Evaluate: (i) `int1/x(logx)^2\ dx` (ii) `intsin^5xcosx\ dx`

Explanation

` (i) I=int1/x(logx)^2\ dx`
taking `logx=t`
`=>1/x dx =dt`
now,`I=int t^2 dt`
`=t^3/3+c`
`=|log x|^3/3+c`

`(ii) intsin^5xcosx\ dx`
taking `sinx=t`
`=>cosxdx=dt`
so, `I=int t^5 dt`
`=t^6/6 +c`
`=sin^6x/6+c`