Thanks. Actually I need to get some expression I can integrate but I forgot the algebra part of it...

These are very different questions. Anon gave a great way to do a whole bunch of things to that equation...but do we want to integrate that? Yikes! I don't want to.

\(\displaystyle \sqrt{4-y^2}(5-y) = 5 \sqrt{4-y^2} - y \sqrt{4-y^2} \)

If we have an integral it would look like

\(\displaystyle \int \sqrt{4-y^2}(5-y) dy = 5 \int \sqrt{4-y^2}dy - \int y \sqrt{4-y^2} dy \)

Let us go through both parts here.

**Part 1**
\(\displaystyle 5 \int \sqrt{4-y^2}dy\)

This is of the form \(\displaystyle \sqrt{a^2 - y^2} \) which is a prime candidate for sin substitution (I will discuss a trick and why we dont need to do this in a moment!)

Let \(\displaystyle y=2sin \theta \) and \(\displaystyle dy= 2cos \theta d \theta \)

\(\displaystyle 5 \int \sqrt{4-4sin^2 \theta }dy = 5 \int 2\sqrt{1 - sin^2 \theta } 2 cos \theta d \theta = 20 \int cos^2 d \theta \)

Remember the double angle formula

\(\displaystyle cos^2 \theta = \frac{ 1+cos2 \theta }{2} \)

\(\displaystyle 20 \int cos^2 d \theta = 20 \int \frac{1}{2} d \theta + 10 \int cos2 \theta d \theta = 10 \theta + 5 sin2 \theta \)

Of course you then switch your limits and evaluate for theta. However, if we don't want to do all of this we can note that

\(\displaystyle 5 \int \sqrt{4-y^2}dy\) is 5 times the area of a semi-circle of radius 2. Therefore,

\(\displaystyle 5 \int \sqrt{4-y^2}dy = 5 \frac{ \pi (2)^2 }{2} = 10 \pi\)

**Part 2**
\(\displaystyle \int y \sqrt{4-y^2} dy \)

In this case let \(\displaystyle p=y^2 \) and \(\displaystyle dp = 2ydy \)

\(\displaystyle \frac{1}{2} \int \sqrt{4-p} dp = - \frac{1}{3} (4-p)^{ \frac{3}{2} } \)

Subbing back in for y,

\(\displaystyle - \frac{1}{3} (4-y^2)^{ \frac{3}{2} } \)

Thus,

\(\displaystyle 5 \int \sqrt{4-y^2}dy - \int y \sqrt{4-y^2} dy = 10 \pi - \frac{1}{3} (4-y^2)^{ \frac{3}{2} } \)