5.3-bo'lim : Noaniq integrallarni almashtirish qoidasi
16. \( \displaystyle \int{{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}}\) 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 sinusning hosilasiga juda o'xshaydi.
\[\frac{d}{{du}}\left( {{{\sin }^{ - 1}}\left( u \right)} \right) = \frac{1}{{\sqrt {1 - {u^2}} }}\]Shunday qilib, keling, integral shunga o'xshash bo'lishi uchun biroz qayta yozaylik.
\[\int{{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}} = \int{{\frac{1}{{\sqrt {4\left( {1 - \frac{9}{4}{w^2}} \right)} }}\,dw}} = \frac{1}{2}\int{{\frac{1}{{\sqrt {1 - \frac{9}{4}{w^2}} }}\,dw}}\]
Maslahat: Integrandning yana bir oz qayta yozilishi bu teskari sinus hosilasiga deyarli o'xshash bo'lishi kerak va o'rnini almashtirish juda aniq bo'lishi kerak.
2-qadam
Keling, integrandni yana bir marta qayta yozaylik.
\[\int{{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}} = \frac{1}{2}\int{{\frac{1}{{\sqrt {1 - {{\left( {\frac{{3w}}{2}} \right)}^2}} }}\,dw}}\]Shu o'rinda quyidagi almashtirish biz uchun ishlayotganini ko'rishimiz mumkin.
\[u = \frac{{3w}}{2}\hspace{0.5in} \to \hspace{0.5in}du = \frac{3}{2}dw\hspace{0.5in} \to \hspace{0.5in}\,dw = \frac{2}{3}du\]3-qadam
O'zgartirishni amalga oshirish va integralni baholash,
\[\int{{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}} = \frac{1}{2}\left( {\frac{2}{3}} \right)\int{{\frac{1}{{\sqrt {1 - {u^2}} }}\,du}} = \frac{1}{3}{\sin ^{ - 1}}\left( u \right) + c\]
Maslahat: Integranddagi asl o'zgaruvchi \(u\) emasligini unutmang!
4-qadam
Nihoyat, asl o'zgaruvchiga qaytishni unutmang!
\[\int{{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{3}{{\sin }^{ - 1}}\left( {\frac{{3w}}{2}} \right) + c}}\]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.