`intsin(logx)dx` का मान ज्ञात कीजिए ।

Explanation

माना `I=int sin(logx)dx= intsin(logx).1dx`
`=sin(logx).int1dx-int{(d)/(dx)sin(logx).int1dx}dx`
`=x sin(logx)-int(cos(logx))/(x).xdx`
`=x sin (logx)-int cos(logx).1dx`
`=x sin (logx)-[cos(logx).int1dx-int{(d)/(dx)cos(logx)int1dx}dx]`
`=x sin(logx)-[x cos (logx)-int(-sin(logx))/(x).xdx]+c_(1)`
`=x sin (logx)-intsin(logx)-x cos(logx)-intsin(logx)dx+c_(1)`
`rArr I=x[sin(logx)-cos(logx)]-I+c_(1)`
`rArr 2I=x[sin(logx)-cos(logx)]+c_(1)`
`rArr I=(x)/(2)[sin (logx)-cos(logx)]+c`
जहाँ `c=(c_(1))/(2)`