichga ko'chirishimiz va keyin biroz soddalashtirishimiz mumkin.

\[\begin{align*}\int{{\tan x\,dx}} & = - \ln \left| {\cos x} \right| + c\\ & = \ln {\left| {\cos x} \right|^{ - 1}} + c\\ & = \ln \frac{1}{{\left| {\cos x} \right|}} + c\\ & = \ln \left| {\sec x} \right| + c\end{align*}\]

Bu odatda tangens integrali uchun berilgan formuladir.

E'tibor bering, biz kotangentni xuddi shunday tarzda birlashtira olamiz.

Bu muammo ham avvaliga almashtirish qoidasi masalalariga tegishli emasdek ko'rinadi. Bu almashtirish qoidasidan foydalanishga imkon beradigan shaklda yozish uchun faqat sekant funktsiyasi ta'rifidan foydalana olmasligimizni payqasak, bu yanada muammoli.

Ushbu muammo bu darajada tez-tez ishlatilmaydigan texnikani talab qiladi, ammo bilish kerak bo'lgan foydali texnikadir. Ba'zan biz yuqori va pastni umumiy atamaga ko'paytirish orqali integralni bajariladigan qilishimiz mumkin. Bu har doim ham ishlamaydi va hatto shunday bo'lganda ham, biz nimani ko'paytirishimiz kerakligi har doim ham aniq emas, lekin u ishlaganda bu juda foydali.

Bu muammoni hal qilish uchun biz ushbu fikrdan qanday foydalanamiz.

\[\int{{\sec y\,dy}} = \int{{\frac{{\sec y}}{1}\frac{{\left( {\sec y + \tan y} \right)}}{{\left( {\sec y + \tan y} \right)}}\,dy}}\]

Birinchidan, biz sekantni kasr sifatida ko'rib chiqamiz va keyin kasrning yuqori va pastki qismini bir xil songa ko'paytiramiz. Ehtimol, nima uchun bu erda buni qilishni xohlashingiz aniq emas, lekin bu bizga almashtirish qoidasidan foydalanishga imkon beradi. Bu qanday ishlashini ko'rish uchun integralni biroz soddalashtiramiz.

\[\int{{\sec y\,dy}} = \int{{\frac{{{{\sec }^2}y + \tan y\sec y}}{{\sec y + \tan y}}\,dy}}\]

Endi biz quyidagi almashtirishdan foydalanishimiz mumkin.

\[u = \sec y + \tan y\hspace{0.25in}du = \left( {\sec y\tan y + {{\sec }^2}y} \right)dy\]

U holda integral,

\[\begin{align*}\int{{\sec y\,dy}} & = \int{{\frac{1}{u}\,du}}\\ & = \ln \left| u \right| + c\\ & = \ln \left| {\sec y + \tan y} \right| + c\end{align*}\]

Ba'zan kasrning yuqori va pastki qismini diqqat bilan tanlangan atama bilan ko'paytirish bizga muammoni hal qilish imkonini beradi. Biroq, bu atama nima bo'lishi kerakligini aniqlash uchun ba'zida biroz o'ylash kerak.

Kosekantni birlashtirish uchun shunga o'xshash jarayondan foydalanishimiz mumkin.

Bu keyingi muammoning o'ziga xos jihati borki, agar e'tibor bermasak, bizni muammoga duchor qilishi mumkin. Kosinusdagi ildiz tufayli quyidagi almashtirishni qo'llash ma'lum ma'noga ega.

\[u = {x^{\frac{1}{2}}} \hspace{0.5in} du = \frac{1}{2}{x^{ - \,\,\frac{1}{2}}}dx\]

Bu erda biz ehtiyot bo'lishimiz kerak. Differensialni qayta yozganimizda, biz olamiz,

\[2du = \frac{1}{{\sqrt x }}dx\]

Maxrajdagi ildiz biz vasvasaga tushgandek \(u\) ga aylanmaydi. Buning o'rniga u differentsialda hal qilinadi.

