`int \ sin(logx)dx`

Explanation

`"Let I"=int sin (logx) dx`
Put logx=t Then `x=e^(t)` `therefore dx=e^(t) dt`.
`therefore I=f(sin t)e^(t) dt=int e^(t)sin t dt`
`=e^(t)(-cos t)-int [e^(t)(-cos t)dt`
`=-e^(t) cos t+int e^(t)cos t dt`
`=-e^(t) cos t+int e^(t)cos t dt-int[(d)/(dt)(e^(t))int cos tdt]dt`
`=-e^(t)cost+e^(t)sin t-int e^(t)sin tdt`
`therefore I=-e^(t)cost+e(t)sin t-I`
`therefore I=-e^(t)cost+e(t)sin t-inte^(t) sin tdt`
`therefore I=-e^(t)cos t+e^(t)sin t-I`
`therefore 2I=-e^(t)cost+e^(t)sint=e^(t)(sin t-cos t)`
`therefore I=(e^(x))/(2)(sin t-cost)+c=(x)/(2)[sin(logx)-cos(log(x)]+c`