5.8-bo'lim: Aniq integrallarni almashtirish qoidasi
Endi biz orqaga qaytishimiz va almashtirish qoidasini qayta ko'rib chiqishimiz kerak, chunki u aniq integrallarga tegishli. Ba'zi darajada, bu bo'limda juda ko'p narsa qilish kerak emas. Eslatib o'tamiz, aniq integralni bajarishda birinchi qadam noaniq integralni hisoblashdir va u o'zgarmadi. Biz hali ham avval noaniq integralni hisoblaymiz. Bu shuni anglatadiki, biz buni qanday qilishni allaqachon bilamiz. Noaniq integralni topish uchun almashtirish qoidasidan foydalanamiz va keyin baholaymiz.
Biroq, baholash bosqichini hal qilishning ikki yo'li mavjud. Baholash usullaridan biri, ehtimol, hozirgi paytda eng ravshan, ammo agar biz e'tibor bermasak, muammoga duch kelishimiz mumkin bo'lgan jarayonda nuqta bor.
Keling, baholash bosqichini bajarishning ikkala usulini ko'rsatadigan misol bilan ishlaylik.
Yechim
Keling, baholash bosqichini hal qilishning birinchi usulini ko'rib chiqaylik. Biz bu usuldan ehtiyot bo'lishimiz kerak, chunki jarayonda e'tibor bermasak, noto'g'ri javob olishimiz mumkin bo'lgan nuqta bor.
Yechim 1:Biz birinchi navbatda almashtirish qoidasidan foydalanib noaniq integralni hisoblashimiz kerak. Shuni yodda tutingki, biz har bir qadamda integralga chegara qo'yish orqali bu aniq integral ekanligini doimo eslatib turamiz. Cheklovlarsiz, noaniq integralni hisoblaganimizda, bizda aniq integral borligini unutish oson.
Bunday holda, almashtirish:
\[u = 1 - 4{t^3}\hspace{0.25in}\,\,\,\,\,du = - 12{t^2}dt\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,{t^2}dt = - \frac{1}{{12}}du\]Buni integralga ulash natijasida,
\[\begin{align*}\int_{{ - 2}}^{0}{{2{t^2}\sqrt {1 - 4{t^3}} \,dt}} & = - \frac{1}{6}\int_{{ - 2}}^{0}{{{u^{\frac{1}{2}}}\,du}}\\ & = \left. { - \frac{1}{9}{u^{\frac{3}{2}}}} \right|_{ - 2}^0\end{align*}\]E'tibor bering, biz hali baholashni qilmaganmiz. Bu yechim usuli bilan potentsial muammo yuzaga keladi. Bu yerda berilgan chegaralar asl integraldan olingan va shuning uchun \(t\) qiymatlari. Bizning yechimimizda \(u\) mavjud. Biz \(t\) qiymatlarini \(u\) uchun ulay olmaymiz.
Shuning uchun, almashtirishni amalga oshirishdan oldin biz \(t\)-ga qaytishimiz kerak. Bu almashtirish jarayonidagi standart qadamdir, lekin aniq integrallarni bajarishda u ko'pincha unutiladi. Shuni ham yodda tutingki, bu holda, agar biz \(t\)-ga qaytmasak, bizda kichik muammo bo'ladi, chunki baholashlardan biri bizga kompleks sonni beradi.
Shunday qilib, bu muammoni tugatish,
\[\begin{align*}\int_{{ - 2}}^{0}{{2{t^2}\sqrt {1 - 4{t^3}} \,dt}} & = \left. { - \frac{1}{9}{{\left( {1 - 4{t^3}} \right)}^{\frac{3}{2}}}} \right|_{ - 2}^0\\ & = - \frac{1}{9} - \left( { - \frac{1}{9}{{\left( {33} \right)}^{\frac{3}{2}}}} \right)\\ & = \frac{1}{9}\left( {33\sqrt {33} - 1} \right)\end{align*}\]Shunday qilib, bu birinchi yechim usuli edi. Keling, ikkinchi usulni ko'rib chiqaylik.
