evaluate: integration of x^2 (square root of (1-2x^3))dx

Let u = 1-2x^3

du = =6 x^2 dx
Your integral is INTEGRAL u^1/2*(-du/6)
= -(1/6)*u^(3/2)/(3/2)
= -(1/9)(1-2x^3)^(3/2)

Check my math

thanks

To evaluate the integral ∫ x^2√(1-2x^3) dx, we can use a method known as u-substitution.

Let's start by identifying a suitable substitution. In this case, we can let u = 1-2x^3.

Next, we need to find the appropriate differential, which is du. To do this, we differentiate both sides of the equation u = 1-2x^3 with respect to x:

du/dx = -6x^2.

Now, we can solve for dx:

dx = du / (-6x^2).

Substituting these values into the integral, we have:

∫ x^2√(1-2x^3) dx = ∫ x^2√u (du / (-6x^2)).

Simplifying this expression, we can cancel out the x^2 terms:

∫ √u du / -6.

Now, we can integrate:

-1/6 * ∫ √u du.

To integrate √u, we can use the power rule of integration. Adding 1 to the exponent of u (which is 1/2) and dividing by the new exponent, we get:

-1/6 * (2/3) * u^(3/2) + C.

Finally, substituting back in for u, we have:

-2/18 * (1-2x^3)^(3/2) + C.

Therefore, the integral ∫ x^2√(1-2x^3) dx evaluates to:

-2/18 * (1-2x^3)^(3/2) + C.