5.7-bo'lim : Aniq integrallarni hisoblash
11. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\[\int_{0}^{\pi }{{\sec \left( z \right)\tan \left( z \right) - 1\,dz}}\]Yechim
Ushbu integral bilan ehtiyot bo'ling. Eslatib o'tamiz,
\[\sec \left( z \right) = \frac{1}{{\cos \left( z \right)}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\tan \left( z \right) = \frac{{\sin \left( z \right)}}{{\cos \left( z \right)}}\]Shuni ham yodda tutingki, \(\cos \left( {\frac{\pi }{2}} \right) = 0\) va \(x = \frac{\pi }{2}\) biz integrallashayotgan \(\left[ {0,\pi } \right]\) oralig'ida va shuning uchun bu intervalda uzluksiz emas.
Shuning uchun bu integralni bajarib bo'lmaydi.
Agar integrand ratsional ifoda sifatida aniq yozilmagan bo'lsa, ko'pincha integrallarda nolga bo'linishning bunday turlarini e'tiborsiz qoldirish oson. Shunday qilib, ehtiyot bo'ling va nolga bo'linish ba'zan integranda "yashirin" bo'lishi mumkinligini unutmang!