Integral bu,

\[\begin{align*}\int{{\frac{{\cos \left( {\sqrt x } \right)}}{{\sqrt x }}\,dx}} & = 2\int{{\cos \left( u \right)\,du}}\\ & = 2\sin \left( u \right) + c\\ & = 2\sin \left( {\sqrt x } \right) + c\end{align*}\]

Ushbu muammo bilan biz o'z o'rnimizni tanlashimiz kerak. Muammo yozilganidek, biz quyidagi almashtirishni ishlatish vasvasasiga tushishimiz mumkin:

\[u = t + {{\bf{e}}^t}\hspace{0.25in}du = \left( {1 + {{\bf{e}}^t}} \right)dt\]

Biroq, bu siz ko'rib turganingizdek ishlamaydi. Differensial integrandning hech bir joyida ko'rinmaydi va biz bu almashtirish bilan barcha \(t\)-larni yo'q qila olmaymiz.

Ushbu muammoni hal qilish uchun biz integralni quyidagicha qayta yozishimiz kerak:

\[\int{{{{\bf{e}}^{t + {{\bf{e}}^t}}}\,dt}} = \int{{{{\bf{e}}^t}{{\bf{e}}^{{{\bf{e}}^t}}}\,dt}}\]

Endi biz almashtirishdan foydalanamiz,

\[u = {{\bf{e}}^t} \hspace{0.5in} du = {{\bf{e}}^t}dt\]

Integral bu,

\[\begin{align*}\int{{{{\bf{e}}^{t + {{\bf{e}}^t}}}\,dt}} & = \int{{{{\bf{e}}^u}\,du}}\\ & = {{\bf{e}}^u} + c\\ & = {{\bf{e}}^{{{\bf{e}}^t}}} + c\end{align*}\]

Ba'zi almashtirishlarni ko'rish juda qiyin bo'lishi mumkin va almashtirishni amalga oshirish uchun biroz soddalashtirish va/yoki qayta yozishingiz kerakligi odatiy hol emas.

Ushbu to'plamdagi so'nggi muammo biz shu paytgacha ishlagan boshqa almashtirish muammolaridan farq qiladi. Ildiz ostida bizda \(x\) dan ko'proq narsa borligini hisobga olsak, almashtirish deyarli bo'lishi kerakligi mantiqiy bo'ladi,

\[u = {x^2} + 1 \hspace{0.5in} du = 2x\,dx\]

Bir qarashda, bu almashtirish uchun ishlamasligi mumkin, chunki bizda ildiz oldida \({x^3}\) mavjud. Biroq, agar biz avval \(2{x^3} = {x^2}\left( {2x} \right)\) ni qayta yozsak, keyin \(2x\) ni integralning oxiriga ko'chirishimiz mumkin, shunda hech bo'lmaganda \(du\) integralda aniq ko'rinadi. Buni amalga oshirish quyidagilarni beradi,

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

Bu haqiqiy muammo. Bizning integrallarimiz ikkita o'zgaruvchiga ega bo'lishi mumkin emas. Odatda bu biz almashtirishni noto'g'ri tanlaganimizni anglatadi. Biroq, bu holda biz almashtirishni quyidagicha qayta yozishimiz mumkin:

\[{x^2} = u - 1\]

va endi biz qolgan \(x\)-larni integralimizdan olib tashlashimiz mumkin. Buni qilish,

\[\begin{align*}\int{{2{x^3}\sqrt {{x^2} + 1} \,dx}} & = \int{{\left( {u - 1} \right){u^{\frac{1}{2}}}\,du}}\\ & = \int{{{u^{\frac{3}{2}}} - {u^{\frac{1}{2}}}\,du}}\\ & = \frac{2}{5}{u^{\frac{5}{2}}} - \frac{2}{3}{u^{\frac{3}{2}}} + c\\ & = \frac{2}{5}{\left( {{x^2} + 1} \right)^{\frac{5}{2}}} - \frac{2}{3}{\left( {{x^2} + 1} \right)^{\frac{3}{2}}} + c\end{align*}\]

Ba'zida biz almashtirishni bir necha marta ishlatishimiz kerak bo'ladi.

Bunday muammo tez-tez paydo bo'lmaydi va ba'zida integralni bajarishning muqobil usullari mavjud bo'ladi. Biroq, ko'pincha integralni bajarishning eng oson usuli bu erda qilgan ishimizni qilishdir.

