Question

`int cot "x".log [log (sin "x")] "dx"` = ____________.

Options

• log (sin x) [log (log (sin x)) − 1] + C

• log (sin x) (log (log (sin x)) + 1] + C

• log (sin x) [log (sin x)) + 1] + C

• log (sin x) [log (sin x) − 1] + C

Answer 1

`int cot "x".log [log (sin "x")] "dx"` = log (sin x) [log (log (sin x)) − 1] + C.

Explanation:

We have,

l = `int cot "x".log [log (sin "x")] "dx"`

Put, log sin x = t = cot x dx = dt

∴ l = `int log "t dt"`

l = `log "t" int "dt" - int (("d" log "t")/"dt") . int "dt". "dt" + "C"`

l = `"t" log "t" - int 1/"t" xx "t dt" + "C"`

l = `"t" log "t" - int "dt" + "C"`

l = t log t − t + C

∴ l = t (log t − 1) + C

l = log (sin x) [log (log (sin x)) − 1] + C