Solving mathematical integral ∫ (x ^ 3 * arccosx) / (sqrt (1-x ^ 2)) DX

Solving mathematical integral ∫ (x ^ 3 * arccosx) / (sqrt (1-x ^ 2)) DX

Let x = cost, have
The original formula = - ∫ TCOS & sup3; t DT

When we know the quadratic equation of one variable x square - ax + a square - 3A = 0, a is a non negative integer (1) and what the value of a is, the equation has two real roots
(2) When a is a number, the equation has two integer roots

x²-ax+a²-3a=0
Δ=a²-4(a²-3a)
=-3a²+12a
Δ≥0 ==> -3a²+12a≥0
==> a²-4a≤0
==> 0≤a≤4
∵ A is a nonnegative integer
∴a=0,1,2,3,4
When a = 0, X & # 178; = 0
When a = 1, X & # 178; - X-2 = 0, the solution is: X1 = 2, X2 = - 1
When a = 2, the equation x & # 178; - 2x-2 = 0 has irrational roots
When a = 3, X & # 178; - 3x = 0, the solution is: X1 = 0, X2 = 3
A = 4, X & # 178; - 4x + 4 = 0, x = 2
When a = 0,1,3,4, the equation has two integer roots

Definite integral of [- (x-a) (X-B)] ^ 1 / 2

1. Find the rule: 13, 16, 19, ()

13、16、19、（22 ）（ 25）.

What are the two natural numbers adjacent to a

A-1t and a + 1

Set of natural numbers set of positive integers set of rational numbers set of integers set of real numbers are represented by those letters?

Natural number set n positive integer set n + rational number set Q integer set Z real number set R '
That's it. Have a good time

There is a string of numbers, the first line is 1,2, the second line is 3,4,5,6, the third line is 7,8,9,10,11,12 What is the fourth number in line 100?
(100-1) * 100 + 1 = 9901, the first number in line 100 is 9901, so the fourth number is 9904

Let's look at the first number of each line, 1,3,7,1 = 0 + 1,3 = 2 + 1,7 = 2 + 4 + 1, the fourth line should be 13 = 2 + 4 + 6 + 1, the fifth line 2 + 4 + 6 + 8 + 1 = 21. That is to say, the number of each line is twice the ordinal number, so the first number of line 100 is 2 + 4 + 6 + 8 +. + 198 + 1 = (2 + 198) * 99 + 1 = 9901, so the fourth number is 9904

Using mathematical induction to prove that the cubic sum of three continuous positive integers can be divided by nine

It is proved that: 1. When n = 1,2,3, it is obvious that the sum 36 is divisible by 9
2. If n = k, the original proposition holds, that is to say
K ^ 3 + (K + 1) ^ 3 + (K + 2) ^ 3 divided by 9
If n = K + 3, then
（k+3）^3+(k+4)^3+(k+5)^3
=【3（k+3+k+5)^2/2】
=9 (K + 4) ^ 2 is obviously an integral multiple of 9

If f (x) is an even function defined on R and satisfies f (x + 2) = - f (x). When 1 is less than or equal to X and less than or equal to 2, f (x) = x squared - 2x
Then f (13 / 2) is equal to

So f [(x + 2) + 2] = - f (x + 2) = f (x), that is, f (x + 4) = f (x)
F (x) is a periodic function with period 4, so f (13 / 2) = f (2.5) = f (- 1.5)
Because f (x) is even function, f (- 1.5) = f (1.5) = 2.25-4.5 = - 0.75 = - 3 / 4

Take a look at the girl next door.

Look at the girl next door