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Evaluate $\int_0^{\frac{\pi}{2}} \log \sin x d x$

Answers (1)

best_answer

Let

$$
\begin{equation*}
\mathrm{I}_1=\int_0^{\frac{\pi}{2}} \log (\sin x) d x \tag{1}
\end{equation*}
$$


Using Property $\mathrm{P}_4$

$$
\int_0^a f(x) d x=\int_0^a f(a-x) d x
$$


$$
\therefore \quad \mathrm{I}_1=\int_0^{\frac{\pi}{2}} \sin \left(\frac{\pi}{2}-x\right) d x
$$


$$
\begin{equation*}
I_1=\int_0^{\frac{\pi}{2}} \log (\cos x) d x \tag{2}
\end{equation*}
$$


Adding (1) and (2) i.e. (1) + (2)

$$
I_1+I_1=\int_0^{\frac{\pi}{2}} \log (\sin x) d x+\int_0^{\frac{\pi}{2}} \log (\cos x) d x
$$

Posted by

Vishal kumar

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