Question

Prove that \[ \int_0^{\pi/2} \log \sin x \, dx = -\frac{\pi}{2} \log 2. \]

Hint:

Use symmetry and integral properties to simplify expressions involving logarithmic trigonometric functions.

Answer 1

Let\(I=\int_0^{\pi/2}\log\sinx\,dx\).Usingthepropertyofdefiniteintegrals: \[ \int_0^af(x)\,dx=\int_0^af(a-x)\,dx, \] wehave: \[ I=\int_0^{\pi/2}\log\sinx\,dx=\int_0^{\pi/2}\log\sin(\pi/2-x)\,dx. \] Since\(\sin(\pi/2-x)=\cosx\),theintegralbecomes: \[ I=\int_0^{\pi/2}\log\cosx\,dx. \] Addingthetwoexpressionsfor\(I\): \[ 2I=\int_0^{\pi/2}(\log\sinx+\log\cosx)\,dx=\int_0^{\pi/2}\log(\sinx\cosx)\,dx. \] Usingtheidentity\(\sinx\cosx=\frac{1}{2}\sin2x\),weget: \[ 2I=\int_0^{\pi/2}\log\left(\frac{1}{2}\sin2x\right)\,dx=\int_0^{\pi/2}\log\frac{1}{2}\,dx+\int_0^{\pi/2}\log\sin2x\,dx. \] Thefirsttermsimplifiesto: \[ \int_0^{\pi/2}\log\frac{1}{2}\,dx=\log\frac{1}{2}\cdot\frac{\pi}{2}=-\frac{\pi}{2}\log2. \] Forthesecondterm,usingthesubstitution\(u=2x\),weget: \[ \int_0^{\pi/2}\log\sin2x\,dx=\frac{1}{2}\int_0^\pi\log\sinu\,du. \] Bysymmetryof\(\sinu\),theintegralevaluatesto\(0\).Thus: \[ 2I=-\frac{\pi}{2}\log2\quad\Rightarrow\quadI=-\frac{\pi}{2}\log2. \]