5.8-bo'lim : Aniq integrallarni almashtirish qoidasi
2. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\[\int_{0}^{{\frac{\pi }{4}}}{{\frac{{8\cos \left( {2t} \right)}}{{\sqrt {9 - 5\sin \left( {2t} \right)} }}\,dt}}\]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,
\[u = 9 - 5\sin \left( {2t} \right)\]2-qadam
Mana bu muammo uchun haqiqiy almashtirish ishi.
\[\begin{align*}du & = - 10\cos \left( {2t} \right)dt\hspace{0.25in}\,\, \to \hspace{0.25in}\cos \left( {2t} \right)dt = - \frac{1}{{10}}du\\ t & = 0:u = 9\hspace{0.75in}t = \frac{\pi }{4}:u = 4\end{align*}\]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.
\[ \int_{0}^{{\frac{\pi }{4}}}{{\frac{{8\cos \left( {2t} \right)}}{{\sqrt {9 - 5\sin \left( {2t} \right)} }}\,dt}} = - \frac{8}{{10}}\int_{9}^{4}{{{u^{ - \,\frac{1}{2}}}\,du}}\]3-qadam
U holda integral,
\[ \int_{0}^{{\frac{\pi }{4}}}{{\frac{{8\cos \left( {2t} \right)}}{{\sqrt {9 - 5\sin \left( {2t} \right)} }}\,dt}} = \left. { - \frac{8}{5}{u^{\frac{1}{2}}}} \right|_9^4 = - \frac{{16}}{5} - \left( { - \frac{{24}}{5}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{8}{5}}}\]