Definite Integration
Symmetric integral with inverse trig
MJAT_TS3_P2
Grade 12
Question:
Let $\displaystyle\int_0^a \frac{\cot^{-1}(e^{-x}-1)+\cot^{-1}(e^x+1)}{\tan^{-1}x+\cot^{-1}x}\cdot\frac{x^2}{x^2+1}\,dx = \frac{\pi}{p}\ln q, \quad (a>0)$
where $p,q\in\mathbb{N}$. Then find the minimum value of $(p+q)$.
Step-by-Step Solution
Key Concept: Note $\cot^{-1}(e^{-x}-1)+\cot^{-1}(e^x+1) = \cot^{-1}(e^{-x}-1)+\tan^{-1}\!\left(\frac{1}{e^x+1}\right)$. Use the identity $\cot^{-1}t + \tan^{-1}t = \pi/2$ and the symmetry $e^{-x}-1$ vs $e^x+1$. Also $\tan^{-1}x+\cot^{-1}x=\pi/2$.
The integral evaluates to $\frac{\pi}{2}\ln 2$. With $p=2$, $q=2$: $p+q=\mathbf{4}$.
Correct Answer: 4