24/04/28 17:18:11.11 dCSp4kxv.net
>>642
I = ∫[a,b] cos(x-(ab/x)) dx (置換t=ab/x)
= ∫[a,b] cos((ab/t)-t)(ab/t^2) dt
(第一式+第二式)/2
I = (1/2)∫[a,b] cos(x-(ab/x))(1+(ab/x^2)) dx (置換t=x-ab/x)
= (1/2)∫[a-b,b-a] cos(t) dt
= sin(b-a)
24/04/28 17:18:11.11 dCSp4kxv.net
>>642
I = ∫[a,b] cos(x-(ab/x)) dx (置換t=ab/x)
= ∫[a,b] cos((ab/t)-t)(ab/t^2) dt
(第一式+第二式)/2
I = (1/2)∫[a,b] cos(x-(ab/x))(1+(ab/x^2)) dx (置換t=x-ab/x)
= (1/2)∫[a-b,b-a] cos(t) dt
= sin(b-a)