很简单的积分 高数1/根号(1+t^2) 对t从0积到1
2019-04-14
很简单的积分 高数
1/根号(1+t^2) 对t从0积到1
优质解答
设t=tanx,则dt=sec²xdx
故 ∫<0,1>dt/(1+t²)=∫<0,π/4>sec²xdx/secx
=∫<0,π/4>secxdx
=∫<0,π/4>cosxdx/cos²x
=∫<0,π/4>d(sinx)/(1-sin²x)
=(1/2)∫<0,π/4>[1/(1+sinx)+1/(1-sinx)]d(sinx)
=(1/2)[ln│(1+sinx)/(1-sinx)│]│<0,π/4>
=(1/2)[ln((1+1/√2)/(1-1/√2))-0]
=(1/2)ln((√2+1)/(√2-1))
=ln(√2+1).
设t=tanx,则dt=sec²xdx
故 ∫<0,1>dt/(1+t²)=∫<0,π/4>sec²xdx/secx
=∫<0,π/4>secxdx
=∫<0,π/4>cosxdx/cos²x
=∫<0,π/4>d(sinx)/(1-sin²x)
=(1/2)∫<0,π/4>[1/(1+sinx)+1/(1-sinx)]d(sinx)
=(1/2)[ln│(1+sinx)/(1-sinx)│]│<0,π/4>
=(1/2)[ln((1+1/√2)/(1-1/√2))-0]
=(1/2)ln((√2+1)/(√2-1))
=ln(√2+1).