If I=∫dxtanxlogcosecx, then I equals
log|logcosecx|+C
log|logcosx|+C
−log|log(cosecx)|+C
None of these
I=−∫cotxlogsinxdx
Putting logsinx=t, we get I=−∫dtt=-log|t|+c
∴ I=-log∥logsinx∣+C