Question

如图，定积分怎么写呀，谢谢

Answer 1

令 x = sinu， 则
I = ∫<下π/4, 上π/2>(cosu)^2du/(sinu)^2
= ∫<π/4, π/2>(cotu)^2du = ∫<π/4, π/2>[(cscu)^2-1]du
= -[cotu+u]<下π/4, 上π/2> = -[0+π/2 -1 - π/4] = 1-π/4
看错题了。

Answer 2

let

x=sinu
dx=cosu du
x=√2/2 , u=π/4
x=1, u=π/2
∫(√2/2->1) [√(1-x^2)/x^6] dx
=∫(π/4->π/2) [ cosu/(sinu)^6] [ cosu du]
=∫(π/4->π/2) (cosu)^2/(sinu)^6 du
=∫(π/4->π/2) [ 1- (sinu)^2]/(sinu)^6 du
=∫(π/4->π/2) [ (cscu)^6- (cscu)^4 ] du
=-∫(π/4->π/2) [ (cscu)^4- (cscu)^2 ] dcotu
=-∫(π/4->π/2) { [ 1+(cotu)^2]^2 - [ 1+(cotu)^2 ] } dcotu
=-∫(π/4->π/2) [ (cotu)^2 + (cotu)^4 ] dcotu
= -[ (1/3)(cotu)^3 + (1/5)(cotu)^5 ] |(π/4->π/2)
= 1/3 +1/5
=8/15