Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade None
Question:
If $I_n = \int \cot^n x dx$, then $I_0 + I_1 + 2(I_2 + I_3 + ... + I_k) + I_9 + I_{10} =$
-\sum_{k=1}^{9} \frac{\cot^k x}{k}
\sum_{k=1}^{9} \frac{\cot^k x}{k!}
\sum_{k=1}^{10} \frac{\cot^k x}{10}
-\sum_{k=1}^{9} k\cot^k x
Step-by-Step Solution
Key Concept: The reduction formula $I_n + I_{n-2} = -\frac{\cot^{n-1}x}{n-1}$ pairs odd and even indexed integrals to produce a telescoping sum structure.
We use the reduction formula: $I_n + I_{n-2} = -\frac{\cot^{n-1}x}{n-1}$ (derived by integrating $\cot^n x = \cot^{n-2}x(\csc^2 x - 1)$). Pairing consecutive terms: $I_0 + I_2 = -\cot x$, $I_1 + I_3 = -\frac{\cot^2 x}{2}$, ..., $I_8 + I_{10} = -\frac{\cot^9 x}{9}$, and $I_9$ stands alone. When we compute $I_0 + I_1 + 2(I_2 + I_3 + ... + I_k) + I_9 + I_{10}$, the coefficient 2 for middle terms combined with the pairing structure yields $-(\cot x + \frac{\cot^2 x}{2} + \frac{\cot^3 x}{3} + ... + \frac{\cot^9 x}{9}) = -\sum_{k=1}^{9} \frac{\cot^k x}{k}$.
Correct Answer: 1