Question
Evaluate:
$\int\limits_0^{\frac{\pi}{2}}$ log sin x dx.
✓
Answer
$\text{I}=\int\limits_0^{\pi/2}\log\sin\text{x dx}=\int\limits_0^{\pi/2}\log \sin\Bigg(\frac{\pi}{2}-\text{x}\Bigg)\text{dx}=\int\limits_0^{\pi/2}\log\cos\text{x dx}$
$\therefore\text{2I}=\int\limits_0^{\pi/2}\log(\text{sin x cos x)dx}=\int\limits_0^{\pi/2}\log\frac{\sin\text{2x}}{2}\text{dx}$
$=\int\limits_0^{\pi/2}\log\sin\text{2x dx}-\int\limits_0^{\pi/2}\log\text{ 2dx}$
$=\frac{1}{2}\times2\int\limits_0^{\pi/2}\log\sin\text{t}\text{ dt}-\frac{\pi}{2}\log2;\Bigg[\text{t = 2x,}\frac{\text{dt}}{2}=\text{dx}\Bigg]$
$=\text{I}-\frac{\pi}{2}\log2$
$\text{I}=-\frac{\pi}{2}\log2$.
$\therefore\text{2I}=\int\limits_0^{\pi/2}\log(\text{sin x cos x)dx}=\int\limits_0^{\pi/2}\log\frac{\sin\text{2x}}{2}\text{dx}$
$=\int\limits_0^{\pi/2}\log\sin\text{2x dx}-\int\limits_0^{\pi/2}\log\text{ 2dx}$
$=\frac{1}{2}\times2\int\limits_0^{\pi/2}\log\sin\text{t}\text{ dt}-\frac{\pi}{2}\log2;\Bigg[\text{t = 2x,}\frac{\text{dt}}{2}=\text{dx}\Bigg]$
$=\text{I}-\frac{\pi}{2}\log2$
$\text{I}=-\frac{\pi}{2}\log2$.
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