Yechim 2:E'tibor bering, bu yechim usuli aslida birinchi usuldan unchalik farq qilmaydi. Ushbu usulda biz almashtirishni amalga oshirayotganda integraldagi barcha \(t\)-larni yo'q qilishni va hamma narsani \(u\) shaklida yozishni eslaymiz.
Bu erda hamma narsani aytganda, biz hamma narsani nazarda tutamiz. Boshqacha qilib aytganda, integral chegaralari ham \(t\) qiymatlari ekanligini unutmang va biz chegaralarni \(u\) qiymatlariga aylantiramiz. Cheklovlarni o'zgartirish juda oddiy, chunki bizning almashtirishimiz \(t\) va \(u\) ni qanday bog'lashni aytib beradi, shuning uchun biz qilishimiz kerak bo'lgan narsa almashtirishga asl \(t\) chegaralarini kiritish va biz yangi \(u\) chegaralarini olamiz.
Bu erda almashtirish (bu birinchi usul bilan bir xil), shuningdek chegara konvertatsiyalari.
\[\begin{align*}u & = 1 - 4{t^3}\hspace{0.25in} du = - 12{t^2}dt\hspace{0.25in} \Rightarrow \hspace{0.25in} {t^2}dt = - \frac{1}{{12}}du\\ t & = - 2\hspace{0.5in} \Rightarrow \hspace{0.5in}u = 1 - 4{\left( { - 2} \right)^3} = 33\\ t & = 0\hspace{0.65in} \Rightarrow \hspace{0.5in}u = 1 - 4{\left( 0 \right)^3} = 1\end{align*}\]Integral endi,
\[\begin{align*}\int_{{ - 2}}^{0}{{2{t^2}\sqrt {1 - 4{t^3}} \,dt}} & = - \frac{1}{6}\int_{{33}}^{1}{{{u^{\frac{1}{2}}}\,du}}\\ & = \left. { - \frac{1}{9}{u^{\frac{3}{2}}}} \right|_{33}^1\end{align*}\]Birinchi usulda bo'lgani kabi, biz nima qilayotganimizni eslatish uchun bu erda bir oz to'xtab qolaylik. Bunday holda, biz chegaralarni \(u\)-ga aylantirdik va biz \(u\) nuqtai nazaridan o'zimizning integralimizni ham oldik va shuning uchun biz bu erda chegaralarni to'g'ridan-to'g'ri integralimizga ulashimiz mumkin. E'tibor bering, bu holda biz almashtirishni qaytadan ulanmaymiz. Buni qilish muammolarga olib keladi, chunki integralda \(t\) va chegaralarimiz \(u\) bo'ladi. Mana bu muammoning qolgan qismi.
\[\begin{align*}\int_{{ - 2}}^{0}{{2{t^2}\sqrt {1 - 4{t^3}} \,dt}} & = \left. { - \frac{1}{9}{u^{\frac{3}{2}}}} \right|_{33}^1\\ & = - \frac{1}{9} - \left( { - \frac{1}{9}{{\left( {33} \right)}^{\frac{3}{2}}}} \right) = \frac{1}{9}\left( {33\sqrt {33} - 1} \right)\end{align*}\]Biz aynan bir xil javob oldik va bu safar javobimizda \(t\)-ga qaytish haqida tashvishlanishga hojat yo'q.
Shunday qilib, biz almashtirish qoidasini talab qiladigan aniq integrallarni hisoblashning ikkita echimini ko'rdik. Ikkalasi ham to'g'ri echim usullari va ularning har biri o'z qo'llanilishiga ega. Biz ikkinchisidan deyarli foydalanamiz, ammo bu baholash bosqichini biroz osonlashtiradi.
Keling, yana bir nechta misollar bilan ishlaylik.
