Question

Вычислите определённый интеграл. Даю 70 баллов ​

Answer 1

$displaystyle int_{piover2}^{3piover2}cos{xover3}mathrm{dx}=3int_{piover2}^{3piover2} cos{xover3}mathrm{dleft({xover3}right)}=3sin{xover3}bigg|_{piover2}^{3piover2}=3left(sin{piover2}-sin{piover6}right)={3over2}\\ 3int_{piover6}^{piover3}{mathrm{dx}oversin^2{(2x)}}={3over2}int_{piover6}^{piover3}{mathrm{d(2x)}oversin^2{(2x)}}=-{3over2}ctg{(2x)}bigg|_{piover6}^{piover3}=-{3over2}left(ctg{2piover3}-ctg{piover3}right)=sqrt{3}\\ int_{-{1}}^{1}{mathrm{dx}over3-2x}=-{1over2}int_{-{1}}^{1}{mathrm{d(3-2x)}over3-2x}=-{1over2}ln{|3-2x|}bigg|_{-{1}}^{1}=-{1over2}(ln{1}-ln{5})={ln{5}over2}\\ int_{0}^{2pi}left(sin{xover6}+cos{(5x)}right)mathrm{dx}=6int_{0}^{2pi}sin{xover6}mathrm{d{xover6}}+{1over5}int_{0}^{2pi}cos{(5x)}mathrm{d(5x)}=(-6cos{xover6}+{1over5}sin{(5x)})|_{0}^{2pi}=3+0=3$