Integal $\int\left(\sqrt{4-9x^2}\right)^3 dx$

Question

$$\int\left(\sqrt{4-9x^2}\right)^3 dx$$
I tried using $t=2/3\sin{x}$ but I get to $\int(\cos{x})^4dx$ which I dont know how can be solved easilly.

Answer 1

$\textbf{Hint:}$

$$\cos(2x)=\cos^2(x)-\sin^2(x)=2\cos^2(x)-1$$

So:

$$\cos^2(x)=\frac{1+\cos(2x)}{2}$$

$$\cos^4(x)=\frac{(1+\cos(2x))^2}{4}$$

The same with $\cos^2(2x)$

$$\cos(4x)=\cos^2(2x)-\sin^2(2x)=2\cos^2(2x)-1$$