Definite Integration — Maths STD 12 Science — Question

$I=\int_0^{\frac{\pi}{4}} \sec ^4 x d x$
$=\int_0^{\frac{\pi}{4}} \sec ^2 x \cdot \sec ^2 x d x$
$=\int_0^{\frac{\pi}{4}}\left(1+\tan ^2 x\right) \sec ^2 x d x$
Put $\tan x=t \Rightarrow \sec ^2 x d x=d t$
When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=1$
$\therefore I=\int_0^1\left(1+t^2\right) d t$
$=\left[t+\frac{t^3}{3}\right]_0^1$
$=1+\frac{1}{3}=\frac{4}{3}$