Question

integrate(ln(sin(x)dx,x=0 to pi)

Answer 1

Let \[I = \int_{0}^{\pi}ln(sinx) dx\] Replacing x with pi-x, we get the same integral. Hence \[I = 2 \int^{\pi/2}_{0}ln(sinx)dx ...... (1)\] replacing, x with pi/2 -x , we see that \[I = 2\int_{0}^{\pi/2}ln(cosx)dx ...... (2)\] Add (1) and (2), we get \[2I = 2\int_{0}^{\pi/2}ln(sinx.cosx)dx = 2\int_{0}^{\pi/2}(ln(sin2x) - ln2 )dx\]
or \[I = \int_{0}^{\pi/2}ln(sin2x)dx - \frac{\pi}{2}ln2\] Now replace 2x with t so that dx = dt/2 and the limits change from 0 to pi. The integral thus transforms to \[I = \int_{0}^{\pi}\frac{1}{2}ln(sint)dt - \frac{\pi}{2}ln2\] Notice that the integral expression is exactly equal to 'I'/2. So,
\[I = \frac{I}{2} - \frac{\pi}{2}ln2 \Rightarrow \frac{I}{2} = -\frac{\pi}{2}ln2\]or \[I = -\pi ln2\]

Answer 2

thanks

Answer 3

\[\int\limits_{0}^{1}\ln (x)\ln (1-x) dx\]

Answer 4

Masila do you still need help with this? I have the answer but its long so I don't wanna waste my time basically lol..