5.3-bo'lim : Noaniq integrallarni almashtirish qoidasi
6. \( \displaystyle \int{{\sec \left( {1 - z} \right)\tan \left( {1 - z} \right)\,dz}}\) ni baholang.
Yechim
Bunday holda, biz o'rnini bosuvchi sifatida quyidagilarni ishlatishimiz kerak.
\[u = 1 - z\]2-qadam
Barcha \(z\)lar \(u\) bilan almashtirilganligiga ishonch hosil qilishimiz kerak, biz differentsialni hisoblashimiz kerak, shunda biz \(dz\) va qolgan \(z\)larni integranddan olib tashlashimiz mumkin.
\[du = - dz\]O'zgartirishga yordam berish uchun keling, buni biroz qayta yozamiz,
\[dz = - du\]3-qadam
O'zgartirishni amalga oshirish va integralni baholash,
\[\int{{\sec \left( {1 - z} \right)\tan \left( {1 - z} \right)\,dz}} = - \int{{\sec \left( u \right)\tan \left( u \right)\,du}} = - \sec \left( u \right) + c\]4-qadam
Nihoyat, asl o'zgaruvchiga qaytishni unutmang!
\[\int{{\sec \left( {1 - z} \right)\tan \left( {1 - z} \right)\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \sec \left( {1 - z} \right) + c}}\]