Evaluate: `intsin(logx)\ dx`

Explanation

`"Let I "= int sin (log x) dx` lt brgt `=int sin (log x). 1 dx`
` =sin(log x) - .int 1 dx - int {(d)/(dx) sin (logx) . int 1 dx } dx`
`= x sin (log x) -int (cos (log x))/(x) .x dx`
` =x sin (log x) - int cos (log x) . 1dx`
` = x sin (log x) - [ cos (log x) . int 1 dx]`
`- int {(d)/(dx) cos (log x) int 1dx }dx`
` =x sin (log x) `
`-[x cos (log x) - int (-sin (log x))/(x) . x dx ] + c_(1)`
`= x sin (log x) - x cos (log x) - int sin (log x) dx + c_(1)`
` rArr " " I=x [sin (log x) - cos (log x) ] -I +c_(1)`
`rArr " "2I =x [sin (log x) - cos (log x) ]+ c_(1)`
` rArr " "I =(x)/(2) [sin (log x) - cos (log x)] +c`.