∫(0,pi/2) 1/(1+ sin^2 x) dx
=∫(0,pi/2) (1/cos^2x)/((1/cos^2x)+ tan^2 x) dx
=∫(0,pi/2) (tan^2x+1)/((tan^2x+1)+ tan^2 x) dx
令tanx=t,x=arctant,dx=1/(1+t^2)*dt t從(0,正無(wú)窮大)
所以
原式=∫(0,pi/2) (t^2+1)/(2t^2+1)*1/(1+t^2)*dt
=∫(0,pi/2) 1/(2t^2+1)*dt
=√2/2 arctan√2t (0,正無(wú)窮大)
=√2/2*π/2
=√2*π/4