∫ [arctan root number (x ＾ 2-1)] / (x ＾ 2) root sign (x ＾ 2-1) DX =

∫ [arctan root number (x ＾ 2-1)] / (x ＾ 2) root sign (x ＾ 2-1) DX =

Let arctan radical x Λ 2-1 be t, which is equivalent to calculating tcost and integrating T. the result is TSINT + cost, Sint = (radical x Λ 2-1) / x, cost = 1 / x, and it will be good to take it in

x> What is 1 D (x-1 under the root of x ^ 2 arctan)

Let X-1 = t, then x = T ^ 2 + 1, t > 0
D (x-1 under x ^ 2 arctan root) = D ((T ^ 2 + 1) ^ 2 arctant) = [2 (T ^ 2 + 1) * 2T * arctant + T ^ 2 + 1) ^ 2 * 1 / (T ^ 2 + 1)] DT
=(t^2+1)*(1+4arctant)dt

It is proved that if x ≥ 1, arctan radical (x ^ 2-1) + arcsin1 / x = π / 2 If we use Rolle's theorem or Lagrange's theorem, Let f (x) = arctan radical (x ^ 2-1) + arcsin1 / X

Let arctan radical (x ^ 2-1) = a arcsin1 / x = B, then Sina = radical (x ^ 2-1) / X cosa = 1 / xsinb = 1 / X CoSb = radical (x ^ 2-1) / xsin (a + b) = sinacosb + cosasinb = radical (x ^ 2-1) / X * radical (x ^ 2-1) / x + 1 / X * 1 / x = 1 = sin π / 2 (a + b) = π / 2

Y = xarcsin (x / 2) + root sign (4-x Square), calculate derivative, please write detailed points,

Y = xarcsin (x / 2) + √ (4-x ^ 2), y '= [xarcsin (x / 2)]' + [√ (4-x ^ 2)] ', = arcsin (x / 2) + X * 1 / 2 * 1 / √ (1-x ^ 2 / 4) + 1 / 2 * (- 2x) * 1 / √ (4-x ^ 2), = arcsin (x / 2) + X / √ (4-x ^ 2) - X / √ (4-x ^ 2), = arcsin (x / 2) is solved

The derivation of arctan radical (x ^ 2-1),

If u = x? - 1, then u '= 2xv = √ u, then V' = 1 / (2 √ U) * u '= 2x / (2 √ U) = x / √ (x? - 1), so let y = arctan √ (x? 2 - 1) = arctanv, then y' = 1 / (1 + V? 2) * V '= 1 / (1 + X? - 1) * x / √ (x? - 1) = 1 / [x √ (x? - 1)]

(1 + TaNx) / (1-tanx) = 3 + 2 root sign 2 Find TaNx

1=tan45°
So 3 + 2 times root 2 = (tan45 ° + TaNx) / (1-tan45 ° TaNx)
According to the formula, 3 + 2 times root sign 2 = Tan (45 ° + x)
Do it yourself

According to the condition, TaNx = radical 3, X belongs to [0,2 π)

60 or 240`
Tan (180 '+ a) = Tana (induction formula)

Write the set of angles X that make the following inequality hold: (1) 1 + TaNx is greater than or equal to 0. (2) radical 3 / 3 is greater than or equal to TaNx less than 1 (important process)

(1) 1 + TaNx > = 0, TaNx > = - 1 = Tan (- π / 4); the solution set is [- π / 4 + K π, π / 2 + K π); K ∈ Z
(2) Tan π / 6 = radical 3 / 3 < = TaNx < 1 = Tan π / 4; the solution set is [π / 6 + K π, π / 4 + K π); K ∈ Z
The basis is that the tangent function is an increasing function in each period

Given TaNx = radical 3, X ∈ (3 π, 7 π / 2,), find the angle X;

TaNx = radical 3 x = k π + π / 3 K ∈ Z
x∈(3π,7π/2)
3π< kπ+π/3

How to express four fifths of the root on the number axis

It's between 0 and + 1. Divide them into 5 parts and write 4 / 5 in the fourth part