Indefinite integral of SiNx of X

Indefinite integral of SiNx of X

The indefinite integral of SiNx / X can not be expressed in the form of elementary function (which can be proved in theory), but the generalized integral of SiNx / x from [0, positive infinity] can be calculated, and its value is π / 2 (which can be calculated by using the knowledge of complex function)
Solve the equations 4x-3y-6 = 0 x ^ 2 + y ^ 2-18x + 45 = 0
3y=4x-6;
y=(4x-6)/3;
x²+(4x-6)²/9-18x+45=0;
9x²+16x²-48x+36-162x+405=0;
25x²-210x+441=0;
(5x-21)²=0;
x=21/5;
y=18/5;
Replace equation 1 with: y = (4x-6) / 3, substitute equation 2 to get: (5x-21) ^ 2 = 0, solve equation 1 to get x = 4.2, substitute equation 1 to get y = 3.6
From 4x-3y-6 = 0, y ^ 2 = (4x / 3-2) ^ 2 is replaced by x ^ 2 + y ^ 2-18x + 45 = 0
The solution of 25X ^ 2-210x + 441 = 0 is x = 21 / 5, and the solution of 4x-3y-6 = 0 is y = 18 / 5
Solving indefinite integral: ∫ DX / SiNx + cosx
∫dx/(sinx+cosx)
=√2∫ dx/(sin(x+π/4)
= √2∫ csc(x+π/4)) dx
=√2ln |csc(x+π/4) - cot(x+π/4)| + C
If the denominator is divided by cosx at the same time, it becomes 1 / (TaNx + 1)
Let t = TaNx, then the original formula of x = aratant is programmed ∫ 1 / (T + 1) * 1 / (T ^ 2 + 1) DT
Then it becomes the indefinite integral of rational fraction, and the split term is 1 / 2 ∫ 1 / (T + 1) - t + 1 / (T ^ 2 + 1) DT
Finally find out the answer, don't forget to add C and use triangle substitution to go back
It is known that x, y, Z are nonnegative numbers, and 3x + 2Y + Z = 5,2x + y-3z = 1. If s = 3x-y-7z, try to find the value range of S
3x+2y+z=5 ①
2x+y-3z=1 ②
① - 2 × 2
7z-x=3
∴z=(x+3)/7 ③
① 3 + 2
11x+7y=16
∴y=(16-11x)/7 ④
Substituting C (3) and C (4) into S = 3x + y-7z, we get
S=3x+(16-11x)/7-(x+3)=(3x-5)/7
Because Z is greater than zero, y is greater than zero
We know 0 ≤ x from z = (x + 3) / 7
From y = (16-11x) / 7, we know that x ≤ 16 / 11
∴0≤x≤16/11
When x = 16 / 11, the maximum value of S is - 1 / 11
When x = 0, the minimum value of S is - 5 / 7
The limit of (1 / sinx-1 / x) is obtained when x tends to 0
Thank you for the steps
When n → 0, the relation of equivalent infinitesimal is SiNx
1 / SiNx 1 / X
lim(n→0)（1/sinx－1/x）
=
lim（n→0）（1/sinx）- lim（n→0）（1/x）
=lim（n→0）（1/x-1/x）
=lim（n→0）0
=0
It is known that x, y, Z are three nonnegative numbers, and satisfy 3x + 2Y + Z = 5,2x + y-3z = 1. Let k = 3x + y-7z, a be the maximum value of K, B be the minimum value of K, and try to find the value of ab
3X+2Y+Z=5,
2X+Y-3Z=1
The solution is x = 7z-3, y = 7-11z
∵ XYZ is three non negative numbers
That is 7z-3 > = 0
y> 7-11z > = 0
z>=0
∴3/7
When X - > 0, the limit of x ^ SiNx?
Such as the title, X is the bottom, SiNx is the index
Find the limit of this formula when X - > 0
Let y = x ^ SiNx, take logarithm, LNY = SiNx, LNX, so LNY = (LNX) / (1 / SiNx), because when x → 0, SiNx ~ x, so when x → 0, limlny = Lim [(LNX) / (1 / SiNx)] = Lim [(LNX) / (1 / x)]. According to lobita's law, limlny = Lim [(LNX) / (1 / x)] = Lim [
Let y = x ^ SiNx
lny=sinx*lnx
=lnx/(1/sinx)
Using the law of lobida
=(1/x)/(-cosx/sin^x)
=-sin^x/xcosx
=2sinxcosx/(cosx-xsinx)
Substitute x = 0 in
=0
So the limit of LNY is 0
So y tends to 1
So the limit of SiNx power of X is 1
ln(x^sinx)=sinx*lnx
Because when X - > 0, SiNx - > 0, LNX - > 0, so SiNx * LNX - > 0 is ln (x ^ SiNx) - > 0
So x ^ SiNx - > 1
X - > 0 LNX - > negative infinity, so the solution of the third floor is wrong~~
X - > 0 can be understood as X - > 0 + and X - > 0-
When X - > 0 + can be solved according to the second floor solution, but when X - > 0 - LNX does not exist, the second floor is not taken into account
1. Because when X - > 0 +
Zero
It is known that x, y and Z are three nonnegative integers satisfying 3x + 2Y + Z = 5 and X + Y-Z = 2. If s = 2x + Y-Z, then the sum of the maximum and minimum of S is___ .
