高等数学积分。1／sin（θ＋pai／4） 从零到pai／2的定积分。求过程。

Question

如题所述

Answer 1

设t=tan(θ/2),则dt=sec(θ/2)*dθ/2=(1+t^2)dθ/2,
sinθ=2t/(1+t^2),cosθ=(1-t^2)/(1+t^2),
∴∫<0,π/2>dθ/sin(θ+π/4)
=2√2∫<0,1>dt/(1+2t-t^2)
=2√2∫<-1,0>du/(2-u^2)
=ln[(√2+u)/(√2-u)]|<-1,0>
=2ln(√2+1），
其中u=t-1.本回答被提问者采纳

Answer 2

先变量替换 θ＋pai／4=t
则积分化为
∫dt/sint 从π/4到3π/4
即∫csctdt
书上有公式 余割的原函数为ln|csct-cott|+C
所以积分结果为0本回答被网友采纳

Answer 3

∫(0,pi/2)1／sin(θ＋pi/4)dθ=∫(pi/4,3pi/4)1/sinθdθ
∫1/sinθdθ=∫{[sin(θ/2)]^2+[cos(θ/2)]^2}/[2sin(θ/2)cos(θ/2)]dθ
=(1/2)∫[sin(θ/2)/cos(θ/2)+cos(θ/2)/sin(θ/2)]dθ
=∫sin(θ/2)/cos(θ/2)d(θ/2)+∫cos(θ/2)/sin(θ/2)d(θ/2)
=∫1/cos(θ/2)d[-cos(θ/2)]+∫1/sin(θ/2)d[sin(θ/2)]
=[-lncos(θ/2)+lnsin(θ/2)]|(pi/4,3pi/4)
=lntan(3pi/8)-lntan(pi/8)
=ln[tan(3pi/8)/tan(pi/8)]
=ln[tan(3pi/8)*cot(pi/8)]
=ln[tan(3pi/8)*tan(pi/2-pi/8)]
=2lntan(3pi/8)

Answer 4

相当于1／sin（θ）从负pai／4到正pai／4积分，奇函数，结果为0

Answer 5

^4dθ在pai/2到-pai/2的定积分也符合3/4*1/2*pai=3pai/8这样一个( π/2) * 3/(4*2) = 3π /16 …… 因为（sinθ^2+cos^