Solve Indefinite Integral: U Substitution

[The Subject]

Homework Statement

Im looking over the notes in my lecture and the prof wrote,
$$\int_{0}^{2} \pi(4x^2-x^4)dx=\frac{64\pi}{15}$$
Im wondering what's the indefinite integral of this equation.

Homework Equations

using u substitution

The Attempt at a Solution

$$\int \pi(4x^2-x^4)dx= \pi \int x^2(4-x^2)dx \\<br /> u = 4 - x^2 \ \ \ \ \ \ \ \ \ \ <br /> -\frac {1}{2} du =xdx \\$$

Im confuse since i have an x^2 but my du=x.

I attempted to also use from u to get,
$$x=\sqrt{4-u} \\<br /> \pi \int \frac{1}{2}u\sqrt{4-u}dx$$
but it seems this made the formula harder to integrate...or am i just giving up too quickly

 

[Samy_A]

Homework Statement

Im looking over the notes in my lecture and the prof wrote,
$$\int_{0}^{2} \pi(4x^2-x^4)dx=\frac{64\pi}{15}$$
Im wondering what's the indefinite integral of this equation.

Homework Equations

using u substitution

The Attempt at a Solution

$$\int \pi(4x^2-x^4)dx= \pi \int x^2(4-x^2)dx \\<br /> u = 4 - x^2 \ \ \ \ \ \ \ \ \ \<br /> -\frac {1}{2} du =xdx \\$$

Im confuse since i have an x^2 but my du=x.

I attempted to also use from u to get,
$$x=\sqrt{4-u} \\<br /> \pi \int \frac{1}{2}u\sqrt{4-u}dx$$
but it seems this made the formula harder to integrate...or am i just giving up too quickly

You are making this far too complicated.

What is $\int x^n dx$, (for $n \geq 1$)?

 

[The Subject]

That makes sense...

I was working on a bunch of u sub equations earlier. I guess I was on a u sub mode

Thanks a lot!