what is integral cotx dx?

Answer 1

thats easy... ln|sinx| +constant
here is the working-
integral(cotx) dx
= integral (cosx / sinx) dx

If we let T = sinx, the dT = cosx dx. That means we have:

integral(cosx / sinx) dx
integral(1 / T) dT
= ln|T| + C

And since T = sin x, we have:

ln|T| + C
= ln|sinx| + C
you can check by differentiating the answer you will get cotx!

Answer 2

If we rewrite cotx as cos x / sin x, then we get:

integral(cotx) dx
= integral (cosx / sinx) dx

If we let u = sinx, the du = cosx dx. That means we have:

integral(cosx / sinx) dx
integral(1 / u) du
= ln|u| + C

And since u = sin x, we have:

ln|u| + C
= ln|sinx| + C

:)

Answer 3

Integral cot x dx = integral cos x / sin x dx = ln I sin x I + C

I would like to correct the people who are writing log. When log is written, it means base ten. The correct answer is ln.

Answer 4

Int. Cotx dx ------------------ (1)

= int. cosx dx / sinx -------- (2)
Put sinx = t , so that
cosx dx = dt
Substitute this value in relation (2) we get

Int dt/t = log I t I = Log IsinxI + c ............... Ans

Answer 5

I = ∫ (cos x / sin x ) dx
I = log sin x + C

Answer 6

log(sinx)+c