A problem of definite integral calculation, The questions are as follows: ∫ (1-cosx) & sup2; multiply by 1-cosx DX under quadratic root It didn't work out, PS: Please list the trigonometric function formulas to be used, and make clear what is set to t by substitution, The value range of X is 0 -- 2 π

A problem of definite integral calculation, The questions are as follows: ∫ (1-cosx) & sup2; multiply by 1-cosx DX under quadratic root It didn't work out, PS: Please list the trigonometric function formulas to be used, and make clear what is set to t by substitution, The value range of X is 0 -- 2 π

x=a(t-sint),y=a(1-cost)
∫{t=0,2π}y²ds
=∫{t=0,2π}a²(1-cost)²√[a²(1-cost)²+a²sin²t]dt
=2a³∫{t=0,2π}(1-cost)²sin(t/2)dt
=16a³∫{u=0,π}(sinu)^5du
=16a³*2*4/5*2/3
=256a³/15

A problem of calculating definite integral
Calculate the definite integral ∫ (3) (0) | (x ^ 2) - 1| DX, where (3) (0) is the upper and lower limit
The main reason is that when you see the absolute value, you feel that you can't start with it,

=∫(1)(0)(1-x^2)dx+∫(3)(1)((x^2)-1)dx
=(x-x ^ 3 / 3) (from 0 to 1) + (x ^ 3 / 3-x) (from 1 to 3)
=2/3+（6-(-2/3)）
=22/3