Question

求积分∫x^2*(1+x^2)^(1/2)（从0积到根号2）

Answer 1

∫(0->√2) x²√(1+x²) dx，x = tanz，dx = sec²z dz
x = 0，z = 0，x = √2，z = arctan√2
=> ∫(0->arctan√2) tan²z * secz * sec²z dz
= ∫(0->arctan√2) (sec²z - 1)sec³z dz
= ∫(0->arctan√2) (sec^5z - sec³z) dz
= (1/4)tanzsec³z - (1/8)tanzsecz - (1/8)ln|secz+tanz|
= 3√6/4 - √6/8 - (1/8)ln(√2+√3)
= (1/8)[5√6 - ln(√2+√3)]

对於∫ sec^5(x) dx 和 ∫ sec³x dx
用降幂公式：∫ sec^n(x) dx = 1/(n-1) * [sinx * sec^(n-1)(x)] + (n-2)/(n-1) * ∫ sec^(n-2)(x) dx