Evaluate ∫π/6π/3 (sin⁡x)dx(sin⁡x)+(cos⁡x)

Given integral I=∫π/6π/3 (sin⁡x)dx(sin⁡x)+(cos⁡x)------(1) or I=∫π/6π/3 (cos⁡x)dx(cos⁡x)+(sin⁡x) Replacing x by π2−x---(2)Adding equations (l ) and (2), we get2I=∫π/6π/3 (sin⁡x)+(cos⁡x)(cos⁡x)+(sin⁡x)dx     =∫π/6π/3 dx=[x]π/6π/3=π/3−π/6=π/6⇒ I=π/12