If t = arcsinx, why x = Sint?

If t = arcsinx, why x = Sint?

t=arcsinx
sint
=sin(arcsinx)
=x
So Sint = X
x=sint
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The derivative of the inverse function is equal to the reciprocal of the derivative of the direct function, and the inverse function of the inverse proportion function y = 1 / X is also y = 1 / x, so their derivatives are not
Should it be the same?

Yes,
x'=1/y'=-x^2=-1/y^2
Change to the expression of inverse function: y '= - 1 / x ^ 2

Find the function y = SiNx (0

I didn't say that it should be enclosed with x = 0 or y = 0, it should be the upper part of the cut
y=sinx=1/2,0

F (x) is a piecewise function, f (x) = xsin (1 / x) + 2, X ≠ 0 and f (x) = k, x = 0. If f (x) is continuous at x = 0, what is k

Continuous at x = 0
LIM (x tends to 0) xsin (1 / x) + 2 = f (0) = k
1 / X goes to infinity
So sin (1 / x) oscillates at [- 1,1], which is bounded
So xsin (1 / x) tends to zero
So 0 + 2 = K

If f (arcsinx) = X-1, then f (π 3) = 0___ ．

Since the function f (arcsinx) = X-1, let t = arcsinx, then x = Sint, so the function f (arcsinx) = X-1 ⇔ f (T) = sint-1, so f (π 3) = sin π 3-1 = 32 & nbsp; - 1

If f (x) = x-arcsinx, and f (a) = - 3 / 5
Then f (- a) =?

F (- x) = - X - arcsin (- x) = - (x-arcsinx) = - f (x) odd function
f(-a) = 3/5

The continuous interval of function f (x) = arcsinx / (x (x + 1)) is

The domain of arcsinx is [- 1,1], X ≠ - 1,0
The continuous interval is (- 1,0), (0,1]

Find the maximum value of the function y = (arcsinx) ^ 2 + 2arcsinx - 3, and find the corresponding x value

Let t = arcsinx, then t ∈ [- π / 2, π / 2]
∴ y=t²+2t-3
=(t+1)²-4
The opening is upward, and the axis of symmetry t = - 1
When t = - 1, y has a minimum value of - 4
In this case, arcsinx = - 1, that is, x = - sin1
When t = π / 2, y has a maximum value of π & # / 4 + π - 3
In this case, arcsinx = π / 2, that is, x = 1

∫arcsinx^2/(1-x^2)^1/2 dx
π/324

Because the derivative of arcsinx is 1 / (1-x ^ 2) ^ 1 / 2, so
∫arcsinx^2/(1-x^2)^1/2 dx
=∫arcsinx^2 d arcsinx
=1/3*arcsinx^3+C
You give the wrong answer, this is indefinite integral, how can there be a specific value?

Solving ∫ (1 + x ^ 2) arcsinx / (x ^ 2 √ (1-x ^ 2)) DX
ditto

Substitution with T = arcsinx
= ∫（1+1/（sint）^2）t dt
=(t ^2)/2~∫ t d cot t
= (t ^2)/2~t cot t + ∫ cot t dt
= (t ^2)/2~t cot t +ln(sint)+C
Just substitute t = arcsinx