5.4-bo'lim : Ko'proq almashtirish qoidasi
13. \( \displaystyle \int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}}\) ni baholang.
Maslahat: "Aniq" almashtirishdan foydalaning va almashtirish bir necha marta va turli xil usullarda ishlatilishi mumkinligini unutmang.
Yechim
Xo'sh, bu erda "aniq" almashtirish, ehtimol,
\[u = 8 + 3{z^4}\hspace{0.25in} \to \hspace{0.25in}du = 12{z^3}dz\hspace{0.25in}\,\,\,\,\,\, \to \hspace{0.25in}\,\,\,\,\,{z^3}dz = \frac{1}{{12}}du\]ammo, bu \({z^7}\) tufayli ishlayotganga o'xshamaydi.
2-qadam
Keling, integrandni tezda qayta yozaylik.
\[\int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} = \int{{{z^4}{z^3}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} = \int{{{z^4}{{\left( {8 + 3{z^4}} \right)}^8}\,{z^3}dz}}\]3-qadam
Endi e'tibor bering, biz barcha \(z\)-larni integrandga o'zgartirishimiz mumkin, bundan tashqari, oldingi \({z^4}\). Biroq, almashtirishdan e'tibor bering, biz \({z^4}\) uchun hal qila olamiz,
\[{z^4} = \frac{1}{3}\left( {u - 8} \right)\]4-qadam
Shu bilan biz endi almashtirishni amalga oshirishimiz va integralni baholashimiz mumkin.
\[\begin{align*}\int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} & = \frac{1}{{12}}\int{{\frac{1}{3}\left( {u - 8} \right){u^8}du}} = \frac{1}{{36}}\int{{{u^9} - 8{u^8}du}} = \frac{1}{{36}}\left( {\frac{1}{{10}}{u^{10}} - \frac{8}{9}{u^9}} \right) + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{36}}\left( {\frac{1}{{10}}{{\left( {8 + 3{z^4}} \right)}^{10}} - \frac{8}{9}{{\left( {8 + 3{z^4}} \right)}^9}} \right) + c}}\end{align*}\]