Given, f(t)=0∫ttan(2n−1)xdx,n∈So, f(t+π)=0∫t+πtan(2n−1)xdx=0∫ttan(2n−1)xdx+t∫t+πtan(2n−1)xdxLet u=x−π⇒du=dxLimit, when x=t,u=t−π andx=t+π,u=tSo, f(t+π)=f(t)+t∫ttan(2n−1)(u+π)duSince, tan(u+π)=tanu=f(t)+t−π∫ttan(2n−1(u)du=f(t)+t−π∫0tan(2n−1)(u)du+0∫ttan(2n−1)(u)duLet v=−u, the dv=−duWhen u=t−π,v=π−t, when u=0,v=0, so we get=f(t)−π−t∫0tan(2n−1)(−v)dv+0∫ttan(2n−1)(u)du=f(t)+π−t∫0tan(2n−1)(v)dv+0∫ttan(2n−1)(u)du=f(t)+π−t∫0tan(2n−1)(v)dv+f(t)=2f(t)−0∫π−ttan(2n−1)(v)dv=2f(t)−0∫πtan(2n−1)(v)dv−π−t∫πtan(2n−1)(v)dvLet w=v−π, then dw=dvWhen v=π−t,w=−t, when v=πw=0=2f(t)−0∫πtan(2n−1)(v)dv−−t∫0tan(2n−1)(w+π)dw2f(t)−0∫πtan(2n−1)(v)dv−−t∫0tan(2n−1)(w)dwLet z=−w, then dz=−dw when w=−t,z=t and w=0,z=0=2f(t)−0∫πtan(2n−1)(v)dv+t∫0tan(2n−1)(−z)dz=2f(t)−0∫πtan(2n−1)(v)dv−0∫ttan(2n−1)(z)dz=2f(t)−0∫πtan(2n−1)(v)dv−f(t)=f(t)−0∫πtan(2n−1)(v)dv=f(t)+0[∵0∫πtan(2n−1)(v)dv=0]=f(t)+f(π)[∵f(π)=0]∴f(t+π)=f(t)+f(π)