Bunday holda, biz logarifmni o'z-o'zidan integrallay olmasligimizni bilamiz va shuning uchun logarifm almashtirishda bo'lishi kerakligi ma'noga ega (umid qilamanki). Mana bu muammoni almashtirish.

\[u = \ln x \hspace{0.5in} du = \frac{1}{x}dx\]

Shunday qilib, integrandning maxrajidagi \(x\) differensial bilan almashtiriladi. Mana bu muammoning integrali.

\[\begin{align*}\int{{\frac{1}{{x\,\ln x}}\,dx}} & = \int{{\frac{1}{u}\,du}}\\ & = \ln \left| u \right| + c\\ & = \ln \left| {\ln x} \right| + c\end{align*}\]

Shunga qaramay, bu erda almashtirish biroz qiyin bo'lib tuyulishi mumkin. Bunday holda, almashtirish:

\[u = 1 + {{\bf{e}}^{2t}}\hspace{0.25in}du = 2{{\bf{e}}^{2t}}dt\hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,{{\bf{e}}^{2t}}dt = \frac{1}{2}du\]

U holda integral,

\[\begin{align*}\int{{\frac{{{{\bf{e}}^{2t}}}}{{1 + {{\bf{e}}^{2t}}}}\,dt}} & = \frac{1}{2}\int{{\frac{1}{u}\,du}}\\ & = \frac{1}{2}\ln \left| {1 + {{\bf{e}}^{2t}}} \right| + c\end{align*}\]

Bunday holda, biz avvalgi masalada ishlatgan almashtirish turidan foydalana olmaymiz. Oldingi misolda almashtirishdan foydalanish uchun pay va maxrajdagi ko'rsatkich bir xil bo'lishi kerak va bu holda ular emas.

Ushbu muammoni to'g'ri almashtirishni ko'rish uchun quyidagilarni e'tiborga oling:

\[{{\bf{e}}^{4t}} = {\left( {{{\bf{e}}^{2t}}} \right)^2}\]

Bundan foydalanib, integralni quyidagicha yozish mumkin:

\[\int{{\frac{{{{\bf{e}}^{2t}}}}{{1 + {{\bf{e}}^{4t}}}}\,dt}} = \int{{\frac{{{{\bf{e}}^{2t}}}}{{1 + {{\left( {{{\bf{e}}^{2t}}} \right)}^2}}}\,dt}}\]

Endi biz quyidagi almashtirishdan foydalanishimiz mumkin.

\[u = {{\bf{e}}^{2t}} \hspace{0.5in} du = 2{{\bf{e}}^{2t}}dt\hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,{{\bf{e}}^{2t}}dt = \frac{1}{2}du\]

U holda integral,

\[\begin{align*}\int{{\frac{{{{\bf{e}}^{2t}}}}{{1 + {{\bf{e}}^{4t}}}}\,dt}} & = \frac{1}{2}\int{{\frac{1}{{1 + {u^2}}}\,du}}\\ & = \frac{1}{2}{\tan ^{ - 1}}\left( u \right) + c\\ & = \frac{1}{2}{\tan ^{ - 1}}\left( {{{\bf{e}}^{2t}}} \right) + c\end{align*}\]

Bu integral ushbu to'plamdagi birinchi masalaga o'xshaydi. Biz teskari sinus funktsiyalarni qanday integrallashni bilmasligimiz sababli, bu bizning almashtirishimiz bo'lishi mumkin. Agar biz buni o'z o'rnimiz sifatida ishlatsak, biz olamiz,

\[u = {\sin ^{ - 1}}\left( x \right) \hspace{0.5in} du = \frac{1}{{\sqrt {1 - {x^2}} }}dx\]

Shunday qilib, integraldagi ildiz almashtirish jarayonida e'tiborga olinadi va bu integraldan barcha \(x\)-larni yo'q qiladi. Shuning uchun bu to'g'ri almashtirish edi.

Integral bu,

\[\begin{align*}\int{{\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\,dx}} & = \int{{u\,du}}\\ & = \frac{1}{2}{u^2} + c\\ & = \frac{1}{2}{\left( {{{\sin }^{ - 1}}x} \right)^2} + c\end{align*}\]