Definite Integration — Maths STD 12 Science — Question

(b) : We have, $\int_0^1 x \cdot \tan ^{-1} x d x$
$
\begin{aligned}
& =\left(\tan ^{-1} x \cdot \frac{x^2}{2}\right)_0^1-\int_0^1 \frac{1}{1+x^2} \cdot \frac{x^2}{2} d x \\
& =\left(\frac{\pi}{4} \cdot \frac{1}{2}-0\right)-\frac{1}{2} \int_0^1 \frac{1+x^2-1}{1+x^2} d x \\
& =\frac{\pi}{8}-\frac{1}{2} \int_0^1\left(1-\frac{1}{1+x^2}\right) d x=\frac{\pi}{8}-\frac{1}{2}\left[x-\tan ^{-1} x\right]_0^1 \\
& =\frac{\pi}{8}-\frac{1}{2}\left[\left(1-\frac{\pi}{4}\right)-(0)\right]=\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}=\frac{\pi}{4}-\frac{1}{2}
\end{aligned}
$