What is the intergral??

Question

What is the intergralor antiderivative of y = (51*sqrt(x))/(sqrt(1-(64*x^3)))?

Answer 1

∫ [51 √x / √(1 - 64x³)] dx =

51 ∫ [√x / √(1 - 64x³)] dx =

let √x = (1/2)³√u →

³√u = 2√x →

u = (2√x)³ = 8√x³ = 8x√x

x = (1/4)³√u²→

x³ = (1/64) u²→

dx = d{(1/4)[u^(2/3)]} = (1/4)(2/3)u^[(2/3) -1] du = (1/6) u^(-1/3) du →

dx = [1 /(6 ³√u)] du

thus, substituting, you get:

51 ∫ [√x /√(1 - 64x³)] dx = 51 ∫ { (1/2)³√u /√{1 - [64(1/64)u²]} } [1 /(6 ³√u)] du =

51 ∫ { (1/2)³√u /{[√(1 - u²)] 6 ³√u } } du =

canceling ³√u and taking out the constants:

(51/12) ∫ [1 /√(1 - u²)] du =

(it's an immediate integral)

(51/12) arcsin u + C

finally, substituting back u = 8x√x, you get:

∫ [51√x /√(1 - 64x³)] dx = (51/12) arcsin(8x√x) + C

I hope it helps...

Bye!