Question

Integration of log (cotx)dx

Answer 1

Answer:

-log|secx|+C

Step-by-step explanation:

we know that

log (cotx)dx = 1/cotx dx

= tanx dx

Now we have to find integration of tanx

we write tanx as sinx/cosx dx

let cosx = t so that sinx dx = - dt

Then

integration of tanx dx = - integration of

dt/t

= - log |t| + C

= - log |cosx| + C

= - log |secx| + C