If g(x)=∫0x cos⁡4tdt,, then g(x+π) equals

g(x+π)=∫0x+π cos⁡4tdt=∫0π cos⁡4tdt+∫πx+π cos⁡4tdt=g(π)+Iwhere I=∫πx+π cos⁡4tdt, Put t=π+θ so thatI=∫0x cos⁡(4π+4θ)dθ=∫0x cos⁡4θdθ=g(x)So g(x+π)=g(π)+g(x) butg(π)=∫0π cos⁡4tdt=14sin⁡4t0=0.∴ g(x+π)=g(x)−g(π) also.