7.2-bo'lim : Trig funksiyalarini o'z ichiga olgan integrallar
9. \( \displaystyle \int_{1}^{3}{{\sin \left( {8x} \right)\sin \left( x \right)\,dx}}\) ni baholang.
Yechim
Bu muammo uchun haqiqatan ham unchalik ko'p narsa yo'q. Biz qilishimiz kerak bo'lgan yagona narsa sinus va kosinusning ko'paytmasini yig'indiga kamaytirish uchun ushbu bo'limda keltirilgan formuladan foydalanishdir. Shunday qilib,
\[ \int_{1}^{3}{{\sin \left( {8x} \right)\sin \left( x \right)\,dx}} = \int_{1}^{3}{{\frac{1}{2}\left[ {\cos \left( {8x - x} \right) - \cos \left( {8x + x} \right)} \right]\,dx}} = \frac{1}{2}\int_{1}^{3}{{\cos \left( {7x} \right) - \cos \left( {9x} \right)\,dx}}\]2-qadam
Endi biz qilishimiz kerak bo'lgan narsa - bu integralni baholash.
\[\begin{align*}\int_{1}^{3}{{\sin \left( {8x} \right)\sin \left( x \right)\,dx}} & = \left. {\frac{1}{2}\left[ {\frac{1}{7}\sin \left( {7x} \right) - \frac{1}{9}\sin \left( {9x} \right)} \right]} \right|_1^3\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{14}}\sin \left( {21} \right) - \frac{1}{{18}}\sin \left( {27} \right) - \frac{1}{{14}}\sin \left( 7 \right) + \frac{1}{{18}}\sin \left( 9 \right) = - 0.0174}}\end{align*}\]Agar siz o'nlik javobni hisoblagan bo'lsangiz, kalkulyatoringiz radianga o'rnatilganligiga ishonch hosil qiling!