7.1-bo'lim : Qismlar bo'yicha integratsiya
4. \( \displaystyle \int{{6{{\tan }^{ - 1}}\left( {\frac{8}{w}} \right)\,dw}}\) ni baholang.
Yechim
Bu erda birinchi qadam \(u\) va \(dv\) ni tanlashdir.
E'tibor bering, agar biz \(dv\) uchun teskari tangensni tanlasak, \(v\)ni olishning yagona yo'li \(dv\)ni integrallashdir, shuning uchun javobni olish uchun javobni bilishimiz kerak va bu biz uchun ishlamaydi. Shuning uchun, teskari tangens uchun yagona haqiqiy tanlov bu \(u\) bo'lishiga ruxsat berishdir.
Mana, \(u\) va \(dv\) uchun tanlovlarimiz.
\[u = 6{\tan ^{ - 1}}\left( {\frac{8}{w}} \right)\hspace{0.5in}dv = \,dw\]\(dw\) ni unutmang! \(dw\) differensialini hali ham \(dv\) ga qo'yish kerak, garchi integralda boshqa hech narsa qolmagan bo'lsa ham.
2-qadam
Keyinchalik, \(du\) (\(u\)ni farqlash orqali) va \(v\) (\(dv\) ni birlashtirib) hisoblashimiz kerak.
\[\begin{align*}u & = 6{\tan ^{ - 1}}\left( {\frac{8}{w}} \right) & \hspace{0.5in} & \to & \hspace{0.25in}du & = 6\frac{{ - \frac{8}{{{w^2}}}}}{{1 + {{\left( {\frac{8}{w}} \right)}^2}}}dw = 6\frac{{ - \frac{8}{{{w^2}}}}}{{1 + \frac{{64}}{{{w^2}}}}}dw\\ dv & = \,dw & \hspace{0.5in} & \to & \hspace{0.25in}v & = w\end{align*}\]3-qadam
Ushbu muammoni hal qilish uchun biz \(du\) da quyidagi tarzda qayta yozishimiz kerak:
\[du = \frac{{ - 48}}{{{w^2} + 64}}dw\]\(u\), \(du\), \(v\) va \(dv\) ni qismlar bo‘yicha integratsiyaga ulash formulasini beradi:
\[ \int{{6{{\tan }^{ - 1}}\left( {\frac{8}{w}} \right)\,dw}} = 6w\tan ^{ - 1}\left( {\frac{8}{w}} \right) + 48\int{{\frac{w}{{{w^2} + 64}}\,dw}}\]4-qadam
Yaxshi, biz olgan yangi integral osonlik bilan bajariladi (\(u = 64 + {w^2}\) almashtirish bilan) va shuning uchun bu masalani hal qilish uchun biz qilishimiz kerak bo'lgan yagona narsa integralni bajarishdir.
\[ \int{{6{{\tan }^{ - 1}}\left( {\frac{8}{w}} \right)\,dw}} = \require{bbox} \bbox[2pt,border:1px solid black]{{6w\tan ^{ - 1}\left( {\frac{8}{w}} \right) + 24\ln \left| {{w^2} + 64} \right| + c}}\]