Evaluate the indefinite integral. Use the Substitution Rule.?

Question

Given: Integral of y^3 * sqrt(2y^4 - 1) dy.

The Substitution Rule states that if u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then the integral of f(g(x))g'(x) dx = the integral of f(u)du.

Answer 1

let 2y^4-1 be x
8y^3dy=dx
y^3dy=dx/8
the integral (1/8)rtxdx
=(1/8)x^3/2(2/3)+C
=(1/12)(2y^4-1)^3/2+C

Answer 2

You start by recognizing that y^3 is the derivative (y^4)/4; So if you set u = (2*y^4 - 1), then du = 8*y^3; take the factor of 8 outside the integral to get

(1/8)∫[1/√u]du; The integral is 1/8 * 2*√u = 1/4 * √[2*y^4 - 1] + C

You can check that this is right by taking the derivative of 1/4 * √[2*y^4 - 1] + C

Answer 3

The trick is which you're able to understand that this is corresponding to the crucial of a million / (a million + x^2) which equals arctan(x). understand the final formulation for a the crucial: crucial( du / (a^2 + u^2) ) = a million/a * arctan(u/a) crucial ( x / (a million + x^4 ) ) We see that u = x^2, and a = a million, consequently du = 2x dx a million/2 * crucial( du / (a million + u^2) ) = a million/2 * a million/a million * artcan(u/a) + c = a million/2 * arctan(x^2) + c

Answer 4

∫y^3 * √(2y^4 - 1) dy
Let u = 2y^4 therefore du = 8y^3 dy
So ∫y^3 * √(2y^4 - 1) dy
= 1/8 *∫√(2y^4 - 1)* 8y^3 dy
= 1/8∫√(u -1)du
= (1/8) * (2/3) * (u -1)^(3/2) + c
= [(2y^4 - 1)^3/2]/12 + c