5.3-bo'lim : Noaniq integrallarni almashtirish qoidasi
15. \( \displaystyle \int{{\frac{6}{{7 + {y^2}}}\,dy}}\) ni baholang.
Maslahat: Bu almashtirishda ehtiyot bo'ling. Integrand biroz tanish ko'rinishi kerak, shuning uchun biz uni yanada tanishroq shaklga solishga harakat qilishimiz kerak.
Yechim
Integrand teskari tangens hosilasiga juda o'xshaydi.
\[\frac{d}{{du}}\left( {{{\tan }^{ - 1}}\left( u \right)} \right) = \frac{1}{{1 + {u^2}}}\]Shunday qilib, keling, integral shunga o'xshash bo'lishi uchun biroz qayta yozaylik.
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \int{{\frac{6}{{7\left( {1 + \frac{1}{7}{y^2}} \right)}}\,dy}} = \frac{6}{7}\int{{\frac{1}{{1 + \frac{1}{7}{y^2}}}\,dy}}\]
Maslahat: Integrandning yana bir oz qayta yozilishi, bu teskari tangens hosilasiga deyarli o'xshash bo'lishi kerak va almashtirish juda aniq bo'lishi kerak.
2-qadam
Keling, integrandni yana bir marta qayta yozaylik.
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \frac{6}{7}\int{{\frac{1}{{1 + {{\left( {\frac{y}{{\sqrt 7 }}} \right)}^2}}}\,dy}}\]Shu o'rinda quyidagi almashtirish biz uchun ishlayotganini ko'rishimiz mumkin.
\[u = \frac{y}{{\sqrt 7 }}\hspace{0.5in} \to \hspace{0.5in}du = \frac{1}{{\sqrt 7 }}dy\hspace{0.5in} \to \hspace{0.5in}\,dy = \sqrt 7 du\]3-qadam
O'zgartirishni amalga oshirish va integralni baholash,
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \frac{6}{7}\left( {\sqrt 7 } \right)\int{{\frac{1}{{1 + {u^2}}}\,du}} = \frac{6}{{\sqrt 7 }}{\tan ^{ - 1}}\left( u \right) + c\]
Maslahat: Integranddagi asl o'zgaruvchi \(u\) emasligini unutmang!
4-qadam
Nihoyat, asl o'zgaruvchiga qaytishni unutmang!
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \frac{6}{7}\left( {\sqrt 7 } \right)\int{{\frac{1}{{1 + {u^2}}}\,du}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{6}{{\sqrt 7 }}{{\tan }^{ - 1}}\left( {\frac{y}{{\sqrt 7 }}} \right) + c}}\]Teskari trig funksiyalarini almashtirish, ularni birinchi marta bajarishni boshlaganingizda, ularni aniqlash biroz qiyin bo'lishi mumkin. Agar siz ularni etarlicha bajarsangiz, ularni aniqlash biroz osonroq bo'ladi.