Question

$\int\limits^ \frac{ \pi }{10} _ \frac{ \pi }{20} {sin(10x+ \frac{ \pi }{3}) } \, dx$

Answer 1

$-1/10*sin(10x+ \pi /3)|^{ \pi /10}_{ \pi /20}=-1/10*cos4 \pi /3+1/10*cos5 \pi /6=$$-1/10*(-1/2)+1/10*(- \sqrt{3} /2)=(1- \sqrt{3} )/20$

Answer 2

$\int\limits^ \frac{ \pi }{10} _ \frac{ \pi }{20} {(sin10x+ \frac{ \pi }{3} )} \, dx =- \frac{1}{10}*cos(10x+ \frac{ \pi }{3} )| _{ \frac{ \pi }{20} } ^{ \frac{ \pi }{10} } =$
$=- \frac{1}{10}*(cos(10* \frac{ \pi }{10}+ \frac{ \pi }{3})-cos(10* \frac{ \pi }{20}+ \frac{ \pi }{3}))=$
$=- \frac{1}{10}*(cos( \pi + \frac{ \pi }{3})-cos( \frac{ \pi }{2}+ \frac{ \pi }{3}))=- \frac{1}{10}*(-cos \frac{ \pi }{3}-(-sin \frac{ \pi }{3} ) )=$
$= \frac{1}{10}*(cos \frac{ \pi }{3}-sin \frac{ \pi }{3} ) = \frac{1}{10}*( \frac{1}{2}- \frac{ \sqrt{3} }{2} )= \frac{1- \sqrt{3} }{20}$