Integral of y^2(2-y^3)^2dy dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 - y^{3}$.

      Then let $du = - 3 y^{2} dy$ and substitute $- \frac{du}{3}$:

      $$\int \frac{u^{2}}{9}\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \left(- \frac{u^{2}}{3}\right)\, du = - \frac{\int u^{2}\, du}{3}$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int u^{2}\, du = \frac{u^{3}}{3}$$

         So, the result is: $- \frac{u^{3}}{9}$

      Now substitute $u$ back in:

      $$- \frac{\left(2 - y^{3}\right)^{3}}{9}$$

   Method #2

   1. Rewrite the integrand:

      $$y^{2} \left(2 - y^{3}\right)^{2} \cdot 1 = y^{8} - 4 y^{5} + 4 y^{2}$$

   2. Integrate term-by-term:

      1. The integral of $y^{n}$ is $\frac{y^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int y^{8}\, dy = \frac{y^{9}}{9}$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \left(- 4 y^{5}\right)\, dy = - 4 \int y^{5}\, dy$$

         1. The integral of $y^{n}$ is $\frac{y^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int y^{5}\, dy = \frac{y^{6}}{6}$$

         So, the result is: $- \frac{2 y^{6}}{3}$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int 4 y^{2}\, dy = 4 \int y^{2}\, dy$$

         1. The integral of $y^{n}$ is $\frac{y^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int y^{2}\, dy = \frac{y^{3}}{3}$$

         So, the result is: $\frac{4 y^{3}}{3}$

      The result is: $\frac{y^{9}}{9} - \frac{2 y^{6}}{3} + \frac{4 y^{3}}{3}$

2. Now simplify:

   $$\frac{\left(y^{3} - 2\right)^{3}}{9}$$

3. Add the constant of integration:

   $$\frac{\left(y^{3} - 2\right)^{3}}{9}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(y^{3} - 2\right)^{3}}{9}+ \mathrm{constant}$$