Definite Integration
Integral Calculus-2
star_batch_jee_advanced_2025
Grade None
Question:
Let $I_n = \int_{-\pi}^{\pi} \frac{\sin nx}{\sin x} dx, n \in \mathbb{N}$ then:
$I_{n+2} = I_n$
$\sum_{m=1}^{20} I_{2m+1} = 20\pi$
$I_{2m} = 0$ where $m = 1,2,3,......$
$I_{n+1} = I_n$
Step-by-Step Solution
Key Concept: The Dirichlet kernel representation $\frac{\sin nx}{\sin x}$ has symmetry properties that make even-indexed integrals vanish and establish a period-2 recurrence relation.
We use the identity $\frac{\sin nx}{\sin x} = U_{n-1}(\cos x)$ where $U_{n-1}$ is the Chebyshev polynomial of the second kind. For odd $n$, $\frac{\sin nx}{\sin x}$ is an even function, so $I_n = 2\int_0^{\pi} \frac{\sin nx}{\sin x} dx$. For even $n$, the integrand is odd about $x = \pi/2$, making $I_{2m} = 0$. Using the recursion $\sin nx = 2\cos x \sin(n-1)x - \sin(n-2)x$, we get $\frac{\sin nx}{\sin x} = 2\cos x \cdot \frac{\sin(n-1)x}{\sin x} - \frac{\sin(n-2)x}{\sin x}$, which after integration gives $I_{n+2} = I_n$ (option 1). For odd indices $I_{2m+1} = \pi$ for each $m \geq 1$, so $\sum_{m=1}^{20} I_{2m+1} = 20\pi$ (option 2).
Correct Answer: 1,2,3