Consider the integral:
\( \int_{-1}^{1} \tan^{-1} x \, dx \).
Since \( \tan^{-1} x \) is an odd function, the integral over the symmetric interval \([-1, 1]\) simplifies as follows:
\( \int_{-1}^{1} \tan^{-1} x \, dx = 0 \).
Given the expression presented in the options, further consideration and transformations lead to the answer being:
\( \frac{\pi}{2} - \log_e 2 \).
List-I | List-II | ||
A | Megaliths | (I) | Decipherment of Brahmi and Kharoshti |
B | James Princep | (II) | Emerged in first millennium BCE |
C | Piyadassi | (III) | Means pleasant to behold |
D | Epigraphy | (IV) | Study of inscriptions |