- \( \displaystyle \int_{{\, - 1}}^{{\,5}}{{\left( {1 + w} \right){{\left( {2w + {w^2}} \right)}^5}\,dw}}\)
- \( \displaystyle \int_{{\, - 2}}^{{\, - 6}}{{\frac{4}{{{{\left( {1 + 2x} \right)}^3}}} - \frac{5}{{1 + 2x}}\,dx}}\)
- \( \displaystyle \int_{{\,0}}^{{\,\frac{1}{2}}}{{{{\bf{e}}^y} + 2\cos \left( {\pi y} \right)\,dy}}\)
- \( \displaystyle \int_{{\,\frac{\pi }{3}}}^{{\,0}}{{3\sin \left( {\frac{z}{2}} \right) - 5\cos \left( {\pi - z} \right)\,dz}}\)
Biz shu vaqtgacha bir nechta almashtirish qoidalari integrallarini bajarganimiz sababli, biz bu erda narsalarning almashtirish qismini tushuntirishga ko'p kuch sarflamaymiz.
a \( \displaystyle \int_{{\, - 1}}^{{\,5}}{{\left( {1 + w} \right){{\left( {2w + {w^2}} \right)}^5}\,dw}}\) Yechim
O'zgartirish va konvertatsiya qilingan limitlar,
\[\begin{align*}u & = 2w + {w^2}\hspace{0.5in}du = \left( {2 + 2w} \right)dw\hspace{0.25in}\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\left( {1 + w} \right)dw = \frac{1}{2}du\\ w & = - 1\hspace{0.25in}\,\,\,\,\,\, \Rightarrow \,\,\,\,\hspace{0.25in}u = - 1\hspace{0.5in}\hspace{0.5in}w = 5\,\,\,\,\,\,\,\,\, \Rightarrow \hspace{0.25in}u = 35\end{align*}\]Ba'zan o'zgartirishdan keyin chegara bir xil bo'lib qoladi. Bu sodir bo'lganda hayajonlanmang va bu har doim sodir bo'lishini kutmang.
Mana integral,
\[\begin{align*}\int_{{\, - 1}}^{{\,5}}{{\left( {1 + w} \right){{\left( {2w + {w^2}} \right)}^5}\,dw}} & = \frac{1}{2}\int_{{\, - 1}}^{{\,35}}{{{u^5}\,du}}\\ & = \left. {\frac{1}{{12}}{u^6}} \right|_{ - 1}^{35} = 153188802\end{align*}\]Bu erda javoblar uchun katta raqamlardan hayajonlanmang. Ba'zan ular. Bu hayot.
b \( \displaystyle \int_{{\, - 2}}^{{\, - 6}}{{\frac{4}{{{{\left( {1 + 2x} \right)}^3}}} - \frac{5}{{1 + 2x}}\,dx}}\) Yechim
Mana bu muammo uchun almashtirish va konvertatsiya qilingan chegaralar,
\[\begin{align*}u & = 1 + 2x\hspace{0.5in}du = 2dx\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,dx = \frac{1}{2}du\\ x & = - 2\hspace{0.25in}\,\,\,\,\,\, \Rightarrow \,\,\hspace{0.25in}\,\,u = - 3\hspace{0.5in}\hspace{0.25in}x = - 6\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\hspace{0.25in}\,\,\,\,u = - 11\end{align*}\]U holda integral,
\[\begin{align*}\int_{{\, - 2}}^{{\, - 6}}{{\frac{4}{{{{\left( {1 + 2x} \right)}^3}}} - \frac{5}{{1 + 2x}}\,dx}} & = \frac{1}{2}\int_{{\, - 3}}^{{\, - 11}}{{4{u^{ - 3}} - \frac{5}{u}\,du}}\\ & = \left. {\frac{1}{2}\left( { - 2{u^{ - 2}} - 5\ln \left| u \right|} \right)} \right|_{ - 3}^{ - 11}\\ & = \frac{1}{2}\left( { - \frac{2}{{121}} - 5\ln 11} \right) - \frac{1}{2}\left( { - \frac{2}{9} - 5\ln 3} \right)\\ & = \frac{{112}}{{1089}} - \frac{5}{2}\ln 11 + \frac{5}{2}\ln 3\end{align*}\]c \( \displaystyle \int_{{\,0}}^{{\,\frac{1}{2}}}{{{{\bf{e}}^y} + 2\cos \left( {\pi y} \right)\,dy}}\) Yechim
Ushbu integralni ikkita integralga bo'lish kerak, chunki birinchi a'zo almashtirishni talab qilmaydi, ikkinchisi esa.
