5.3-bo'lim : Noaniq integrallarni almashtirish qoidasi
13. \( \displaystyle \int{{10\sin \left( {2x} \right)\cos \left( {2x} \right)\sqrt {{{\cos }^2}\left( {2x} \right) + 5} \,dx}}\) ni baholang.
Maslahat: Eslatib o'tamiz, agar integralda (yoki atamaning bir qismida) "aniq" ichki funktsiyaga ega bo'lgan atama mavjud bo'lsa, hech bo'lmaganda "ichki" funksiya bizga kerak bo'lgan almashtirish bo'lishi ehtimoli bor.
Yechim
Bunday holda, biz o'rnini bosuvchi sifatida quyidagilarni ishlatishimiz kerak.
\[u = {\cos ^2}\left( {2x} \right) + 5\]
Maslahat: Esda tutingki, almashtirishdan so'ng integraldagi barcha asl o'zgaruvchilar \(u\) bilan almashtirilishi kerak.
2-qadam
Barcha \(x\)lar \(u\) bilan almashtirilganligiga ishonch hosil qilishimiz kerak, biz differentsialni hisoblashimiz kerak, shunda biz \(dx\) va qolgan \(x\)larni integranddan olib tashlashimiz mumkin.
\[du = - 4\cos \left( {2x} \right)\sin \left( {2x} \right)dx\]O'zgartirishga yordam berish uchun keling, buni biroz qayta yozamiz,
\[\begin{align*}du = - 4\cos \left( {2x} \right)\sin \left( {2x} \right)dx & = - 2\left( 2 \right)\left( {\frac{5}{5}} \right)\cos \left( {2x} \right)\sin \left( {2x} \right)dx\,\,\\ & \hspace{1.0in}\Rightarrow \,\,\,\,10\cos \left( {2x} \right)\sin \left( {2x} \right)dx = - \frac{5}{2}du\end{align*}\]3-qadam
O'zgartirishni amalga oshirish va integralni baholash,
\[\int{{10\sin \left( {2x} \right)\cos \left( {2x} \right)\sqrt {{{\cos }^2}\left( {2x} \right) + 5} \,dx}} = - \frac{5}{2}\int{{{u^{\frac{1}{2}}}\,du}} = - \frac{5}{3}{u^{\frac{3}{2}}} + c\]
Maslahat: Integranddagi asl o'zgaruvchi \(u\) emasligini unutmang!
4-qadam
Nihoyat, asl o'zgaruvchiga qaytishni unutmang!
\[\int{{10\sin \left( {2x} \right)\cos \left( {2x} \right)\sqrt {{{\cos }^2}\left( {2x} \right) + 5} \,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{5}{3}{{\left( {{{\cos }^2}\left( {2x} \right) + 5} \right)}^{\frac{3}{2}}} + c}}\]