$\displaystyle\int^{\pi/4}_0\sec^4xdx$.

We are given $I=\displaystyle \int_{0}^{\pi/4}{\sec^4x dx}$

$I=\displaystyle \int_{0}^{\pi/4}{\sec^{2}.\sec^2x dx}$ ( $\therefore$ we know that $\sec^2\theta=1+\tan^2\theta$)

$I=\displaystyle \int_{0}^{\pi/4}{\sec^2x.(1+\tan^2x) dx}$

Let, take $\tan x=t\Rightarrow \sec^2xdx=dt$

( differentiating both sides)

now, when $x=0\Rightarrow t=\tan 0=0$

& when $x=\pi/4\Rightarrow t=\tan \pi/4=1$

$I=\displaystyle \int_{0}^{1}{(1+t^2)dt}=\displaystyle \int_{0}^{1}{dt}+\displaystyle \int_{0}^{1}{t^2 dt}$

$=[t]_{0}^{1}+\left[\dfrac{t^3}{3}\right]_{0}^{1}$

$=(1-0)+(1/3-0)$

$I=1+\dfrac{1}{3}=\dfrac{4}{3}$