Integration

Let f(x)=1/[2+(tanx)^2]= (cosx)^2/[2(cosx)^2+(sinx)^2], then
f(x) is a periodic function with period π, and
∫[π/4~π] 1/[2+(tanx)^2] dx
=∫[0~π] f(x) dx - ∫[0~π/4] f(x) dx
=∫[-π/2~π/2] f(x) dx-∫[0~π/4] f(x) dx [ since the period of f(x) is π ]
=2∫[0~π/2] f(x) dx - ∫[0~π/4] f(x) dx [sub. x=arctan(t)]
=2∫[0~∞] 1/[(2+t^2)(1+t^2)] dt -∫[0~1] 1/[(2+t^2)(1+t^2)] dt
=2∫[0~∞] [1/(1+t^2)- 1/(2+t^2)] dt-∫[0~1] [1/(1+t^2)- 1/(2+t^2)] dt
=2[arctan(t)-(1/√2) arctan(t/√2)]|[0~∞]-[arctan(t)-(1/√2)arctan(t/√2)]|[0~1]
=2[π/2-(1/√2)π/2]-[π/4-(1/√2)arctan(1/√2)]
=(3/4 - 1/√2)π+(1/√2)arctan(1/√2)