$\int\limits_0^{\pi / 2} \frac{a \sin x+b \cos x}{\sin x+\cos x} d x=$

$(a+b) \frac{\pi}{4}$

$I=\int\limits_0^{\pi / 2} \frac{a \sin x+b \cos x}{\sin x+\cos x} d x$

$=\int\limits_0^{\pi / 2} \frac{a \sin \left(\frac{\pi}{2}-x\right)+b \cos \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x$

$\Rightarrow I=\int\limits_0^{\pi / 2} \frac{a \cos x+b \sin x}{\cos x+\sin x} d x$

(1) + (2) gives

$\Rightarrow 2 I=\int\limits_0^{\pi / 2} \frac{(a+b)(\sin x+\cos x}{\sin x+\cos x} d x=(a+b) \int\limits_0^{\pi / 2} d x$

$\Rightarrow I=(a+b) \frac{\pi}{4}$

Hence (4) is the correct answer.