Question

Evaluate \[ \int_0^{\pi/2} \ln(\sin x) \, dx \]

Hint:

Standard result: $\int_0^{\pi/2} \ln(\sin x) dx = -\tfrac{\pi}{2}\ln 2$. Similarly, $\int_0^{\pi/2} \ln(\cos x) dx = -\tfrac{\pi}{2}\ln 2$.

Answer 1

Step 1: Use property of definite integrals.
Let \[ I = \int_0^{\pi/2} \ln(\sin x)\, dx \] Using property: \[ I = \int_0^{\pi/2} \ln(\cos x)\, dx \]

Step 2: Add the two results.
\[ 2I = \int_0^{\pi/2} [\ln(\sin x) + \ln(\cos x)] dx \] \[ 2I = \int_0^{\pi/2} \ln(\sin x \cos x)\, dx \]

Step 3: Simplify.
\[ \sin x \cos x = \tfrac{1}{2}\sin 2x \] So, \[ 2I = \int_0^{\pi/2} \ln\left(\tfrac{1}{2}\sin 2x\right)\, dx \]

Step 4: Substitution.
Put $2x = t \;\Rightarrow\; dx = dt/2$, limits $0 \to \pi$. \[ 2I = \frac{1}{2}\int_0^{\pi} \ln\left(\tfrac{1}{2}\sin t\right) dt \] \[ = \frac{1}{2}\left[\int_0^{\pi} \ln(\sin t)\, dt - \int_0^{\pi} \ln 2 \, dt\right] \]

Step 5: Evaluate.
\[ \int_0^{\pi} \ln(\sin t)\, dt = 2\int_0^{\pi/2} \ln(\sin t)\, dt = 2I \] So, \[ 2I = \frac{1}{2}(2I - \pi \ln 2) \] \[ 4I = 2I - \pi \ln 2 \;\;\Rightarrow\;\; 2I = -\pi \ln 2 \] \[ I = -\frac{\pi}{2}\ln 2 \]

Final Answer: \[ \boxed{-\tfrac{\pi}{2}\ln 2} \]