5.4-bo'lim : Ko'proq almashtirish qoidasi
12. \( \displaystyle \int{{\frac{{7x + 2}}{{\sqrt {1 - 25{x^2}} }}\,dx}}\) ni baholang.
Yechim
Aytganimizdek, biz bu muammoni hal qila olmaymiz. Ushbu integralni bajarish uchun biz integralni quyidagi tarzda qayta yozishimiz kerak.
\[\int{{\frac{{7x + 2}}{{\sqrt {1 - 25{x^2}} }}\,dx}} = \int{{\frac{{7x}}{{\sqrt {1 - 25{x^2}} }}\,dx}} + \int{{\frac{2}{{\sqrt {1 - 25{x^2}} }}\,dx}}\]2-qadam
Endi ikkinchi integral teskari sinus bo'lishi mumkin (garchi biz bu integralni qayta yozishimiz kerak bo'lsa ham) va birinchisi oddiy almashtirish biz uchun ishlaydi kabi ko'rinadi.
Shunday qilib, ikkinchi integralni qayta yozish.
\[\int{{\frac{{7x + 2}}{{\sqrt {1 - 25{x^2}} }}\,dx}} = \int{{\frac{{7x}}{{\sqrt {1 - 25{x^2}} }}\,dx}} + 2\int{{\frac{1}{{\sqrt {1 - {{\left( {5x} \right)}^2}} }}\,dx}}\]3-qadam
Endi bizga har bir integral uchun almashtirish kerak bo'ladi. Bu erda har bir integral uchun kerak bo'ladigan almashtirishlar.
\[u = 1 - 25{x^2}\hspace{0.5in}v = 5x\,\,\,\,\,\,\left( {{\mbox{so }}{v^2} = 25{x^2}} \right)\]4-qadam
Bu erda almashtirish uchun differentsial ish.
\[du = - 50x\,dx\hspace{0.25in} \to \hspace{0.25in}x\,dx = - \frac{1}{{50}}du\hspace{0.5in}dv = 5dx\hspace{0.25in} \to \hspace{0.25in}dx = \frac{1}{5}dv\]Endi almashtirishlarni bajarish va integrallarni baholash,
\[\begin{align*}\int{{\frac{{7x + 2}}{{\sqrt {1 - 25{x^2}} }}\,dx}} & = - \frac{7}{{50}}\int{{{u^{ - \,\frac{1}{2}}}\,du}} + \frac{2}{5}\int{{\frac{1}{{\sqrt {1 - {v^2}} }}\,dv}} = - \frac{7}{{25}}{u^{\frac{1}{2}}} + \frac{2}{5}{\sin ^{ - 1}}\left( v \right) + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{7}{{25}}{{\left( {1 - 25{x^2}} \right)}^{\frac{1}{2}}} + \frac{2}{5}{{\sin }^{ - 1}}\left( {5x} \right) + c}}\end{align*}\]Integralni baholagandan so'ng asl o'zgaruvchiga qaytishni unutmang!