Definite Integration — Maths STD 12 Science — Question

 (b) : Let $I=\int_0^{\pi / 4} x \sec ^2 x d x$
Integrating by parts, we get
$
\begin{aligned}
I & =[x \tan x]_0^{\pi / 4}-\int_0^{\pi / 4} \tan x d x \\
& =[x \tan x]_0^{\pi / 4}+[\log |\cos x|]_0^{\pi / 4} \\
& =\frac{\pi}{4} \tan \frac{\pi}{4}-0+\log \left|\cos \frac{\pi}{4}\right|-\log |\cos 0| \\
& =\frac{\pi}{4} \times 1+\log \left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}+\log (1)-\log \sqrt{2}=\frac{\pi}{4}-\log (\sqrt{2})
\end{aligned}
$