大一高数.求详解
2个回答

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