Indefinite Integral

Question

I'm having problems with starting this integral, I'm not looking for the solution, but the first step/two would be appreciated.

Integral:

Evaluate

$$ \int dx \sqrt{x(4-x)} $$

Answer 1

Hint: $ \displaystyle \int \sqrt{ 4 - 4 + 4x - x^2} dx$

$ = \displaystyle \int \sqrt{ 2^2 - (x-2)^2}dx$

Now use the formula - $ \displaystyle \int \sqrt{a^2 - x^2} dx$

This should be enough

Answer 2

Hint:

Since $x(4-x)=4x-x^2=4-(x^2-4x+4)=2^2-(x-2)^2$ we can set $x-2=2\sin t$ so

$$\int\sqrt{x(4-x)}dx=\int 2\cos t(2\cos t)dt=4\int\cos^2 t dt$$

Answer 3

Sub $x=4 \sin^2{t}$, $dx = 8 \sin{t} \cos{t} \, dt$, so the integral becomes

$$32 \int dt \sin^2{t} \cos^2{t} $$

That should be enough of a hint.