How can you integrate [find anti-deriv of] cot x ln (sin x) dx?

Answer 1

It's just (1/2) [ln (sin x)]^2. (See method below, at ***)

CHECK: By the chain rule, the derivative of this is:

(1/2) 2 [ln (sin x)] (1/sin x) cos x = cot x ln (sin x). QED

If this is hard to see, let y = (1/2) [ln (sin x)]^2 = (1/2) z^2,
so that

z = ln (sin x), and:

Let t = sin x so that z = ln t

Then dy/dx = (dy/dz) * (dz/dt) * (dt/dx), that is:

dy/dx = z*(1/t)*cos x = ln(sin x)*(1/sin x)*cos x = cot x ln(sin x).

IT CHECKS OUT !

*** How did I obtain this? I immediately recognized that d [ln (sin x)] is (cos x) / (sin x), i.e. cot x. But I stlll needed a "ln (sin x)" in the end result, multiplying cot x. I'd get that if I started with a multiple of
[ln sin x)]^2, and the obvious multiple was (1/2), to cancel out the "2" from differentiating the "square."

And that was it: (1/2) [ln (sin x)]^2.

A more ordered way is to make a substitution such as u = sin x, du = cos x dx (which already looks quite promising), or w = ln (sin x), dw = cot x dx (even more promising!), etc. Both of these are part way there. But British high-school experience, even though > 50 years ago, enabled me to "sniff out" the solution in the way I've indicated, without actually making any subsidiary substitution along the way. It's an almost intuitive way of applying the principle(s) behind the chain rule, in real time.

Live long and prosper.

Answer 2

Write u = ln sin x ; du = cot x dx to obtain the answer

(1/2) u^2 + C = (1/2) (ln sin x)^2 + C.

Answer 3

Integral (cot(x) ln(sin(x))dx

I'm going to rearrange the terms to prove a point.

Integral (ln(sin(x)) cot(x) dx)

Let u = ln(sin(x))
du = 1/sin(x) [cos(x)] dx. This can be rewritten as

du = cos(x)/sin(x) dx, OR
du = cot(x) dx

Note that we now have our cot(x) dx, and can replace it entirely by du. Substituting accordingly, we have

Integral (u) du

And now, integrating using the reverse power rule,

(1/2)u^2 + C

Replacing u = ln(sin(x)), we have

(1/2) [ln(sinx)]^2 + C

Answer 4

u=sin x
du=cos x dx

anti ln(sin x) * (1/sin x)* cos x dx
= anti ln(u) * (1/u) * du

v=ln(u)
dv=1/u du

= anti v dv
= v^2 / 2 + C
= ln(u)^2 / 2 + C
= ln(sin x)^2 / 2 + C

Answer 5

cosx/sinx=cotx Use u substitution the position u=sinx du=cosx dx The essential will become a million/u du The anti-spinoff of this in lnu + C Plug your u fee back in to get ln(sinx) + C

Answer 6

is there an anti-derivative of natural log???

Answer 7

1/2log^2(sin(x))