The value of the integral $$\int_0^{\pi/2} \frac{60\sin(6x)}{\sin x} dx$$ is equal to

We need to evaluate $$\int_0^{\pi/2} \frac{60\sin 6x}{\sin x}\,dx$$.

We use the telescoping identity: $$\frac{\sin 2nx}{\sin x} = 2\cos x + 2\cos 3x + 2\cos 5x + \ldots + 2\cos(2n-1)x$$. This can be verified by multiplying both sides by $$\sin x$$ and using the product-to-sum formula $$2\sin x \cos(2k-1)x = \sin 2kx - \sin(2k-2)x$$, which telescopes to $$\sin 2nx$$.

With $$2n = 6$$ (so $$n = 3$$), we get:

$$\frac{\sin 6x}{\sin x} = 2\cos x + 2\cos 3x + 2\cos 5x$$

Now we integrate:

$$\int_0^{\pi/2} \frac{60\sin 6x}{\sin x}\,dx = 60\int_0^{\pi/2} (2\cos x + 2\cos 3x + 2\cos 5x)\,dx$$

$$= 120\left[\sin x\right]_0^{\pi/2} + 120\left[\frac{\sin 3x}{3}\right]_0^{\pi/2} + 120\left[\frac{\sin 5x}{5}\right]_0^{\pi/2}$$

Evaluating each term: $$\sin(\pi/2) = 1$$, $$\sin(3\pi/2) = -1$$, and $$\sin(5\pi/2) = 1$$. So:

$$= 120(1) + 120 \cdot \frac{-1}{3} + 120 \cdot \frac{1}{5}$$

$$= 120 - 40 + 24 = 104$$

Hence, the correct answer is 104.