اوجد طول منحنى الدالة د(س) = س² فى الفترة [0 ، 1]

 :
let x = ½ tan(u)
dx =½ sec²(u) du
∫sqrt(4x² + 1) dx

from 0 to sqrt(1)

½∫sqrt(tan²u + 1) sec²(u) du
= ½∫sec³(u) du
لاحظ اننا ذكرنا تكامل
الصيغة sec^n(u)
ll
= ½[½sec(u) tan(u) +½ ln|sec(u)+tan(u)|]

= ¼[sec(u) tan(u) + ln|sec(u)+tan(u)| ]

but x = ½tan(u)
tan(u) = 2x
sec(u) = sqrt(4x² + 1 )  ... by substitution

¼[2x sqrt(4x² + 1 ) + ln|2x+sqrt(4x² + 1 )|]
from 0 to 1

= ¼[2sqrt(5) + ln|2+sqrt(5)] ≈ 1.4789