Answer
$\dfrac{\pi^2}{36}+\dfrac{\sqrt 2}{2}-1$
Work Step by Step
We know that $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$
where $C$ is a constant of proportionality
and $\int \sin x dx=\cos x +c$
Thus:
$\int_{0}^{\pi/4} (x-\sin x) dx=[\dfrac{x^{1+1}}{1+1}-\cos x]_0^{\pi/4}$
or, $=\dfrac{1}{2}((\pi/4)^2-0)-(\cos (\pi/4)-\cos 0)$
or $=\dfrac{\pi^2}{36}+\dfrac{\sqrt 2}{2}-1$
