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प्रश्न
`int cot "x".log [log (sin "x")] "dx"` = ____________.
पर्याय
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
MCQ
रिकाम्या जागा भरा
उत्तर
`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
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