考研，高数，积分题，语文画线部分怎么弄？如何使tanx=t，然后我们可以找到下面的代入公式？背部

tanx = t，

∴sinx = 2t/(1+t^2)

dx = dt/(1+t^2)

∴∫(0,π/2)(sinx)^2/[1+(sinx)^2)

=∫(0，+∞)t^2/[2(t^2)+1][(t^2)+1]dt

注意到:

t^2 =[2(t^2)+1]-[(t^2)+1]

因此:

原积分= ∫ (0，+∞){[2(T2)+1]-[(T2)+1]}/[2(T2)+1][(T2)

=∫(0，+∞)[2(t^2)+1]/[2(t^2)+1][(t^2)+1]dt-

∫(0，+∞)[(t^2)+1]/[2(t^2)+1][(t^2)+1]dt

=∫(0，+∞)1/[(t^2)+1]dt-∫(0，+∞) 1/[2(t^2)+1]dt

=反正切|(0，+∞) - (1/√2)∫(0，+∞)1/[(√2t)^2+1]d(√2t)

=[反正切- (1/√2)反正切√2t ]|(0，+∞)

=[1-(1/√2)]π