5.8-bo'lim : Aniq integrallarni almashtirish qoidasi
3. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\[ \int_{\pi }^{0}{{\sin \left( z \right){{\cos }^3}\left( z \right)\,dz}}\]Yechim
Biz qilishimiz kerak bo'lgan birinchi qadam - almashtirishni amalga oshirish.
Shu nuqtada siz almashtirishlar bilan juda qulay bo'lishingiz kerak. Agar almashtirishlar sizga mos kelmasa, siz almashtirish bo'limlariga qaytib, u erda ba'zi muammolarni hal qilishingiz kerak.
Ushbu muammoning o'rnini bosish,
\[u = \cos \left( z \right)\]2-qadam
Mana bu muammo uchun haqiqiy almashtirish ishi.
\[\begin{align*}du & = - \sin \left( z \right)dz\hspace{0.25in}\,\, \to \hspace{0.25in}\sin \left( z \right)dz = - du\\ z & = \pi :u = - 1\hspace{0.75in}z = 0:u = 1\end{align*}\]Ushbu bo'lim uchun eslatmalarda qilganimizdek, biz integralni bajargandan so'ng orqaga almashtirish bilan shug'ullanmaslik uchun chegaralarni \(u\) ga aylantiramiz.
Bu erda almashtirishdan keyingi integral.
\[ \int_{\pi }^{0}{{\sin \left( z \right){{\cos }^3}\left( z \right)\,dz}} = - \int_{{ - 1}}^{1}{{{u^3}\,du}}\]3-qadam
U holda integral,
\[ \int_{\pi }^{0}{{\sin \left( z \right){{\cos }^3}\left( z \right)\,dz}} = \left. { - \frac{1}{4}{u^4}} \right|_{ - 1}^1 = - \frac{1}{4} - \left( { - \frac{1}{4}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{0}\]