Using Beta Gamma function, show that :$\int_0^{\frac {\pi}{6}} \cos^2 (6\theta).\sin^4 (3\theta) d\theta$

Question

Using Beta Gamma function, show that :$\int_0^{\frac {\pi}{6}} \cos^2 (6\theta)\cdot\sin^4 (3\theta) d\theta$

My Attempt

$$\int_0^{\frac {\pi}{6}} \cos^2 (6\theta) \cdot \sin^4 (3\theta) d\theta$$
Put $3\theta=t$
$$3d\theta=dt$$
$$d\theta=\frac {dt}{3}$$
When $\theta=0$, $t=0$
When $\theta=\frac {\pi}{6}$, $t=\frac {\pi}{2}$
Now,
$$=\int_0^{\frac {\pi}{2}} \cos^2 (2t)\cdot\sin^4 (t) \frac {dt}{3}$$

Answer 1

Hint:

Use $\cos2t=1-2\sin^2t$

$$\cos^2(2t)=(1-2\sin^2t)^2=?$$

$$\beta\left(m,n\right)=2\int_0^{\pi/2}\sin^{2m-1}t\cos^{2n-1}t\ dt$$