Definite Integration — Maths STD 12 Science — Question

(b) : Let $I=\int_0^{\pi / 4} \sec ^4 x d x$
$
I=\int_0^{\pi / 4} \sec ^2 x \cdot \sec ^2 x \cdot d x=\int_0^{\pi / 4}\left(1+\tan ^2 x\right) \sec ^2 x d x
$
Let $z=\tan x$, at $x=0, z=0$ and at $x=\pi / 4, z=1$
$
\begin{aligned}
& \frac{d z}{d x}=\sec ^2 x \Rightarrow d z=\sec ^2 x d x \\
& I=\int_0^1\left(1+z^2\right) d z=\left[z+\frac{z^3}{3}\right]_0^1=1+\frac{1}{3}=\frac{4}{3}
\end{aligned}
$