5.4-bo'lim : Ko'proq almashtirish qoidasi
6. \( \displaystyle \int{{20{{\bf{e}}^{2 - 8w}}\sqrt {1 + {{\bf{e}}^{2 - 8w}}} \, + 7{w^3} - 6\,\,\sqrt[3]{w}\,dw}}\) ni baholang.
Yechim
Shubhasiz, birinchi atama almashtirishga muhtoj, ikkinchi va uchinchi shartlar esa kerak emas. Shunday qilib, biz birinchi navbatda integralni quyidagicha ajratishimiz kerak.
\[\int{{20{{\bf{e}}^{2 - 8w}}\sqrt {1 + {{\bf{e}}^{2 - 8w}}} \, + 7{w^3} - 6\,\,\sqrt[3]{w}\,dw}} = \int{{20{{\bf{e}}^{2 - 8w}}\sqrt {1 + {{\bf{e}}^{2 - 8w}}} \,dw}} + \int{{7{w^3} - 6\,\,\sqrt[3]{w}\,dw}}\]2-qadam
Birinchi integral uchun zarur bo'lgan almashtirish:
\[u = 1 + {{\bf{e}}^{2 - 8w}}\]Agar siz asosiy almashtirish mexanikasidan mamnun bo'lmasangiz, oldingi bo'limda ba'zi muammolar bilan shug'ullanishingiz kerak, chunki biz ushbu bo'limda asoslar haqida batafsil ma'lumot bermaymiz. Ushbu bo'limdagi muammolar almashtirishning asosiy mexanikasi bilan ancha qulay bo'lganlar uchun mo'ljallangan va bir nechta "ilg'or" almashtirishlarni o'z ichiga oladi.
3-qadam
Bu erda almashtirish uchun differentsial ish.
\[du = - 8{{\bf{e}}^{2 - 8w}}dw\hspace{0.25in} \to \hspace{0.25in}\,{{\bf{e}}^{2 - 8w}}dw = - \frac{1}{8}du\]Endi almashtirishlarni bajarish va integrallarni baholash,
\[\begin{align*}\int{{20{{\bf{e}}^{2 - 8w}}\sqrt {1 + {{\bf{e}}^{2 - 8w}}} \, + 7{w^3} - 6\,\,\sqrt[3]{w}\,dw}} & = - \frac{{20}}{8}\int{{{u^{\frac{1}{2}}}\,du}} + \int{{7{w^3} - 6{w^{\frac{1}{3}}}\,dw}}\\ & = - \frac{5}{3}{u^{\frac{3}{2}}} + \frac{7}{4}{w^4} - \frac{9}{2}{w^{\frac{4}{3}}} + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{5}{3}{{\left( {1 + {{\bf{e}}^{2 - 8w}}} \right)}^{\frac{3}{2}}} + \frac{7}{4}{w^4} - \frac{9}{2}{w^{\frac{4}{3}}} + c}}\end{align*}\]Integralni baholagandan so'ng asl o'zgaruvchiga qaytishni unutmang!