Question

Вычислите определённый интеграл. Даю 70 баллов ​

Answer 1

$\displaystyle \int_{\pi\over2}^{3\pi\over2}\cos{x\over3}\mathrm{dx}=3\int_{\pi\over2}^{3\pi\over2} \cos{x\over3}\mathrm{d\left({x\over3}\right)}=3\sin{x\over3}\bigg|_{\pi\over2}^{3\pi\over2}=3\left(\sin{\pi\over2}-\sin{\pi\over6}\right)={3\over2}\\\\ 3\int_{\pi\over6}^{\pi\over3}{\mathrm{dx}\over\sin^2{(2x)}}={3\over2}\int_{\pi\over6}^{\pi\over3}{\mathrm{d(2x)}\over\sin^2{(2x)}}=-{3\over2}ctg{(2x)}\bigg|_{\pi\over6}^{\pi\over3}=-{3\over2}\left(ctg{2\pi\over3}-ctg{\pi\over3}\right)=\sqrt{3}\\\\ \int_{-{1}}^{1}{\mathrm{dx}\over3-2x}=-{1\over2}\int_{-{1}}^{1}{\mathrm{d(3-2x)}\over3-2x}=-{1\over2}\ln{|3-2x|}\bigg|_{-{1}}^{1}=-{1\over2}(\ln{1}-\ln{5})={\ln{5}\over2}\\\\ \int_{0}^{2\pi}\left(\sin{x\over6}+\cos{(5x)}\right)\mathrm{dx}=6\int_{0}^{2\pi}\sin{x\over6}\mathrm{d{x\over6}}+{1\over5}\int_{0}^{2\pi}\cos{(5x)}\mathrm{d(5x)}=(-6\cos{x\over6}+{1\over5}\sin{(5x)})|_{0}^{2\pi}=3+0=3$