integral ln sin

$\int_{0}^{\frac{\pi}{2}} ln(sin(x))dx$

---

We know,
$sin x = cos( \frac{\pi}{2}-x)$
$I = \int_{0}^{\frac{\pi}{2}} ln(sin(x))dx \\ = \int_{0}^{\frac{\pi}{2}} ln( cos( \frac{\pi}{2}-x))dx \\ (u = \frac{\pi}{2}-x, du = -dx) \\ = \int_\frac{\pi}{2}^{0}ln(cos(u))-du\\ = \int_{0}^{\frac{\pi}{2}}ln(cos(u))du$
So,
$I = \int_{0}^{\frac{\pi}{2}} ln(sin(x))dx = \int_{0}^{\frac{\pi}{2}}ln(cos(u))du$
$2I = \int_{0}^{\frac{\pi}{2}} ln(sin(x))dx + \int_{0}^{\frac{\pi}{2}}ln(cos(x))dx\\ =\int_{0}^{\frac{\pi}{2}} ln(sin(x))+ln(cos(x))dx\\ =\int_{0}^{\frac{\pi}{2}} ln(sin(x)cos(x))dx \\ =\int_{0}^{\frac{\pi}{2}} ln(\frac{1}{2} sin(2x))dx \\ = \int_{0}^{\frac{\pi}{2}} ln(\frac{1}{2})dx + \int_{0}^{\frac{\pi}{2}} ln(sin(2x))dx \\ =\frac{\pi}{2}ln(\frac{1}{2})+ \int_{0}^{\frac{\pi}{2}} ln(sin(2x))dx \\ =-\frac{\pi}{2}ln(2)+ \int_{0}^{\frac{\pi}{2}} ln(sin(2x))dx \\ (u=2x, du=2dx)$

$=-\frac{\pi}{2}ln(2)+ \frac{1}{2}\int_{0}^{\pi} ln(sin(u))du$
The ln area of Sin from 0 to pi is 2 times of the area of sin from 0 to pi/2.
$=-\frac{\pi}{2}ln(2)+ \int_{0}^{\frac{\pi}{2}} ln(sin(u))du \\ 2I=-\frac{\pi}{2}ln(2)+ I \\ \therefore I = \int_{0}^{\frac{\pi}{2}} ln(sin(x))dx = -\frac{\pi}{2}ln(2)$

Author
crazyj7@gmail.com
Written with StackEdit.