7.1-bo'lim : Qismlar bo'yicha integratsiya
3. \( \displaystyle \int{{\left( {3t + {t^2}} \right)\sin \left( {2t} \right)\,dt}}\) ni baholang.
Yechim
Bu erda birinchi qadam \(u\) va \(dv\) ni tanlashdir. Biz \(u\) va \(dv\) ni tanlamoqchimiz, shunda biz \(du\) va \(v\) ni hisoblab, hamma narsani Qismlar bo‘yicha integratsiya formulasiga ulaganimizda, biz oladigan yangi integral biz qila oladigan yoki hech bo‘lmaganda integral bo‘lib qoladi, uni hal qilish osonroq bo‘ladi.
Shuni hisobga olib, \(u\) va \(dv\) uchun quyidagi tanlovlar biz uchun ishlashi kerak.
\[u = 3t + {t^2}\hspace{0.5in}dv = \sin \left( {2t} \right)\,dt\]2-qadam
Keyinchalik, \(du\) (\(u\)ni farqlash orqali) va \(v\) (\(dv\) ni birlashtirish orqali) hisoblashimiz kerak.
\[\begin{align*}u & = 3t + {t^2} & \hspace{0.5in} & \to & \hspace{0.25in}du & = \left( {3 + 2t} \right)dt\\ dv & = \sin \left( {2t} \right)\,dt & \hspace{0.5in} & \to & \hspace{0.25in}v & = - \frac{1}{2}\cos \left( {2t} \right)\end{align*}\]3-qadam
\(u\), \(du\), \(v\) va \(dv\) ni Qismlar bo'yicha integratsiyaga ulash formulasini beradi:
\[ \int{{\left( {3t + {t^2}} \right)\sin \left( {2t} \right)\,dt}} = - \frac{1}{2}\left( {3t + {t^2}} \right)\cos \left( {2t} \right) + \frac{1}{2}\int{{\left( {3 + 2t} \right)\cos \left( {2t} \right)\,dt}}\]4-qadam
Endi, yangi integral hali ham biz faqat Calculus I texnikasi bilan qila oladigan integral emas. Biroq, biz qismlar bo'yicha boshqa integratsiyani amalga oshirishimiz mumkin bo'lgan narsa va \(t\)-ning quvvati bir marta kamayganligi sababli biz to'g'ri yo'nalishda ketyapmiz.
Shunday qilib, yangi integral uchun \(u\) va \(dv\) uchun tanlovlar.
\[\begin{align*}u & = 3 + 2t & \hspace{0.5in} & \to & \hspace{0.25in}du & = 2dt\\ dv & = \cos \left( {2t} \right)\,dt & \hspace{0.5in} & \to & \hspace{0.25in}v & = \frac{1}{2}\sin \left( {2t} \right)\end{align*}\]5-qadam
OK, hozir qilishimiz kerak bo'lgan yagona narsa \(u\) va \(dv\) ning yangi tanlovlarini 3-bosqichda olingan yangi integralga ulash va muammoni hal qilishdir.
\[\begin{align*}\int{{\left( {3t + {t^2}} \right)\sin \left( {2t} \right)\,dt}} & = - \frac{1}{2}\left( {3t + {t^2}} \right)\cos \left( {2t} \right) + \frac{1}{2}\left[ {\frac{1}{2}\left( {3 + 2t} \right)\sin \left( {2t} \right) - \int{{\sin \left( {2t} \right)\,dt}}} \right]\\ & = - \frac{1}{2}\left( {3t + {t^2}} \right)\cos \left( {2t} \right) + \frac{1}{2}\left[ {\frac{1}{2}\left( {3 + 2t} \right)\sin \left( {2t} \right) + \frac{1}{2}\cos \left( {2t} \right)} \right] + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{2}\left( {3t + {t^2}} \right)\cos \left( {2t} \right) + \frac{1}{4}\left( {3 + 2t} \right)\sin \left( {2t} \right) + \frac{1}{4}\cos \left( {2t} \right) + c}}\end{align*}\]