答：
设x=√2sint,x=1,t=π/4；x=√2,t=π/2
原式
=(π/4→π/2) ∫√2costd(√2sint)
=(π/4→π/2) 2∫ (cost)^2 dt
=(π/4→π/2) ∫ (cos2t+1) dt
=(π/4→π/2) [t+(1/2)sin2t]
=π/2+0-(π/4+1/2)
=π/4-1/2
=(π-2)/4
答：
设x=√2sint,x=1,t=π/4；x=√2,t=π/2
原式
=(π/4→π/2) ∫√2costd(√2sint)
=(π/4→π/2) 2∫ (cost)^2 dt
=(π/4→π/2) ∫ (cos2t+1) dt
=(π/4→π/2) [t+(1/2)sin2t]
=π/2+0-(π/4+1/2)
=π/4-1/2
=(π-2)/4