The value of $\int\limits_{-1}^1 \frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right) d x$, is
Answer & explanation
Correct answer: option 3
We have,
$\frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right)=\frac{d}{d x}\left(\cot ^{-1} x\right)=-\frac{1}{1+x^2}$
∴ $\int\limits_{-1}^1 \frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right) d x=\int\limits_{-1}^1-\frac{1}{1+x^2} d x=-2 \int\limits_0^1 \frac{1}{1+x^2} d x$
$\Rightarrow \int\limits_{-1}^1 \frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right) d x=-2\left[\tan ^{-1} x\right]_0^1=-2\left(\frac{\pi}{4}\right)=-\frac{\pi}{2}$