If $I_1=\int\limits_0^{π/2}ln(\sin x)dx;I_2=\int\limits_{-π/4}^{π/4}ln(\sin x+\cos x)dx$, then:
Answer & explanation
Correct answer: option 1
$I_1=\int\limits_0^{π/2}ln(\sin x)dx=\frac{-π}{2}ln\,2$
$I_2=\int\limits_{-π/4}^{π/4}ln[\sqrt{2}\sin x(x+\frac{π}{4})]dx=\int\limits_{-π/4}^{π/4}ln\sqrt{2}dx+\int\limits_{0}^{π/2}ln(\sin x)dx=\frac{π}{2}ln\sqrt{2}-\frac{π}{2}ln\,2=\frac{-π}{4}ln2$
$I_2=\frac{I_1}{2}$