7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

$\text { Put } x \quad =\frac{1}{ t } dx =-\frac{1}{ t ^2} dt$

$I =-\int \limits_2^{1 / 2} \frac{\tan ^{-1} \frac{1}{ t }}{\frac{1}{ t }} \cdot \frac{1}{ t ^2} dt =-\int \limits_2^{1 / 2} \frac{\tan ^{-1} \frac{1}{ t }}{ t } dt$

$I =\int \limits_{1 / 2}^2 \frac{\cot ^{-1} t }{ t } dt =\int \limits_{1 / 2}^2 \frac{\cot ^{-1} x }{ x } dx \ldots \ldots(ii)$

Add both equation

$2 I =\int \limits_{1 / 2}^2 \frac{\tan ^{-1} x +\cot ^{-1} x }{ x } dx =\frac{\pi}{2} \int \limits_{1 / 2}^2 \frac{ dx }{ x }=\frac{\pi}{2}(\ell \ln 2)_{1 / 2}^2$

$=\frac{\pi}{2}\left(\ell n 2-\ell n \frac{1}{2}\right)=\pi \ell n 2$

$I =\frac{\pi}{2} \ell n 2$