Solving inequality 2x-1-9x + 2 ≤ 1

Solving inequality 2x-1-9x + 2 ≤ 1

4x-2-9x－2≤6
-5x≤10
x≥-2

Solving inequality: 2x-1 / 3-9x + 2 / 6 ≤ 1
2x-1 / 3 - 9x + 2 / 6 ≤ 1, so it's not easy to see the three items

(2x-1)/3-(9x+2)/6≤1
Multiply both sides by 6
2(2x-1)-(9x+2)≤6
4x-2-9x-2≤6
-5x≤10
x≥2

The inequality of solution about X: (1) (x-3) (X-2) greater than 0 (2) 2x-1 / 5x-6 less than 0

1、
（x-3）（x-2）＞0
Greater than two sides:
So:
X > 3 or x < 2
2. (2x-1) (5x-6) < 0
Multiply both sides by (2x-1) & #, the direction of unequal sign remains unchanged
So:
（2x-1）（5x-1）＜0
Less than middle
1 / 5 x 1 / 2
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∫ (upper Pai, lower 0) radical (SiNx sin ^ 3x) DX

If x ∈ [0, Π / 2], then √ (SiNx sin ^ 3x) = cosx · √ SiNx;
If x ∈ [Π / 2, Π], then √ (SiNx sin ^ 3x) = - cosx ·√ SiNx
simple form
＝∫(0～∏/2)cosx√sinx dx - ∫(∏/2～∏)cosx√sinx dx
＝∫(0～∏/2)√sinx d(sinx) - ∫(∏/2～∏)√sinx d(sinx)
＝(2/3)·(sinx)^(2/3)|(0～∏/2) - (2/3)·(sinx)^(2/3)|(∏/2~∏)
＝(2/3)·[（1-0）-（0-1）]
＝4/3

F (x) = cos (cosx) and G (x) = sin (SiNx)·

The period of F (x), G (x) is 2 π

An integral problem: ∫ Sinh ((LNX)) sin ((LNX)) DX
If it's not convenient, you can write it on paper and upload pictures. It's also acceptable,

Q: ∫sinh((lnx))sin((lnx)) dx
Let u = ln(x)
∴ du = (1/x) dx
Note that e^(u)sinh((lnx))
= e^(u) [ e^ln(x) - e^-ln(x) ] / 2
= e^(u)(e^(u) - e^(-u)) /2
= (1/2) (e^(2u) -1)
∫ sinh((lnx))sin((lnx)) dx
= ∫ e^(u)sin(u)sinh(u)du (add in e^u, so you can utilize above formula)
= ∫ (1/2)(e^(2u)-1)sin(u) du (substitute in above formula)
= 1/2 ∫ (e^(2u)sin(u) - sin(u))du
= 1/2 ∫ e^(2u)sin(u)du - 1/2 ∫ sin(u)du
You have two choices here:
1) Use integration by parts to solve the first integral, ∫ e^(2u)sin(u)du
2) Use the formula, ∫ e^(αu)sin(βu)du = e^(αu)(-β cos(βu) + αsin(βu)) / (α²+ β²)、
I chose the latter one.
= 1/5 e^(2u)sin(u) - 1/10 e^(2u)cos(u) - 1/2 ∫ sin(u) u
= 1/5 e^(2u)sin(u) - 1/10 e^(2u)cos(u) + cos(u)/2 + C
Now substitude u = ln(x) back into the equation.
= 1/5 x²sin(ln(x)) - 1/10 x²cos(ln(x)) + 1/2 cos(ln(x)) + C
= (1/10)(2x²sin(ln(x)) - (x²-5)cos(ln(x)) + C
∴ (1/10)(2x²sin(ln(x)) - (x²-5)cos(ln(x)) + C

Solution of indefinite integral (1-sin ^ 3 * x) DX

$（1-sin^3*x）dx=$dx-$(sinx)^3dx
=x+$(sinx)^2dcosx
=x+$(1-(cosx)^2)dcosx
=x+$dcosx-$(cosx)^2dcosx
=x+cosx-(1/3)(cosx)^3+C

Solving indefinite integral: ∫ xcos (4x ^ 2 + 5) DX

∫xcos(4x^2 +5)dx
let 4x^2+5=t
dt=d(4x^2+5)=4d(x^2)=4*2xdx=8xdx so dx=[dt/8x]
∫xcos(4x^2 +5)dx=∫xcostdt/8x=1/8∫costdt=1/8sint
=1/8(4x^2+5)
After writing for such a long time, there is no reward

Finding indefinite integral ∫ 2x + 3 / √ (4x ^ 2-4x + 5) DX

Can be divided into two, as shown in the figure, the differential. Economic mathematics team to help you solve, please adopt

How does sin (45 + 2x) sin (45-2x) = 1 / 4 turn into sin (45 + 2x) cos (45 + 2x)
How does sin (45 + 2x) sin (45-2x) = 1 / 4 become sin (45 + 2x) cos (45 + 2x) = 1 / 4

Because sin (45 ° + 2x) sin (45 ° - 2x) = 1 / 4
So sin (45 ° + 2x) cos [90 ° - (45 ° - 2x)]
=sin(45°+2x)cos(45°+2x)
=1/4