The process of ∫ f '(LNX) / xdx = SiNx + C to find f (x)

The process of ∫ f '(LNX) / xdx = SiNx + C to find f (x)

∫ f'(lnx)/x dx = sinx + C
f'(lnx)/x = cosx
f'(lnx) = xcosx
f'(x) = e^x * cos(e^x)
f(x) = ∫ e^x * cos(e^x) dx = ∫ cos(e^x) d(e^x)
f(x) = sin(e^x) + C

Given the function f (x) = SiNx + LNX, then the value of F (1) is?
There are four options a / cos1-1 B, 1-cos1 C, 1 + cos1 d-1-cos1

f（1）=sin1+ln1
=sin1+0
=sin1

Given x > 1, find x > LNX

F (x) = x-lnx, the derivative is f '(x) = 1-1 / x > 0, increasing, so x-lnx > F (1) > 0, so x > LNX

The decreasing interval of the function y = SiNx (0 ≤ x ≤ 2 π) is

The decreasing interval of y = SiNx (0 ≤ x ≤ 2 π) is (π / 2,3 π / 2)

Monotone decreasing interval of function y = SiNx Λ 2

I don't know how to type symbols. I'll just talk about the idea. The upstairs derivation idea is good, but the derivation is wrong. It should be sin2x, and then let the derivative be less than zero, [4 / PAI + KPAI, 4 / 3pai + KPAI] can also be solved by drawing, roughly drawing, as long as the less than zero is turned up, the important thing is the abscissa, not the ordinate

Let x ∈ [0,2 π], then the interval in which the functions y = SiNx and y = cosx decrease simultaneously is

SiNx decrement is 23 quadrants
The decrease of cosx is 12 quadrants
So it's the second quadrant
So it's (π / 2, π)

For the monotone decreasing interval of function y = | SiNx |
For the function y = | SiNx |
(1) Monotone decreasing interval
(2) Odd function or even function (((process)))) thank you!

(1) Function subtraction interval [K π + π / 2, K π + π]
(2) The function y = | SiNx | is even

What is the monotone decreasing interval of function y = 1 / 2 (SiNx plus | SiNx |)?

The period of the function is 2 π
As long as we expand from [0,2 π]
When x ∈ [0, π], y = SiNx increases monotonically,
The left half increases monotonically and the right half decreases monotonically;
When x ∈ (π, 2 π],
Y = 0 is not monotone
The above conclusion is actually the second quadrant monotonic reduction,
That is, the range of single adjustment and subtraction is:
【π/2+2kπ,π+2kπ】

Given FX = 2Sin (x + π / 6) - 2cox, X ∈ [π / 2, π] if SiNx = 4 / 5, find the value of function FX
Seeking process

fx=2sin(x+π/6)-2cosx
=2(sinxcosπ/6+cosxsinπ/6)-2cosx
=2(√3/2sinx+1/2cosx)-2cosx
=√3sinx-cosx
Because SiNx = 4 / 5, and X ∈ [π / 2, π]
So cosx = - √ (1-sinx ^ 2)
So FX = (4 √ 3 + 3) / 5

The known function FX = SiNx + TaNx
Given the function FX = SiNx + TaNx, the arithmetic sequence an with the number of terms 27 satisfies that an belongs to (- π / 2, π / 2), and the tolerance D ≠ 0. If FA1 + fa2 +. Fa27 = 0, then when k =?, FAK = 0

Because FX = SiNx + TaNx is an odd function in the interval (- π / 2, π / 2)
And an is the arithmetic sequence of interval (- π / 2, π / 2)
Therefore, FA1 + fa2 +. Fa27 must have
fa1=-fa27
fa2=-fa26
……
So to make
fa1+fa2+.fa27=0
There must be fa14 = 0