Question

定积分根号下2-x^2的原函数怎么求？ 上限是根号2，下限是1

Answer 1

∫(1->√2)√(2-x^2)dx

令x=√2cost
√(2-x^2)=√2sint
dx=-√2sintdt
先不忙管上下限
∫√2sint*-√2sintdt

=∫(-2sin^2t) dt
=∫(cos2t-1)dt
=∫cos2tdt-∫dt
=sin2t /2 -t
=sintcost-t
=1/2*√2sint*√2cost -t
=1/2*√(2-x^2)*x-arccos(x/√2)

∴∫(1->√2)√(2-x^2)dx
=1/2*√(2-x^2)*x-arccos(x/√2) |(1->√2)
=[1/2*√(2-2)*√2-arccos(√2/√2)]-[1/2*√(2-1)*1-arccos(1/√2)]
=(0-0)-[1/2-π/4]
=π/4-1/2

Answer 2

令x=√2sint
则原式=∫(π/4→π/2)√2cost*√2costdt
=∫(π/4→π/2)2cos^2(t)dt
=∫(π/4→π/2)(cos(2t)+1)dt
=sin(2t)/2|(π/4→π/2)+t|(π/4→π/2)
=0-1/2+π/4=π/4-1/2