Question

定积分求解问题

请给出详细步骤，多谢

Answer 1

裂项，

凑微分，

方法如下图所示，请作参考，祝学习愉快：

Answer 2

∫1/(cosx)^3 dx = ∫cosx/(cosx)^4 dx
=-∫dsinx/(1-(sinx)^2)^2
=∫dt/(1-t^2)^2
令1/(1-t^2)^2=（a+bt)/(1-t)^2 +(c +dt)/（1+t)^2

=[(a+bt)(1+t)^2 +(c+dt)(1-t)^2]/(1-t^2)^2
=[(b+d)t^3 +(2b+a+c-2d)t^2 +(b+2a+c-2d)t +a+c]/(1-t^2)^2
b+d=0 => d=-b

2b+a+c-2d=0 => 4b +a+c=0

b+2a+c-2d =0 => 3b+2a+c =0

a+c =1 => a+c=1
b= -1/4
a=-1/4
d=1/4
c=3/4
原积分=1/4∫(3+4t)/(1+t)^2 -(1+t)/(1-t)^2 dt
=1/4∫4/(t+1) - 1/(t+1)^2 -2/(1-t)^2 +1/(1-t) dt
=ln(t+1) +1/(t+1) +1/(1-t) - ln(1-t)

然后带人上下限即可