Q70 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\int_\limits{-1}^1 \dfrac{d}{d x}\left(\tan ^{-1} \dfrac{1}{x}\right) d x \text { is }$
- A$\dfrac{\pi}{2}$
- Correct answerB$-\dfrac{\pi}{2}$
- C$\dfrac{\pi}{4}$
- D$-\dfrac{\pi}{4}$
Solution
Let $f(x) = \tan^{-1} \left( \dfrac{1}{x} \right)$. The integral is $\int_{-1}^{1} f'(x) dx$.
The function $f(x)$ is discontinuous at $x = 0$. Therefore, the integral must be evaluated as the sum of two improper integrals:
$\int_{-1}^{0} f'(x) dx + \int_{0}^{1} f'(x) dx = \lim_{t \to 0^{-}} [f(t) - f(-1)] + \lim_{t \to 0^{+}} [f(1) - f(t)]$.
We know that $\lim_{t \to 0^{-}} \tan^{-1} \left( \dfrac{1}{t} \right) = -\dfrac{\pi}{2}$ and $\lim_{t \to 0^{+}} \tan^{-1} \left( \dfrac{1}{t} \right) = \dfrac{\pi}{2}$.
Also, $f(-1) = \tan^{-1}(-1) = -\dfrac{\pi}{4}$ and $f(1) = \tan^{-1}(1) = \dfrac{\pi}{4}$.
Substituting these values:
$\left( -\dfrac{\pi}{2} - (-\dfrac{\pi}{4}) \right) + \left( \dfrac{\pi}{4} - \dfrac{\pi}{2} \right) = \left( -\dfrac{\pi}{2} + \dfrac{\pi}{4} \right) + \left( \dfrac{\pi}{4} - \dfrac{\pi}{2} \right) = -\dfrac{\pi}{4} - \dfrac{\pi}{4} = -\dfrac{\pi}{2}$.
Answer: $-\dfrac{\pi}{2}$
The function $f(x)$ is discontinuous at $x = 0$. Therefore, the integral must be evaluated as the sum of two improper integrals:
$\int_{-1}^{0} f'(x) dx + \int_{0}^{1} f'(x) dx = \lim_{t \to 0^{-}} [f(t) - f(-1)] + \lim_{t \to 0^{+}} [f(1) - f(t)]$.
We know that $\lim_{t \to 0^{-}} \tan^{-1} \left( \dfrac{1}{t} \right) = -\dfrac{\pi}{2}$ and $\lim_{t \to 0^{+}} \tan^{-1} \left( \dfrac{1}{t} \right) = \dfrac{\pi}{2}$.
Also, $f(-1) = \tan^{-1}(-1) = -\dfrac{\pi}{4}$ and $f(1) = \tan^{-1}(1) = \dfrac{\pi}{4}$.
Substituting these values:
$\left( -\dfrac{\pi}{2} - (-\dfrac{\pi}{4}) \right) + \left( \dfrac{\pi}{4} - \dfrac{\pi}{2} \right) = \left( -\dfrac{\pi}{2} + \dfrac{\pi}{4} \right) + \left( \dfrac{\pi}{4} - \dfrac{\pi}{2} \right) = -\dfrac{\pi}{4} - \dfrac{\pi}{4} = -\dfrac{\pi}{2}$.
Answer: $-\dfrac{\pi}{2}$
Hence, the correct answer is option: 'B'.