在2*arctan2 到π/4 上对函数 1/【（1-cosX）（sinX）^2】求定积分,求详解

设t=tan(x/2),则dx=2dt/(1+t²),sin(x/2)=t/√(1+t²),cos(x/2)=1/√(1+t²)
于是,原式=1/4∫(2,tan(π/8))(1+2/t²+1/t^4)dt
=1/4(t-2/t-(1/3)/t³)│(2,tan(π/8))
=1/4(tan(π/8)-2/tan(π/8)-(1/3)/tan³(π/8)-2+1+1/24)
=1/4((√2-1)-2(√2+1)-(1/3)(√2+1)³-23/24)
=1/4(-8√2/3-225/24)
=-2√2/3-225/96.