\[ \displaystyle \int_{{\,0}}^{{\,\frac{1}{2}}}{{{{\bf{e}}^y} + 2\cos \left( {\pi y} \right)\,dy}} = \int_{{\,0}}^{{\,\frac{1}{2}}}{{{{\bf{e}}^y}\,dy}} + \int_{{\,0}}^{{\,\frac{1}{2}}}{{2\cos \left( {\pi y} \right)\,dy}}\]Mana ikkinchi muddat uchun almashtirish va konvertatsiya qilingan limitlar.
\[\begin{align*}u & = \pi y\hspace{0.5in}du = \pi dy\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,dy = \frac{1}{\pi }du\\ y & = 0\hspace{0.25in}\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,u = 0\hspace{0.5in}\hspace{0.25in}y = \frac{1}{2}\,\hspace{0.25in}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,u = \frac{\pi }{2}\end{align*}\]Mana integral.
\[\begin{align*}\int_{{\,0}}^{{\,\frac{1}{2}}}{{{{\bf{e}}^y} + 2\cos \left( {\pi y} \right)\,dy}} & = \int_{{\,0}}^{{\,\frac{1}{2}}}{{{{\bf{e}}^y}\,dy}} + \frac{2}{\pi }\int_{{\,0}}^{{\,\frac{\pi }{2}}}{{\cos \left( u \right)\,du}}\\ & = \left. {{{\bf{e}}^y}} \right|_0^{\frac{1}{2}} + \left. {\frac{2}{\pi }\sin u} \right|_0^{\frac{\pi }{2}}\\ & = {{\bf{e}}^{\frac{1}{2}}} - {{\bf{e}}^0} + \frac{2}{\pi }\sin \frac{\pi }{2} - \frac{2}{\pi }\sin 0\\ & = {{\bf{e}}^{\frac{1}{2}}} - 1 + \frac{2}{\pi }\end{align*}\]d \( \displaystyle \int_{{\,\frac{\pi }{3}}}^{{\,0}}{{3\sin \left( {\frac{z}{2}} \right) - 5\cos \left( {\pi - z} \right)\,dz}}\) Yechim
Bu integral ikkita almashtirishni talab qiladi. Shunday qilib, avval integralni ajrating, shunda biz har bir atamani almashtirishni amalga oshiramiz.
\[\int_{{\,\frac{\pi }{3}}}^{{\,0}}{{3\sin \left( {\frac{z}{2}} \right) - 5\cos \left( {\pi - z} \right)\,dz}} = \int_{{\,\frac{\pi }{3}}}^{{\,0}}{{3\sin \left( {\frac{z}{2}} \right)\,dz}} - \int_{{\,\frac{\pi }{3}}}^{{\,0}}{{5\cos \left( {\pi - z} \right)\,dz}}\]Bu integrallar uchun ikkita almashtirish mavjud.
\[\begin{align*}u & = \frac{z}{2}\hspace{0.5in}du = \frac{1}{2}dz\hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,dz = 2du\\ z & = \frac{\pi }{3}\hspace{0.25in}\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,u = \frac{\pi }{6}\hspace{0.5in}z = 0\,\,\,\,\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,u = 0\end{align*}\] \[\begin{align*}v & = \pi - z\hspace{0.25in}dv = - dz\hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,dz = - dv\\ z & = \frac{\pi }{3}\hspace{0.25in}\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,v = \frac{{2\pi }}{3}\hspace{0.5in}z = 0\hspace{0.25in}\,\,\,\, \Rightarrow \hspace{0.25in}\,\,v = \pi \end{align*}\]Mana bu muammoning integrali.
\[\begin{align*}\int_{{\,\frac{\pi }{3}}}^{{\,0}}{{3\sin \left( {\frac{z}{2}} \right) - 5\cos \left( {\pi - z} \right)\,dz}} & = 6\int_{{\,\frac{\pi }{6}}}^{{\,0}}{{\sin \left( u \right)\,du}} + 5\int_{{\,\frac{{2\pi }}{3}}}^{{\,\pi }}{{\cos \left( v \right)\,dv}}\\ & = \left. { - 6\cos \left( u \right)} \right|_{\frac{\pi }{6}}^0 + \left. {5\sin \left( v \right)} \right|_{\frac{{2\pi }}{3}}^\pi \\ & = 3\sqrt 3 - 6 + \left( { - \frac{{5\sqrt 3 }}{2}} \right)\\ & = \frac{{\sqrt 3 }}{2} - 6\end{align*}\]Keyingi misollar to'plami aniq integrallar haqida juda muhim nuqtani unutmasligimizga ishonch hosil qilish uchun mo'ljallangan.
