Evaluate `int sin(logx)dx`.

Explanation

Put logx=t so that `x=e^(t)and(1)/(x)=dtordx=e^(t)dt`.
`:.intsin(logx)dx=inte^(t)sintdt` . . . (i)
Now, `inte^(t)sintdt=e^(t)(-cost)-inte^(t)*(-cost)dt" [integrating by parts]"`
`=-e^(t)cost+inte^(t)costdt`
`=-e^(t)cost+{:[e^(t)sint-inte^(t)sintdt]:}` [integrating `e^(t)` cos t by parts]
`=-e^(t)cost+e^(t)sint-inte^(t)sintdt`.
`:.2inte^(t)sintdt=-e^(t)tcost+e^(t)sint`
`orinte^(t)sintdt=(1)/(2)(-e^(t)cost+e^(t)sint)+C`.
Putting this value in (i), we get
`intsin(logx)dx=inte^(t)sintdt`
`=(1)/(2)(-e^(t)cost+e^(t)sint)+C`
`=(1)/(2)[-xcos(logx)+xsin(logx)]+C`
`=-(1)/(2)xcos(logx)+(1)/(2)xsin(logx)+C`.