integrate (cotx)?

Answer 1

from standard tables,

int(cotx)dx =ln(sinx)+C,
where C is a constant

this is easy to prove

let cotx=cosx/sinx and t=sinx
dt/dx=cosx,giving dx=dt/cosx
hence,
int(cotx)dx=int(cosx/sinx)dx
=int(cosx/(t))dt/cosx
(substituting the above values)
=int(1/t)dt
=lnt+C,{standard integral for 1/t
is lnt+a constant}
=ln(sinx)+C{t=sinx,stated above}

therefore,
int(cotx)dx=ln(sinx)+C

i hope that this helps

Answer 2

ln sin(x) + c

cot(x) = cos(x)/sin(x) and noticing that the numerator is the derivative of the denominator means that the integral is the ln of the denominator.

It's similar for sec and cosec although integrating these two is a bit more tricky.

A classic example is 1/x where the integral is ln x + c. When you differentiate x you get 1 ==> the numerator is the derivative of the denominator.

Answer 3

integration is the necessary route to the salvation of society!
OK cot(x)= cos(x)/sin (x) by defn of inv tan function
easy way.. u=sinx..du=cosxdx ...differentiating
Integral becomes integral(cosx/cosx*sinx)
cancel, and subst u=sinx again..
integral du/u= log(base e)u
so integral (cotx)=log(sinx) QED! (10 points??)

Answer 4

This is really not a difficult problem at all.

Remember, cot x = cos x / sin x. So, what you have here is:

(1 / sin x ) (cos x)

Let u = sin x. Then du = cos x dx

What you have here is the old familiar integral of (1/ u) du:

integral of (1/ u) du = ln u + c.

Back substituting, ln u + c = ln (sin x) + c.

Answer 5

Int(cot x)dx = int (cosx/sinx) dx
Let u = sinx and thus du = cosx dx

Int (cot x) dx = int (cosx dx) /sinx
int (cot x) dx = int (1/u) du = log u + c = log(sinx) + c

Note a standard integral is int(1/u) = log u ie if top line is derivative of bottom line , answer is log(bottom line)

Thus in your question cosx is derivative of sin x thus answer is log (sinx)

Answer 6

ln sin