7.2-bo'lim : Trig funksiyalarini o'z ichiga olgan integrallar
3. \( \displaystyle \int{{{{\cos }^4}\left( {2t} \right)\,dt}}\) ni baholang.
Yechim
Bu erda e'tibor berish kerak bo'lgan birinchi narsa shundaki, bizda faqat teng ko'rsatkichlar mavjud va shuning uchun biz bu integralni biz qila oladigan integralga kamaytirish uchun yarim burchakli va ikki burchakli formulalardan foydalanishimiz kerak.
Bundan tashqari, bizda integranda sinuslar yo'qligidan hayajonlanmang. Ba'zan bizda integranda ikkala trig funksiyasi ham bo'lmaydi. Bu biz ushbu bo'limda ishlatgan usullardan foydalana olmasligimizni anglatmaydi.
Shunday qilib, keling, ushbu muammoni quyidagicha boshlaylik.
\[ \int{{{{\cos }^4}\left( {2t} \right)\,dt}} = \int{{{{\left( {{{\cos }^2}\left( {2t} \right)} \right)}^2}\,dt}}\]2-qadam
Endi biz yarim burchak formulasidan foydalanishimiz mumkin,
\[\int{{{{\cos }^4}\left( {2t} \right)\,dt}} = \int{{{{\left[ {\frac{1}{2}\left( {1 + \cos \left( {4t} \right)} \right)} \right]}^2}\,dt}} = \int{{\frac{1}{4}\left( {1 + 2\cos \left( {4t} \right) + {{\cos }^2}\left( {4t} \right)} \right)\,dt}}\]3-qadam
Yarim burchak formulasini uchinchi muddatda yana bir marta ishlatishimiz kerak,
\[\begin{align*}\int{{{{\cos }^4}\left( {2t} \right)\,dt}} & = \frac{1}{4}\int{{1 + 2\cos \left( {4t} \right) + \frac{1}{2}\left[ {1 + \cos \left( {8t} \right)} \right]\,dt}}\\ & = \frac{1}{4}\int{{\frac{3}{2} + 2\cos \left( {4t} \right) + \frac{1}{2}\cos \left( {8t} \right)\,dt}}\end{align*}\]4-qadam
Endi biz faqat integralni baholashimiz kerak.
\[\int{{{{\cos }^4}\left( {2t} \right)\,dt}} = \frac{1}{4}\left( {\frac{3}{2}t + \frac{1}{2}\sin \left( {4t} \right) + \frac{1}{{16}}\sin \left( {8t} \right)} \right) + c = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{8}t + \frac{1}{8}\sin \left( {4t} \right) + \frac{1}{{64}}\sin \left( {8t} \right) + c}}\]