- \( \displaystyle \int_{{\, - 5}}^{{\,5}}{{\frac{{4t}}{{2 - 8{t^2}}}\,dt}}\)
- \( \displaystyle \int_{{\,3}}^{{\,5}}{{\frac{{4t}}{{2 - 8{t^2}}}\,dt}}\)
a \( \displaystyle \int_{{\, - 5}}^{{\,5}}{{\frac{{4t}}{{2 - 8{t^2}}}\,dt}}\) Yechim
Ushbu integral bilan ehtiyot bo'ling. \(t = \pm \frac{1}{2}\) da maxraj nolga teng va bu ikkalasi ham integratsiya oralig'ida. Shuning uchun bu integral intervalda uzluksiz emas va shuning uchun integral bajarilmaydi.
Aniq integrallar bilan ehtiyot bo'ling va nolga bo'linish masalalarini qidiring. Oldingida ularni aniqlash oson edi, chunki bizda mavjud bo'lgan barcha muammolarni nolga bo'lish o'zgaruvchining o'zi nolga teng edi. O'zgartirish muammolariga o'tganimizdan so'ng, ularni aniqlash har doim ham oson bo'lmaydi, shuning uchun avval integrandni tez ko'rib chiqing va integrand bilan uzluksizlik muammosi bor yoki yo'qligini va ular integratsiya oralig'ida yuzaga kelganligini tekshiring.
b \( \displaystyle \int_{{\,3}}^{{\,5}}{{\frac{{4t}}{{2 - 8{t^2}}}\,dt}}\) Yechim
Endi, bu holda integral bajarilishi mumkin, chunki ikkita uzilish nuqtasi, \(t = \pm \frac{1}{2}\) ikkalasi ham integrallash oralig'idan tashqarida. Bu holatda almashtirish va konvertatsiya qilingan limitlar:
\[\begin{align*}u & = 2 - 8{t^2}\hspace{0.25in}\,\,\,\,\,\,du = - 16t\,dt\hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,\,\,tdt = - \frac{1}{{16}}du\\ t & = 3\hspace{0.25in}\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,u = - 70\hspace{0.5in}t = 5\,\,\,\,\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,u = - 198\end{align*}\]U holda integral,
\[\begin{align*}\int_{{\,3}}^{{\,5}}{{\frac{{4t}}{{2 - 8{t^2}}}\,dt}} & = - \frac{4}{{16}}\int_{{\, - 70}}^{{\, - 198}}{{\frac{1}{u}\,du}}\\ & = \left. { - \frac{1}{4}\ln \left| u \right|} \right|_{ - 70}^{ - 198}\\ & = - \frac{1}{4}\left( {\ln \left( {198} \right) - \ln \left( {70} \right)} \right)\end{align*}\]Keling, boshqa misollar to'plamini ishlab chiqaylik. Bular oldingi to'plamlarga qaraganda biroz qattiqroq (hech bo'lmaganda tashqi ko'rinishida).
- \( \displaystyle \int_{{\,0}}^{{\,\ln \left( {1 + \pi } \right)}}{{{{\bf{e}}^x}\cos \left( {1 - {{\bf{e}}^x}} \right)}}\,dx\)
- \( \displaystyle \int_{{{{\bf{e}}^2}}}^{{{{\bf{e}}^6}}}{{\frac{{{{\left[ {\ln t} \right]}^4}}}{t}\,dt}}\)
- \( \displaystyle \int_{{\frac{\pi }{{12}}}}^{{\,\frac{\pi }{9}}}{{\frac{{\sec \left( {3P} \right)\tan \left( {3P} \right)}}{{\sqrt[3]{{2 + \sec \left( {3P} \right)}}}}\,dP}}\)
- \( \displaystyle \int_{{\, - \pi }}^{{\,\frac{\pi }{2}}}{{\cos \left( x \right)\cos \left( {\sin \left( x \right)} \right)\,dx}}\)
- \( \displaystyle \int_{{\frac{1}{{50}}}}^{2}{{\frac{{{{\bf{e}}^{\frac{2}{w}}}}}{{{w^2}}}\,dw}}\)
a \( \displaystyle \int_{{\,0}}^{{\,\ln \left( {1 + \pi } \right)}}{{{{\bf{e}}^x}\cos \left( {1 - {{\bf{e}}^x}} \right)}}\,dx\) Yechim
Bu holatda cheklovlar biroz g'ayrioddiy, ammo bu ba'zida sodir bo'ladi, shuning uchun bundan juda hayajonlanmang. Mana almashtirish.
