如图，高数题

令 u = x^2,
原式 I = (1/2)∫√[(1-u)/(1+u)]du,
令 t = √[(1-u)/(1+u)], 则 u = (1-t^2)/(1+t^2), du = -4tdt/(1+t^2)^2,
l = -2∫ t^2dt/(1+t^2)^2 = -2∫ [1/(1+t^2) - 1/(1+t^2)^2]dt
= -2arctant + 2∫dt/(1+t^2)^2, (令 t = tanv)
2∫dt/(1+t^2)^2 = 2∫dv/(secv)^2 = 2∫(cosv)^2dv = ∫(1+cos2v)dv
= v + (1/2)sin2v = v + sinvcosv = arctant + t/(1+t^2) + C
则 I = -arctant + t/(1+t^2) + C
= √[(1-x^2)/(1+x^2)] + (1/2)√(1-x^4) + C

cos2x=cos²x-sin²x=2cos²x-1=1-2sin²x
所以1-cos2x=1-(1-2sin²x)=2sin²x
原式=∫x/2sin²x dx
=1/2*∫x/sin²xdx
=1/2*∫xcsc²xdx
=-1/2*∫xdcotx
=-1/2*xcotx+1/2*∫cotxdx
=-1/2xcotx+1/2∫cosx/sinxdx
=-1/2xcotx+1/2∫1/sinxdsinx
=-1/2xcotx+1/2ln|sinx|+c