Definite Integration
King's Rule — Inverse Cot Integral
nta_pyq_2026_jan
Grade 12
Question:
If $\displaystyle\int_0^1 4\cot^{-1}(1-2x+4x^2)\,dx = a\tan^{-1}(2)-\log_e(5)$, where $a,b\in\mathbf{N}$, then $(2a+b)$ is equal to _____
Step-by-Step Solution
Key Concept: Write $\cot^{-1}(1-2x+4x^2)=\cot^{-1}(2x-1)-\cot^{-1}(2x)$ (using identity). Apply King's rule $x\to1-x$ and add to get $2I=\int_0^1(\pi-2\cot^{-1}(2x))dx$.
$a=4$, $b=1$. $2a+b=9$.
Correct Answer: 9