Answer
$\dfrac{\pi^2}{32}+\dfrac{\sqrt 2}{2}-1$
Work Step by Step
Consider $f(x)=\int_{0}^{\pi/4} (x-\sin x) dx$
Use formula such as: $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$
This implies that $[\dfrac{x^{1+1}}{1+1}-\cos x]_0^{\pi/4}=\dfrac{1}{2}(\dfrac{\pi}{4})^2-(\cos (\dfrac{\pi}{4})-\cos 0)=\dfrac{\pi^2}{36}-(\cos (\dfrac{\pi}{4})-1)$
Hence, $f(x)=\int_{0}^{(\pi/4)} (x-\sin x) dx=\dfrac{\pi^2}{32}+\dfrac{\sqrt 2}{2}-1$
