The value of ∫02π cos2n⁡xcos2n⁡x+sin2n⁡xdx is

Here f(x)=cos2n⁡xcos2n⁡x+sin2n⁡x for which f(x)=f(2π−x)⇒ I=∫02π cos2n⁡xcos2n⁡x+sin2n⁡xdx=2∫0π cos2n⁡xcos2n⁡x+sin2n⁡xdx Again, f(x)=f(π−x)⇒ I=4∫0π/2 cos2n⁡xcos2n⁡x+sin2n⁡xdx---(1)=4∫0π/2 cos2n⁡π2−xcos2n⁡π2−x+sin2n⁡π2−xdx (by property IV) =4∫0π/2 sin2n⁡xsin2n⁡x+cos2n⁡xdx-------(2) Adding (1) and (2), we have 2I=4∫0π/2 1dx⇒I=π