$\underset{0}{\overset{\pi /2}{\int }}\frac{200\sin x+100\cos x}{\sin x+\cos x}dx=$

The given integral is $I={\int }_0^{\frac{\pi }{2}}\frac{200\sin x+100\cos x}{\sin x+\cos x}dx$

Use the property that ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx$

Hence, 

$\begin{array}{rcl}I+I& =& {\int }_0^{\frac{\pi }{2}}\frac{200\sin x+100\cos x}{\sin x+\cos x}dx+{\int }_0^{\frac{\pi }{2}}\frac{200\cos x+100\sin x}{\cos x+\sin x}dx\\ & =& {\int }_0^{\frac{\pi }{2}}\frac{300\sin x+300\cos x}{\sin x+\cos x}dx\\ & =& {\int }_0^{\frac{\pi }{2}}300dx\\ & =& 300\frac{\pi }{2}\\ & =& 150\pi \end{array}$

Therefore, $I=75\pi$