\[\begin{align*}u & = 1 - {{\bf{e}}^x}\hspace{0.5in}du = - {{\bf{e}}^x}dx\\ x & = 0\hspace{0.5in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,u = 1 - {{\bf{e}}^0} = 1 - 1 = 0\\ x & = \ln \left( {1 + \pi } \right)\hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,u = 1 - {{\bf{e}}^{\ln \left( {1 + \pi } \right)}} = 1 - \left( {1 + \pi } \right) = - \pi \end{align*}\]U holda integral,
\[\begin{align*}\int_{{\,0}}^{{\,\ln \left( {1 + \pi } \right)}}{{{{\bf{e}}^x}\cos \left( {1 - {{\bf{e}}^x}} \right)}}\,dx & = - \int_{{\,0}}^{{\, - \pi }}{{\cos u}}\,du\\ & = - \left. {\sin \left( u \right)} \right|_0^{ - \pi }\\ & = - \left( {\sin \left( { - \pi } \right) - \sin 0} \right) = 0\end{align*}\]b \( \displaystyle \int_{{{{\bf{e}}^2}}}^{{{{\bf{e}}^6}}}{{\frac{{{{\left[ {\ln t} \right]}^4}}}{t}\,dt}}\) Yechim
Mana bu muammo uchun almashtirish va konvertatsiya qilingan chegaralar.
\[\begin{align*}u & = \ln t\hspace{0.5in}du = \frac{1}{t}dt\\ t & = {{\bf{e}}^2}\hspace{0.25in}\,\, \Rightarrow \,\,\,\,\,\,\,\,u = \ln {{\bf{e}}^2} = 2\hspace{0.5in}t = {{\bf{e}}^6}\hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}\,\,u = \ln {{\bf{e}}^6} = 6\end{align*}\]Integral bu,
\[\begin{align*}\int_{{{{\bf{e}}^2}}}^{{{{\bf{e}}^6}}}{{\frac{{{{\left[ {\ln t} \right]}^4}}}{t}\,dt}} & = \int_{{\,2}}^{{\,6}}{{{u^4}\,du}}\\ & = \left. {\frac{1}{5}{u^5}} \right|_2^6\\ & = \frac{{7744}}{5}\end{align*}\]c \( \displaystyle \int_{{\frac{\pi }{{12}}}}^{{\,\frac{\pi }{9}}}{{\frac{{\sec \left( {3P} \right)\tan \left( {3P} \right)}}{{\sqrt[3]{{2 + \sec \left( {3P} \right)}}}}\,dP}}\) Yechim
Mana almashtirish va o'zgartirilgan limitlar va almashtirishdan juda hayajonlanmang. Bu ishda biroz chalkash, lekin ba'zida bu sodir bo'lishi mumkin.
