Evaluate the integral I=∫_0^(π/2)ln⁡sinx dx

I=∫_0^(π/2)ln⁡sin(π/2-x)  dx =∫_0^(π⁄2)ln⁡cosx  dx
Adding to get:
2I=∫_0^(π⁄2)(ln⁡sinx+ln⁡cosx )  dx=∫_0^(π⁄2)ln⁡(sin⁡x  cos⁡x )  dx=∫_0^(π⁄2)ln⁡(sin⁡2x/2)  dx
∫_0^(π⁄2)ln⁡(sin⁡2x )  dx-∫_0^(π⁄2)ln⁡2  dx=∫_0^(π⁄2)ln⁡(sin⁡2x )  dx-π/2  ln⁡2
=I_1-π/2  ln⁡2→(1)
In   I_1  put  2x=y  →dx=1/2 dy
∴I_1=∫_0^(π⁄2)ln⁡(sin⁡2x )  dx=1/2 ∫_0^πln⁡siny   dy=∫_0^(π/2)ln⁡siny   dy=∫_0^(π/2)ln⁡sinx   dx=I
Substituting in (1), we get:
2I=I-π/2  ln⁡2
∴I=-π/2  ln⁡2=π/2  ln⁡〖1/2〗
Also  ∫_0^(π⁄2)ln⁡cosx  dx=π/2  ln⁡〖1/2〗