5.4-bo'lim : Ko'proq almashtirish qoidasi
10. \( \displaystyle \int{{{{\sec }^2}\left( {2t} \right)\left[ {9 + 7\tan \left( {2t} \right) - {{\tan }^2}\left( {2t} \right)} \right]\,dt}}\) ni baholang.
Yechim
Ushbu integral birinchi qarashda biroz qo'rqinchli bo'lishi mumkin. Buni amalga oshirish uchun biz shuni ta'kidlashimiz kerakki, \(\tan \left( x \right)\) ning hosilasi \({\sec ^2}\left( x \right)\) va biz \({\sec ^2}\left( {2t} \right)\) integrandning qolgan qismidan ko'p marta borligini va bu qism faqat konstantalar va tangenslarni o'z ichiga oladi.
Demak, almashtirish o'shanda bo'lganga o'xshaydi,
\[u = \tan \left( {2t} \right)\]Agar siz asosiy almashtirish mexanikasidan mamnun bo'lmasangiz, oldingi bo'limda ba'zi muammolar bilan shug'ullanishingiz kerak, chunki biz ushbu bo'limda asoslar haqida batafsil ma'lumot bermaymiz. Ushbu bo'limdagi muammolar almashtirishning asosiy mexanikasi bilan ancha qulay bo'lganlar uchun mo'ljallangan va bir nechta "ilg'or" almashtirishlarni o'z ichiga oladi.
2-qadam
Bu erda almashtirish uchun differentsial ish.
\[du = 2{\sec ^2}\left( {2t} \right)dt\hspace{0.25in} \to \hspace{0.25in}\,\,\,\,\,{\sec ^2}\left( {2t} \right)dt = \frac{1}{2}du\]Endi almashtirishni amalga oshirish va integrallarni baholash,
\[\begin{align*}\int{{{{\sec }^2}\left( {2t} \right)\left[ {9 + 7\tan \left( {2t} \right) - {{\tan }^2}\left( {2t} \right)} \right]\,dt}} & = \frac{1}{2}\int{{9 + 7u - {u^2}du}} = \frac{1}{2}\left( {9u + \frac{7}{2}{u^2} - \frac{1}{3}{u^3}} \right) + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}\left( {9\tan \left( {2t} \right) + \frac{7}{2}{{\tan }^2}\left( {2t} \right) - \frac{1}{3}{{\tan }^3}\left( {2t} \right)} \right) + c}}\end{align*}\]Integralni baholagandan so'ng asl o'zgaruvchiga qaytishni unutmang!