7.3-bo'lim : Triglarni almashtirish
14. \( \displaystyle \int{{\frac{1}{{\sqrt {9{x^2} - 36x + 37} }}\,dx}}\) ni baholash uchun trigni almashtirishdan foydalaning.
Yechim
Bu erda qilishimiz kerak bo'lgan birinchi narsa ko'phaddagi kvadratni to'ldirishdir, bu biz trig almashtirishdan foydalanishimiz mumkin bo'lgan shaklga keltiramiz.
\[\begin{align*}9{x^2} - 36x + 37 & = 9\left( {{x^2} - 4x + \frac{{37}}{9}} \right) = 9\left( {{x^2} - 4x + 4 - 4 + \frac{{37}}{9}} \right) = 9\left[ {{{\left( {x - 2} \right)}^2} + \frac{1}{9}} \right]\\ & = 9{\left( {x - 2} \right)^2} + 1\end{align*}\]Integral endi,
\[\int{{\frac{1}{{\sqrt {9{x^2} - 36x + 37} }}\,dx}} = \int{{\frac{1}{{\sqrt {9{{\left( {x - 2} \right)}^2} + 1} }}\,dx}}\]Endi biz trigni almashtirishni davom ettirishimiz mumkin.
2-qadam
Bizga quyidagi trigni almashtirish kerak bo'ladi.
\[x - 2 = \frac{1}{3}\tan \left( \theta \right)\]Keyin ildizni yo'q qilaylik.
\[\sqrt {9{{\left( {x - 2} \right)}^2} + 1} = \sqrt {\tan {{\left( \theta \right)}^2} + 1} = \sqrt {{{\sec }^2}\left( \theta \right)} = \left| {\sec \left( \theta \right)} \right|\]Keyinchalik, biz noaniq integralni bajarayotganimiz uchun sekant musbat deb faraz qilamiz va shuning uchun biz olish uchun mutlaq qiymat satrlarini tashlab qo'yishimiz mumkin,
\[\sqrt {9{{\left( {x - 2} \right)}^2} + 1} = \sec \left( \theta \right)\]O'zgartirishga tayyorgarlikning yakuniy bosqichi uchun, keling, differentsialni ham hisoblaylik, shuning uchun uni almashtirishda ishlatishni unutmang!
\[\left( 1 \right)dx = \frac{1}{3}{\sec ^2}\left( \theta \right)\,d\theta \hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}dx = \frac{1}{3}{\sec ^2}\left( \theta \right)\,d\theta \]Eslatib o'tamiz, bu erda biz qilishimiz kerak bo'lgan narsa almashtirishning o'ng va chap tomonlari uchun differentsialni hisoblashdir.
3-qadam
Endi haqiqiy almashtirishni qilaylik.
\[\int{{\frac{1}{{\sqrt {9{x^2} - 36x + 37} }}\,dx}} = \int{{\frac{1}{{\sec \left( \theta \right)}}\left( {\frac{1}{3}{{\sec }^2}\left( \theta \right)} \right)\,d\theta }} = \frac{1}{3}\int{{\sec \left( \theta \right)\,d\theta }}\]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.
4-qadam
Endi biz integralni baholashimiz kerak. Mana o'sha ish.
\[\int{{\frac{1}{{\sqrt {9{x^2} - 36x + 37} }}\,dx}} = \frac{1}{3}\ln \left| {\sec \left( \theta \right) + \tan \left( \theta \right)} \right| + c\]E'tibor bering, bu triglarni almashtirish integrallaridan biri bo'lib, to'liq baholash uchun trig funktsiyalarini ko'p manipulyatsiya qilishni talab qilmaydi. Bu erda biz qilishimiz kerak bo'lgan yagona narsa, biz oldingi bo'limda \(\sec \left( \theta \right)\) integralini aniqlaganimiz va natijani bu erda qayta ishlatishimiz edi.
5-qadam
Yakuniy qadam sifatida biz \(x\)-ga qaytishimiz kerak. Buning uchun bizga tez uchburchak kerak bo'ladi. Mana o'sha ish.
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Bizda mavjud almashtirishdan, \[\tan \left( \theta \right) = \frac{{3\left( {x - 2} \right)}}{1}\,\,\,\,\,\left( { = \frac{{{\mbox{opp}}}}{{{\mbox{adj}}}}} \right)\]To'g'ri uchburchakdan biz olamiz, \[\sec \left( \theta \right) = \sqrt {9{{\left( {x - 2} \right)}^2} + 1} \] |
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U holda integral,
\[\int{{\frac{1}{{\sqrt {9{x^2} - 36x + 37} }}\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{3}\ln \left| {\sqrt {9{{\left( {x - 2} \right)}^2} + 1} + 3\left( {x - 2} \right)} \right| + c}}\]