can anyone solve the following indef integration?

Question

integrate the following function:

f(x)=log(sin x)

please specify the steps??

Answer 1

Integrate[Log[Sin[x]], x] = -(x*Log[1 - E^((2*I)*x)]) + x*Log[Sin[x]] + (I/2)*(x^2 + PolyLog[2, E^((2*I)*x)])

Answer 2

So sorry. This integral cannot be expressed in terms
of elementary functions! The answer involves
a complex argument of the dilogarithm function.
Let's try to work the integral and see where our luck runs out.
First let u = sin x, x = arcsin(u), dx = du/sqrt(1-x^2).
So we have to integrate log(u) du/sqrt(1-x^2).
Now we can do this integral by parts.
Let
U = log(u), dV = 1/sqrt(1-x^2)
dU = 1/u du, V = arcsin(u)

So int[ log(sin x) dx] reduces to
log(u) arcsin(u) - int[ arcsin(u) du/u]
and here's where our luck runs out, for it can be
proved that this last integral is not elementary.
Rather than go into the proof here, let me
refer you to the Google group sci.math.
There are several excellent articles by
Matthew Wiener on the subject of elementary
integrals, and in one of these he proves the result
for f(x) = arcsin(x)/x.
BTW, nb2020's suggestion(below) is a good one, but
you run into the same dead end:
His integral reduces to int [x cot x dx],
which is also nonelementary!

Answer 3

You will have to use integration by parts since you have log by itself.
Integral(udv) = uv + Integral(vdu)
Pick u = log(sinx) and dv = 1
So du = cosx/sinx and v = x

I am sure u can plug that in and figure out the rest. If not let me know!

Answer 4

fixing ? a million / e^(2x+a million) dx it really is amazingly easy. save in concepts your exponent guidelines! X ^ - a million = a million / X in reality... any variety to the skill of a adverse variety is an same because the reciprocal of that variety to the skill of an same valuable variety: hence e^-(2x + a million) = a million / e^(2x + a million) This reality simplifies our necessary a great deal: we've correct the following necessary to remedy: ? e^-(2x + a million) dx the following we in basic terms use the opposite chain rule: ? e^-(2x + a million) dx = [e^-(2x + a million)]/-2 + C= -a million/[2e^(2x + a million)] + C The inverse chain rule in basic terms contains integrating the outer function and dividing by employing the spinoff of the interior function. i desire this helped :D

Answer 5

that's a difficult one and i dont remember my Calc AB lol