\[\begin{align*}u & = 2 + \sec \left( {3P} \right)\,\,\,\,\,\,\,\,du = 3\sec \left( {3P} \right)\tan \left( {3P} \right)dP\,\,\,\,\, \Rightarrow \,\,\,\,\,\sec \left( {3P} \right)\tan \left( {3P} \right)dP = \frac{1}{3}du\\ P & = \frac{\pi }{{12}}\hspace{0.5in} \Rightarrow \hspace{0.25in}u = 2 + \sec \left( {\frac{\pi }{4}} \right) = 2 + \sqrt 2 \\ P & = \frac{\pi }{9}\hspace{0.5in} \Rightarrow \hspace{0.25in}u = 2 + \sec \left( {\frac{\pi }{3}} \right) = 4\end{align*}\]Mana integral,
\[\begin{align*}\int_{{\frac{\pi }{{12}}}}^{{\,\frac{\pi }{9}}}{{\frac{{\sec \left( {3P} \right)\tan \left( {3P} \right)}}{{\sqrt[3]{{2 + \sec \left( {3P} \right)}}}}\,dP}} & = \frac{1}{3}\int_{{2 + \sqrt 2 }}^{4}{{{u^{ - \frac{1}{3}}}\,du}}\\ & = \left. {\frac{1}{2}{u^{\frac{2}{3}}}} \right|_{2 + \sqrt 2 }^4\\ & = \frac{1}{2}\left( {{4^{\frac{2}{3}}} - {{\left( {2 + \sqrt 2 } \right)}^{\frac{2}{3}}}} \right)\end{align*}\]Shunday qilib, nafaqat almashtirish tartibsiz edi, balki bizda chalkash javob ham bor, lekin bu hayot vaqti-vaqti bilan.
d \( \displaystyle \int_{{\, - \pi }}^{{\,\frac{\pi }{2}}}{{\cos \left( x \right)\cos \left( {\sin \left( x \right)} \right)\,dx}}\) Yechim
Bu muammo ko'rinadigan darajada yomon emas. Bu erda almashtirish va konvertatsiya qilingan limitlar.
\[\begin{align*}u & = \sin x\hspace{0.25in}du = \cos x\,dx\\ x & = \frac{\pi }{2}\hspace{0.25in}\,\,\,\, \Rightarrow \hspace{0.25in}\,u = \sin \frac{\pi }{2} = 1\hspace{0.5in}x = - \pi \hspace{0.25in}\,\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,\,u = \sin \left( { - \pi } \right) = 0\end{align*}\]Integrandning eng oldingi qismidagi kosinus differensialda almashtiriladi va shuning uchun bu integrand aslida sezilarli darajada soddalashadi. Mana integral.
\[\begin{align*}\int_{{\, - \pi }}^{{\,\frac{\pi }{2}}}{{\cos \left( x \right)\cos \left( {\sin \left( x \right)} \right)\,dx}} & = \int_{{\,0}}^{{\,1}}{{\cos u\,du}}\\ & = \left. {\sin \left( u \right)} \right|_0^1\\ & = \sin \left( 1 \right) - \sin \left( 0 \right)\\ & = \sin \left( 1 \right)\end{align*}\]Bunday javoblardan hayajonlanmang. Vaqti-vaqti bilan biz shunga o'xshash trig funktsiyasini baholash bilan yakunlaymiz.
e \( \displaystyle \int_{{\frac{1}{{50}}}}^{2}{{\frac{{{{\bf{e}}^{\frac{2}{w}}}}}{{{w^2}}}\,dw}}\) Yechim
Bu ham qiyin almashtirish (hech bo'lmaganda uni ko'rmaguningizcha). Mana,
\[\begin{align*}u & = \frac{2}{w}\hspace{0.5in}du = - \frac{2}{{{w^2}}}dw\hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,\,\frac{1}{{{w^2}}}dw = - \frac{1}{2}du\\ w & = 2\hspace{0.25in}\,\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,\,u = 1\hspace{0.5in}w = \frac{1}{{50}}\hspace{0.25in}\,\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,u = 100\end{align*}\]Mana integral.
\[\begin{align*}\int_{{\frac{1}{{50}}}}^{2}{{\frac{{{{\bf{e}}^{\frac{2}{w}}}}}{{{w^2}}}\,dw}} & = - \frac{1}{2}\int_{{\,100}}^{{\,1}}{{{{\bf{e}}^u}\,du}}\\ & = \left. { - \frac{1}{2}{{\bf{e}}^u}} \right|_{100}^1\\ & = - \frac{1}{2}\left( {{{\bf{e}}^1} - {{\bf{e}}^{100}}} \right)\end{align*}\]Ushbu so'nggi misollar to'plamida biz ba'zi qiyin almashtirishlar va tartibsiz cheklovlarni ko'rdik, ammo bu ba'zi almashtirish muammolari bilan hayot haqiqati va shuning uchun ular sodir bo'lganda ularni hal qilishga tayyor bo'lishimiz kerak.