Find the indefinite integral x ∫ (1-x ^ 2) / X DX under the root sign of X ∫ (1-x ^ 2) / X * x DX under the root sign

Find the indefinite integral x ∫ (1-x ^ 2) / X DX under the root sign of X ∫ (1-x ^ 2) / X * x DX under the root sign

∫(1-x^2)/x^(3/2) dx
=∫[x^(-3/2) - x^(1/2)] dx
= -2x^(1/2) - (2/3)x^(3/2) + c

Find the indefinite integral (x ^ 2-A ^ 2) DX under the root sign

Answer: (x / 2) √ (x) ² - a ²) - (a ²/ 2)ln|x + √(x ² - a ²)| + C
Let x = a * secz, DX = a * secztanz DZ, assuming x > a
∫ √(x ² - a ²) dx
= ∫ √(a ² sec ² z - a ²) * (a * secztanz dz)
= a ² ∫ tan ² z * secz dz
= a ² ∫ (sec ² z - 1) * secz dz
= a ² ∫ sec ³ z dz - a ² ∫ secz dz
= a ² M - a ² N
M = ∫ sec ³ z dz = ∫ secz dtanz
= secztanz - ∫ tanz dsecz
= secztanz - ∫ tanz * (secztanz dz)
= secztanz - ∫ (sec ² z - 1) * secz dz
= secztanz - M + N
2M = secztanz + N => N = (1/2)secztanz + N/2
Original formula = (a) ²/ 2)secztanz + a ² N/2 - a ² N
= (a ²/ 2)secztanz - (a ²/ 2)∫ secz dz
= (a ²/ 2)secztanz - (a ²/ 2)ln|secz + tanz| + C
= (a ²/ 2)(x/a)[√(x ² - a ²)/ a] - (a ²/ 2)ln|x/a + √(x ² - a ²)/ a| + C
= (x/2)√(x ² - a ²) - (a ²/ 2)ln|x + √(x ² - a ²)| + C

Indefinite integral ∫ DX under the root sign (x + 1 / x-1), and the root sign includes the whole fraction I set the global root sign T. to 4 ∫ (1 / (T ^ 2 + 1) + 1 / (t Λ 2 + 1) ^ 2) DT e. I've got the wrong number. The plus sign is minus. It's not plus. It's t-square-1. No one found it

∫ ((x + 1) / (x-1)) ^ 0.5dx multiply up and down (x + 1) = ∫ (x + 1) / (x ^ 2-1) ^ 0.5dx = ∫ X / (x ^ 2-1) ^ 0.5dx + ∫ 1) / (x ^ 2-1) ^ 0.5dx = 1 / 2 ∫ 1 / (x ^ 2-1) ^ 0.5d (x ^ 2-1) + ∫ 1 / (x ^ 2-1) ^ 0.5dx = (x ^ 2-1) ^ 0.5 + ln (x + (x ^ 2-1) ^ 0.5) + C

The indefinite integral DX of (1-x ^ 2) DX / x ^ 2 under the root sign / the indefinite integral DX of 1-x ^ 2 under the root sign shall be solved by the second transformation method,

∫ (√(1-x^2) /x^2) dxletx= sinydx= cosy dy ∫ (√(1-x^2) /x^2) dx= ∫ (coty)^2 dy= ∫ [(cscy)^2 - 1] dy= -coty - y + C= - √(1-x^2) /x - arcsinx + C ∫ (1/√(1-x^2)) dxletx = sinydx = cosy dy∫ (1...

Interaction of complex number and trigonometric function RT…… How did you convert it?

Click the figure below

How is the complex exponential function defined? I'm learning complex change. Is it derived from Euler's formula? Or is Euler's formula derived from it?

Let the exponential function of Z = x + iy complex be defined as e ^ z = e ^ x (cosy + isiny), which can be regarded as the derivation of Euler formula. A proof of e ^ iy = cosy + isiny Euler formula is to consider the expansion of power series. For the proof process of e ^ IX = cosx + isinx, please refer to the contribution of our delegation 522 http://zhidao.baidu.com/question/33...