Find the limit cost ^ 2DT / X when x approaches to 0, where cost ^ 2DT is the definite integral of line X and line 0

Find the limit cost ^ 2DT / X when x approaches to 0, where cost ^ 2DT is the definite integral of line X and line 0

Okay!

Limit limx → 2 ∫ (lower limit 2, upper limit x) cost ^ 2DT / (x ^ 3-8)=

This is the 0 / 0 type limit. The L'Hospital Rule can be used. The molecule is a variable upper limit integral, and its derivative is derived according to the variable upper limit integral
lim(x→2) ∫(2,x) cos²tdt/(x³-8)
=lim(x→2) cos²x/(3x²)
=(cos²2)/12

Reciprocal of the definite integral of cost / T in [square of X, 1]

The derivative of the definite integral of cost / T in [square of X, 1]
=2xcosx²/x²
=2cosx²/x

Let f (x) = ∫ (1, x ^ 3) Sint / TDT, find ∫ (0,1) x ^ 2F (x) DX (if f (x) = ∫ (1, x ^ n) Sint / TDT, then ∫ (0,1) x ^ (x-1) f (x) DX

Obviously, f (1) = 0; from the basic theorem of calculus, we know that f '(x) = sin (x ^ 3) / x ^ 3 * 3x ^ 2 = 3sin (x ^ 3) / X
So ∫ (0,1) x ^ 2F (x) DX
=∫(0,1)f(x)d（x^3/3）
=X ^ 3 * f (x) / 3 | upper limit 1 lower limit 0 - ∫ (0,1) x ^ 3 * f '(x) / 3DX
=-∫(0,1) x^2sin(x^3)dx
=Cos (x ^ 3) / 3 | upper limit 1 lower limit 0
=(cos1-1)/3.
The result is (COS 1-1) / n