Bizga quyidagi trigni almashtirish kerak bo'ladi.

\[z + 3 = 7\sin \left( \theta \right)\]

Keyin ildizni yo'q qilaylik.

\[{\left( {49 - {{\left( {z + 3} \right)}^2}} \right)^{\frac{3}{2}}} = {\left[ {\sqrt {49 - \left( {z + 3} \right)^{2}} } \right]^3} = {\left[ {\sqrt {49 - 49{{\sin }^2}\left( \theta \right)} } \right]^3} = {\left[ {7\sqrt {{{\cos }^2}\left( \theta \right)} } \right]^3} = 343{\left| {\cos \left( \theta \right)} \right|^3}\]

Keyinchalik, biz noaniq integralni bajarayotganimiz sababli, biz kosinusni ijobiy deb hisoblaymiz va shuning uchun biz olish uchun mutlaq qiymat satrlarini tashlab qo'yishimiz mumkin,

\[{\left( {49 - {{\left( {z + 3} \right)}^2}} \right)^{\frac{3}{2}}} = 343{\cos ^3}\left( \theta \right)\]

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

\[\left( 1 \right)dz = 7\cos \left( \theta \right)\,d\theta \hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}dz = 7\cos \left( \theta \right)\,d\theta \]

Eslatib o'tamiz, bu erda biz qilishimiz kerak bo'lgan narsa almashtirishning o'ng va chap tomonlari uchun differentsialni hisoblashdir.

Endi haqiqiy almashtirishni qilaylik.

\[\int{{\frac{{{{\left( {z + 3} \right)}^5}}}{{{{\left( {40 - 6z - {z^2}} \right)}^{\frac{3}{2}}}}}\,dz}} = \int{{\frac{{16807{{\sin }^5}\left( \theta \right)}}{{343{{\cos }^3}\left( \theta \right)}}\left( {7\cos \left( \theta \right)} \right)\,d\theta }} = 343\int{{\frac{{{{\sin }^5}\left( \theta \right)}}{{{{\cos }^2}\left( \theta \right)}}\,d\theta }}\]

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.

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

\[\begin{align*}\int{{\frac{{{{\left( {z + 3} \right)}^5}}}{{{{\left( {40 - 6z - {z^2}} \right)}^{\frac{3}{2}}}}}\,dz}} & = 343\int{{\frac{{{{\left[ {1 - {{\cos }^2}\left( \theta \right)} \right]}^2}}}{{{{\cos }^2}\left( \theta \right)}}\sin \left( \theta \right)\,d\theta }}\hspace{0.25in}\hspace{0.25in}u = \cos \left( \theta \right)\\ & = - 343\int{{\frac{{{{\left[ {1 - {u^2}} \right]}^2}}}{{{u^2}}}\,du}} = - 343\int{{{u^{ - 2}} - 2 + {u^2}\,du}}\\ & = - 343\left( { - {u^{ - 1}} - 2u + \frac{1}{3}{u^3}} \right) + c\\ & = - 343\left( { - \frac{1}{{\cos \left( \theta \right)}} - 2\cos \left( \theta \right) + \frac{1}{3}{{\cos }^3}\left( \theta \right)} \right) + c\end{align*}\]

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.

Yakuniy qadam sifatida biz \(z\)-ga qaytishimiz kerak. Buning uchun bizga tez uchburchak kerak bo'ladi. Mana o'sha ish.

Bizda mavjud almashtirishdan, \[\sin \left( \theta \right) = \frac{{z + 3}}{7}\,\,\,\,\,\left( { = \frac{{{\mbox{opp}}}}{{{\mbox{hyp}}}}} \right)\] To'g'ri uchburchakdan biz olamiz, \[\cos \left( \theta \right) = \frac{{\sqrt {49 - {{\left( {z + 3} \right)}^2}} }}{7}\]

U holda integral,

\[\int{{\frac{{{{\left( {z + 3} \right)}^5}}}{{{{\left( {40 - 6z - {z^2}} \right)}^{\frac{3}{2}}}}}\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{2401}}{{\sqrt {49 - {{\left( {z + 3} \right)}^2}} }} + 98\sqrt {49 - {{\left( {z + 3} \right)}^2}} - \frac{{{{\left( {49 - {{\left( {z + 3} \right)}^2}} \right)}^{\frac{3}{2}}}}}{3} + c}}\]