$\int\limits_0^1ln(\sin\frac{πx}{2})dx$ is equal to:
Answer & explanation
Correct answer: option 2
$\int\limits_0^1ln(\sin\frac{πx}{2})dx$ ⇒ Let $\frac{πx}{2}=x⇒dx=\frac{2}{π}dx$
$⇒I=\frac{2}{π}\int\limits_0^{π/2}ln\,\sin x\,dx=\frac{2}{π}(-\frac{π}{2}ln\,2)=-ln\, 2$