# U substitution

## 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 whats the indefinite integral of this equation.

## Homework Equations

using u substitution

## The Attempt at a Solution

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

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

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

Samy_A
Homework Helper

## 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 whats the indefinite integral of this equation.

## Homework Equations

using u substitution

## The Attempt at a Solution

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

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

I attempted to also use from u to get,
$$x=\sqrt{4-u} \\ \pi \int \frac{1}{2}u\sqrt{4-u}dx$$
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##)?

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!