1. ## Integral Problem ?

I have difficulty to solve this problem...plz help me

2. use $\sin (x) = 2 \sin (\frac{x}{2}) \cos (\frac{x}{2})$

$\displaystyle \int_0^{\frac{\pi}{2}}\ln (\sin(x))\;dx = \int_0^{\frac{\pi}{2}} \ln(2\sin(x/2)\cos(x/2))\dx$

$\displaystyle = \int_0^{\frac{\pi}{2}} \ln(2)\;dx+\int_0^{\frac{\pi}{2}} \ln(sin(x/2))\;dx+\int_0^{\frac{\pi}{2}} \ln(\cos(x/2))\;dx$

$let\;u=\frac{x}{2} \rightarrow dx = 2du$

$\displaystyle \int_0^{\frac{\pi}{2}} \ln(\sin(x)) = \dfrac{\pi\ln(2)}{2}+ 2 \int_0^{\frac{\pi}{4}} \ln(\sin(u))\;du+2\int_0^{\frac{\pi}{4}}\ln(\cos(u ))\;du$........(I)

In the last integral Let $u=\frac{\pi}{2}-t$

$\displaystyle \int_0^{\frac{\pi}{4}}\ln(\cos(u))\;du= -\int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \ln(\sin(t)\;dt = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln(\sin(t)\;dt$

now substitute this back in (I) to finish..

3. Originally Posted by harish21
use $\sin (x) = 2 \sin (\frac{x}{2}) \cos (\frac{x}{2})$

$\displaystyle \int_0^{\frac{\pi}{2}}\ln (\sin(x))\;dx = \int_0^{\frac{\pi}{2}} \ln(2\sin(x/2)\cos(x/2))\dx$

$\displaystyle = \int_0^{\frac{\pi}{2}} \ln(2)\;dx+\int_0^{\frac{\pi}{2}} \ln(sin(x/2))\;dx+\int_0^{\frac{\pi}{2}} \ln(\cos(x/2))\;dx$

$let\;u=\frac{x}{2} \rightarrow dx = 2du$

$\displaystyle \int_0^{\frac{\pi}{2}} \ln(\sin(x)) = \dfrac{\pi\ln(2)}{2}+ 2 \int_0^{\frac{\pi}{4}} \ln(\sin(u))\;du+2\int_0^{\frac{\pi}{4}}\ln(\cos(u ))\;du$........(I)

In the last integral Let $u=\frac{\pi}{2}-t$

$\displaystyle \int_0^{\frac{\pi}{4}}\ln(\cos(u))\;du= -\int_{\frac{-\pi}{2}}^{\frac{\pi}{4}} \ln(\sin(t)\;dt = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln(\sin(t)\;dt$

now substitute this back in (I) to finish..

Very nice, indeed! As I knew it took at least twice more calculations. Just a little observation: in the last line, first integral from the right,

the lower limit must be $\frac{\pi}{2}$ , not minust this.

Tonio

4. great work