Maths Solve the Following Question.(2 Marks)

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

$\int_0^{\frac{\pi}{2}} \frac{1}{1+\cos x} d x$
Solving the integral without limits,
$\begin{aligned} & \int \frac{1}{1+\cos x} d x \\ & =\int \frac{1}{2 \cos ^2\left(\frac{x}{2}\right)} d x \\ & =\frac{1}{2} \int \sec ^2\left(\frac{x}{2}\right) d x \\ & =\frac{1}{2}\left[\frac{\tan \left(\frac{x}{2}\right)}{\frac{1}{2}}\right]+C \\ & =\tan \left(\frac{x}{2}\right)+C\end{aligned}$
Substituting the limits,we get
$\begin{aligned} & =\left[\tan \left(\frac{x}{2}\right)\right]_0^{\frac{\pi}{2}} \\ & =\left[\tan \left(\frac{\pi}{4}\right)-\tan 0\right] \\ & =1\end{aligned}$