Keling, birinchi navbatda ildizni yo'q qilish uchun almashtirishdan foydalanamiz.

\[\sqrt {2 + 9{z^2}} = \sqrt {2 + 2{{\tan }^2}\left( \theta \right)} = \sqrt 2 \sqrt {{{\sec }^2}\left( \theta \right)} = \sqrt 2 \left| {\sec \left( \theta \right)} \right|\]

Xo'sh, bu holda bizda \(z\) limitlari bor va shuning uchun biz \(\theta \) uchun limitlarni olishimiz mumkin, bu bizga mutlaq qiymat satrlarini yo'q qilish uchun tangensning ijobiy yoki salbiy ekanligini aniqlashga imkon beradi.

Shunday qilib, keling, \(\theta \) uchun ba'zi limitlarni olaylik.

\[\begin{align*}z & = 1:\,\,\,\,\,1 = \frac{{\sqrt 2 }}{3}\tan \left( \theta \right)\hspace{0.25in} \to \hspace{0.25in}\tan \left( \theta \right) = \frac{3}{{\sqrt 2 }}\hspace{0.25in} \to \hspace{0.25in}\theta = {\tan ^{ - 1}}\left( {\frac{3}{{\sqrt 2 }}} \right) = 1.1303\\ z & = 4:\,\,\,\,4 = \frac{{\sqrt 2 }}{3}\tan \left( \theta \right)\hspace{0.25in} \to \hspace{0.25in}\tan \left( \theta \right) = \frac{{12}}{{\sqrt 2 }}\hspace{0.25in} \to \hspace{0.25in}\theta = {\tan ^{ - 1}}\left( {\frac{{12}}{{\sqrt 2 }}} \right) = 1.4535\end{align*}\]

Shunday qilib, ushbu muammo uchun \(\theta \) \(1.1303 \le \theta \le 1.4535\) diapazonida va ular birinchi kvadrantda. Birinchi kvadrantda biz bilamizki, kosinus va shuning uchun sekant ijobiydir va shuning uchun biz shunchaki mutlaq qiymat chiziqlarini tashlab qo'yishimiz mumkin. Bu beradi,

\[\sqrt {2 + 9{z^2}} = \sqrt 2 \sec \left( \theta \right)\]

O'zgartirishga tayyorgarlikning yakuniy bosqichi uchun, keling, differentsialni ham hisoblaylik, shuning uchun uni almashtirishda ishlatishni unutmang!

\[dz = \frac{{\sqrt 2 }}{3}{\sec ^2}\left( \theta \right)\,d\theta \]

Endi haqiqiy almashtirishni qilaylik.

\[\begin{align*}\int_{1}^{4}{{2{z^5}\sqrt {2 + 9{z^2}} \,dz}} & = \int_{{1.1303}}^{{1.4535}}{{2{{\left( {\frac{{\sqrt 2 }}{3}} \right)}^5}{{\tan }^5}\left( \theta \right)\left( {\sqrt 2 \sec \left( \theta \right)} \right)\,\left( {\frac{{\sqrt 2 }}{3}{{\sec }^2}\left( \theta \right)} \right)d\theta }}\\ & = \frac{{16\sqrt 2 }}{{729}}\int_{{1.1303}}^{{1.4535}}{{{{\tan }^5}\left( \theta \right){{\sec }^3}\left( \theta \right)d\theta }}\end{align*}\]

Oldingi bosqichda biz hisoblagan differensialni almashtirishni unutmang. Bu, ehtimol, triglarni almashtirishda eng keng tarqalgan xatodir. Differensialni unutish muammoni sezilarli darajada o'zgartirishi mumkin, ko'pincha integralni baholash juda qiyin bo'ladi.

Shuningdek, podstansiyani ishga tushirgandan so'ng biz \(y\) chegaralarini \(\theta \) chegaralariga almashtirganimizga e'tibor bering. Bu keyingi bosqichda yordam beradi.

Endi biz integralni baholashimiz kerak. Mana o'sha ish.

\[\begin{align*}\int_{1}^{4}{{2{z^5}\sqrt {2 + 9{z^2}} \,dz}} & = \frac{{16\sqrt 2 }}{{729}}\int_{{1.1303}}^{{1.4535}}{{{{\left[ {{{\sec }^2}\left( \theta \right) - 1} \right]}^2}{{\sec }^2}\left( \theta \right)\,\,\tan \left( \theta \right)\sec \left( \theta \right)d\theta }}\\ & = \frac{{16\sqrt 2 }}{{729}}\int_{{\sec \left( {1.1303} \right)}}^{{\sec \left( {1.4535} \right)}}{{{{\left[ {{u^2} - 1} \right]}^2}{u^2}du}}\hspace{0.75in}u = \sec \left( \theta \right)\\ & = \frac{{16\sqrt 2 }}{{729}}\int_{{2.3452}}^{{8.5440}}{{{u^6} - 2{u^4} + {u^2}du}}\\ & = \left. {\frac{{16\sqrt 2 }}{{729}}\left[ {\frac{1}{7}{u^7} - \frac{2}{5}{u^5} + \frac{1}{3}{u^3}} \right]} \right|_{2.3452}^{8.5440} = \require{bbox} \bbox[2pt,border:1px solid black]{{14178.20559}}\end{align*}\]

Trig funksiyalarini o'z ichiga olgan integrallarni baholash uchun biz tez-tez bajarishimiz kerak bo'lgan integralning barcha "standart" manipulyatsiyalarini unutmang. Agar siz ularni eslamasangiz, oldingi bo'limga qaytib, ularni yaxshi o'rganish uchun ba'zi amaliy muammolarni hal qilishingiz kerak bo'ladi.

Har bir trigni almashtirish muammosi trig funksiyalarini o'z ichiga olgan integralgacha kamayadi va ularning ko'pchiligini baholash uchun integralni biroz manipulyatsiya qilish kerak bo'ladi.

Shuni ham yodda tutingki, biz har bir almashtirishdagi chegaralarni "yangi" o'zgaruvchining chegaralariga aylantirganimiz uchun javobimiz bo'yicha hech qanday almashtirish ishlarini bajarishimiz shart emas edi!