Let $f=\int_0^{\frac{\pi}{2}} \frac{\sum_{n=0}^4 \frac{\pi \pi}{4}+x}{\cos x+\sin x} d r$
$\operatorname{since} \sum_{x=1}^4\left(\frac{\operatorname{ex}}{4}+x\right)=(0+1+2+3+4) \frac{\pi}{4}+(1+1+1+1+1) x$ $=\frac{5 \pi}{2}+5 x=5\left(\frac{\pi}{2}+x\right)$
Hence, $f=\int_0^{\frac{\pi}{2}} \frac{5\left(\frac{\pi}{2}+x\right) d x}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right)}$
$\Rightarrow I=\frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}+x\right) \sec \left(x-\frac{\pi}{4}\right) d x$ ...(i)
Replacing $(x)$ by $\left(\frac{\pi}{2}-x\right)$ we get
$$
I=\frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}}\left[\frac{\pi}{2}+\left(\frac{\pi}{2}-x\right)\right] \sec \left[\left(\frac{\pi}{2}-x\right)-\frac{\pi}{4}\right] d x
$$
$I=\frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}}[\pi-x] \sec \left(x-\frac{\pi}{4}\right) d x$ ...(ii)
$\{\because \sec (-\alpha)=\sec \alpha\}$
Adding equations (i) \& (ii), we get
$$
2 I=\frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}}\left(\frac{3 \pi}{2}\right) \sec \left(x-\frac{\pi}{4}\right) d x
$$
$\begin{aligned} & \Rightarrow I=\frac{15 \pi}{4 \sqrt{2}}\left[\log \left|\sec \left(x-\frac{\pi}{4}\right)+\tan \left(x-\frac{\pi}{4}\right)\right|\right]_0^{\frac{\pi}{2}} \\ & \Rightarrow I=\frac{15 \pi}{4 \sqrt{2}} \log \left|\frac{\sqrt{2}+1}{\sqrt{2}-1}\right|=\frac{15 \pi}{4 \sqrt{2}} \log \left|(\sqrt{2}+1)^2\right| \\ & \Rightarrow I=\frac{15 \pi}{2 \sqrt{2}} \log |\sqrt{2}+1|\end{aligned}$