Question

Evaluate: \( \int_{0}^{\pi} \frac{\sin 2px}{\sin x} dx \), \( p \in N \).

Hint:

Using reduction formulas or recurrence relations can be helpful for integrals involving parameters. In this case, by relating \( I_p \) to \( I_{p-1} \), we found a pattern. Remember the values of trigonometric functions at multiples of \( \pi \): \( \sin(n\pi) = 0 \) for any integer \( n \).

Answer 1

We need to evaluate the integral \( I_p = \int_{0}^{\pi} \frac{\sin 2px}{\sin x} dx \), where \( p \in N \). We can use the identity \( \sin(A+B) - \sin(A-B) = 2 \cos A \sin B \). Let \( A = (2p-1)x \) and \( B = x \), then \( A+B = 2px \) and \( A-B = (2p-2)x \). So, \( \sin 2px - \sin (2p-2)x = 2 \cos (2p-1)x \sin x \). Therefore, \( \frac{\sin 2px}{\sin x} = \frac{\sin (2p-2)x}{\sin x} + 2 \cos (2p-1)x \). Integrating from \( 0 \) to \( \pi \): $$I_p = \int_{0}^{\pi} \frac{\sin (2p-2)x}{\sin x} dx + \int_{0}^{\pi} 2 \cos (2p-1)x dx$$ $$I_p = I_{p-1} + \left[ \frac{2 \sin (2p-1)x}{2p-1} \right]_{0}^{\pi}$$ $$I_p = I_{p-1} + \frac{2 \sin (2p-1)\pi}{2p-1} - \frac{2 \sin 0}{2p-1}$$ Since \( p \in N \), \( 2p-1 \) is an integer, so \( \sin (2p-1)\pi = 0 \). Also, \( \sin 0 = 0 \). Thus, \( I_p = I_{p-1} \). This shows that the value of the integral is independent of \( p \). Let's evaluate for \( p = 1 \): $$I_1 = \int_{0}^{\pi} \frac{\sin 2x}{\sin x} dx = \int_{0}^{\pi} \frac{2 \sin x \cos x}{\sin x} dx = \int_{0}^{\pi} 2 \cos x dx$$ $$= [2 \sin x]_{0}^{\pi} = 2 \sin \pi - 2 \sin 0 = 2(0) - 2(0) = 0$$ Since \( I_p = I_{p-1} \) and \( I_1 = 0 \), it follows that \( I_p = 0 \) for all \( p \in N \).