Finding indefinite integral (x ^ 3) * sqrt (a ^ 2-x ^ 2)

Finding indefinite integral (x ^ 3) * sqrt (a ^ 2-x ^ 2)

Let x = asint = > DX = acost DT
Then the original integral becomes: ∫ (asint) ^ 3 √ A & # 178; - A & # 178; sin & # 178; t × acost DT
=a^4∫(sint)^3cos²t dt
=a^4∫(sint)^3-(sint)^5 dt
∵ (sint)^3=(-1/4)sin3t+(3/4)sint
(sint)^5=(1/16)sin5t-(5/16)sin3t+(5/8)sint
∴a^4∫(sint)^3-(sint)^5 dt
＝a^4∫(-1/16)sin5t+(1/16)sin3t+(1/8)sint dt
＝a^4((-1/80)cos5t+(1/48)cos3t+(1/8)cost +C)

Sqrt (x ^ 2 + 1) SiNx for indefinite integral

This function is an odd function, and the definite integral is 0 on the symmetric interval about the origin, but the original function may not be expressed by elementary function

Let y = f (x) be a solution of the differential equation y ″ - 2Y ′ + 4Y = 0, and f (x0) > 0, f ′ (x0) = 0, then f (x) is ()
A. There is a maximum B. there is a minimum C. monotone increase in a neighborhood D. monotone decrease in a neighborhood

Since y = f (x) is a solution of the differential equation y ″ - 2Y ′ + 4Y = 0, f ″ (x) - 2F ′ (x) + 4f (x) = 0. Because f (x0) > 0, f ′ (x0) = 0, f ″ (x0) = 2F ′ (x0) - 4f (x0) = - 4f (x0) < 0; F ″ (x0) ＜ 0. According to the extreme value theorem of function of one variable, f (x) has a maximum at x0

Let y = f (x) be a solution of the equation y '' + 2Y '+ 4Y = 0. If f (X.) > 0 and f' (X.) = 0, then f (x) is at point X
① Get a maximum or a minimum. 2. Monotonic increase or decrease in a certain field

A lot of freshmen forget, but read carefully, it seems that this topic is very simple, free, and then do

Given the sequence an, even if the arithmetic sequence is an equal ratio sequence, it is proved that the sequence is a non-zero constant sequence

a,b,c
a+c=2b
a/b=b/c=k
---------------
a=bk=ckk
b=ck
ckk+c=2ck
kk+1=2k
(k-1)*(k-1)=0
So a = b = C

The monotone increasing interval of function y = ln (x ^ 2-1) is

That is: y '> 0,2x / (x ^ 2-1) > 0, and x ^ 2-1 > 0, so x > 0, because x ^ 2-1 > 0, so x > 1, that is: (1, + infinity) is an increasing interval

The problem of ternary linear equation in grade one of junior high school
【1】
｛3x-y+z=3
2x+y-3z=11
x+y+z=12
【2】
｛5x-4y+4z=13
2x+7y-3z=19
3x+2y-z=18

3x-y+z=3①
2x+y-3z=11②
x+y+z=12③
① + 2, 5x-2z = 14, 5
① In addition, 4x + 2Z = 15, 6
⑤ The solution is x = 29 / 9, and the solution is x = 29 / 9
Y = 19 / 18 (8), substituting into (1), we get z = - 139 / 18
Well, I'll tell you the method. First, I'll use the addition and subtraction elimination method to eliminate an unknown number in the ternary equation system, and turn it into a binary linear equation system. Then I can solve the binary linear equation system. The second one is to try to do it by myself!

1+3+6+10+15+21+28+… +N summation formula

Original formula = 1 + (1 + 2) + (1 + 2 + 3) + +(1+2+3+…… +n)1+2+3+…… +N = n (n + 1) / 2 = n / 2 + n ^ 2 / 2, so the original formula = (1 + 2 + 3 +...) +n)/2+(1^2+2^2+3^2+…… +n^2)/21^2+2^2+3^2+…… +N ^ 2 = n (n + 1) (2n + 1) / 6, so the original formula = [n (n + 1) / 2] / 2 + [n (n + 1) (2n + 1) / 6] / 2 = n (n

X-6 / X to the second power + 1 * x to the second power + X / X to the second power - 36

=x-1/6x²+x²+x-36
=5/6x²+2x-36

How to do if there is a constant in the denominator of linear equation with one variable? Do you want to multiply the least common multiple of the denominator together?

Yes
This is one of the most common mistakes for students in solving linear equation of one variable. Every term must be multiplied by the least common multiple, and the term without denominator (including the term with constant formula and integer coefficient) must be multiplied