Method 1: to make s take the maximum, 2x + y the maximum, z the minimum, ∵ x, y, Z are three non negative integers, ∵ z = 0, solve the equations 3x + 2Y = 5x + y = 2, the solution is: x = 1y = 1, ∵ s the maximum = 2 × 1 + 1-0 = 3; to make s take the minimum, the equations 3x + 2Y + Z = 5 (1) x + Y-Z = 2 (2), (1) + (2) are 4x + 3y
When x tends to zero, find the limit of the square of X multiplied by one-third of the square of SiNx
Wrong number in the picture. It's going to zero
When x approaches 0, sin (1 / x) = 1 / x, so the formula is equal to 1
Given that x, y and Z are all nonnegative numbers, satisfying x + Y-Z = 1, x + 2Y + 3Z = 4, w = 3x + 2Y + Z, find the maximum and minimum of X
Wrong. It should be
Given that x, y and Z are all nonnegative numbers, satisfying x + Y-Z = 1, x + 2Y + 3Z = 4, w = 3x + 2Y + Z, find the maximum and minimum of W
The maximum is 6 and the minimum is 3.2
2*（x+y-z)=2x+2y-2z=1*2=2
(2x+2y-2z)+(x+2y+3z)=3x+4y+z=2+4=6
W = 3x + 2Y + Z = (3x + 4Y + Z) - 2Y = 6-2y, when y is the minimum, W is the maximum, and Y is the minimum of 0 (the equation can be solved, x = 1.75, z = 0.75), then w is the minimum of 6-2 * 0 = 6.
When y is the maximum, W is the minimum. X + Y-Z = 1. From this formula, it can be seen that when x is the minimum, y is the maximum and X is the minimum, which can be solved by substituting into the equation, y = 1.4,... Expansion
2*（x+y-z)=2x+2y-2z=1*2=2
(2x+2y-2z)+(x+2y+3z)=3x+4y+z=2+4=6
W = 3x + 2Y + Z = (3x + 4Y + Z) - 2Y = 6-2y, when y is the minimum, W is the maximum, and Y is the minimum of 0 (the equation can be solved, x = 1.75, z = 0.75), then w is the minimum of 6-2 * 0 = 6.
When y is the largest, W is the smallest. X + Y-Z = 1. From this formula, it can be seen that when x is the smallest, y is the largest and X is the smallest. If x is 0 (which can be solved by substituting into the equation, y = 1.4, z = 0.4), then w is the smallest 6-2 * 1.4 = 3.2.
So the maximum W is 6 and the minimum W is 3.2. Put it away
2*（x+y-z)=2x+2y-2z=1*2=2
(2x+2y-2z)+(x+2y+3z)=3x+4y+z=2+4=6
W = 3x + 2Y + Z = (3x + 4Y + Z) - 2Y = 6-2y, when y is the minimum, W is the maximum, and Y is the minimum of 0 (the equation can be solved, x = 1.75, z = 0.75), then w is the minimum of 6-2 * 0 = 6.
When y is the maximum, W is the minimum. X + Y-Z = 1. From this formula, it can be seen that when x is the minimum, y is the maximum and X is the minimum, which can be solved by substituting into the equation, y = 1.4,... Expansion
2*（x+y-z)=2x+2y-2z=1*2=2
(2x+2y-2z)+(x+2y+3z)=3x+4y+z=2+4=6
W = 3x + 2Y + Z = (3x + 4Y + Z) - 2Y = 6-2y, when y is the minimum, W is the maximum, and Y is the minimum of 0 (the equation can be solved, x = 1.75, z = 0.75), then w is the minimum of 6-2 * 0 = 6.
When y is the largest, W is the smallest. X + Y-Z = 1. From this formula, we can see that when x is the smallest, y is the largest and X is the smallest. If x is 0 (which can be solved by substituting into the equation, y = 1.4, z = 0.4), then w is the smallest 6-2 * 1.4 = 3.2
x+y-z=13x+3y-3z=3
We obtain the maximum value of Y + 5Y = 7
y=(7-4x)/5
In this paper, we give the formula 2x + 2y-2z = 2
X + 2Y + 3Z = 4minus z = (3 + x) / 5
A kind of
w=3x+2y+z=3x+2*(7-4x)/5+(3+x)/5=2x+3.2
A kind of
2.5W
The maximum value of W is 7