Question

\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]

Answer 1

\[Let\, I = \int_0^\pi \log\left( 1 - \cos x \right) d x\]
\[ = \int_0^\pi \log\left( 2 \sin^2 \frac{x}{2} \right) dx\]
\[ = \int_0^\pi \log2 dx + 2 \int_0^\pi \log \sin\frac{x}{2} dx\]
\[ Let, t = \frac{x}{2} \text{in the secong integral . then } dt = \frac{1}{2}dx\]
\[\text{When }x \to 0 ; t \to 0\text{ and } x \to \pi ; t \to \frac{\pi}{2}\]
\[I = \log2 \left[ x \right]_0^\pi + 4 \int_0^\frac{\pi}{2} \log \sin t dt\]
\[ = \pi\ log2 + 4 \times \left( - \frac{\pi}{2}\log2 \right) ...............\left[\text{Where, }\int_0^\frac{\pi}{2} \log \sin t dt = - \frac{\pi}{2}\log2 \right]\]
\[ = - \pi \log2\]