Question

Find the integral of $\int [\sin x-\cos x]dx$ from o to $\frac{\pi }{2}$.

Answer 1

${\int }_0^{\pi /2}[\sin x-\cos x]dx$
$={\int }_0^{\pi /2}\left[\sqrt{2}(\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x)\right]dx$
$={\int }_0^{\pi /2}\left[\sqrt{2}\sin (x-\frac{\pi }{4})\right]dx$
$={\int }_0^{\pi /4}-1dx+{\int }_{\pi /4}^{\pi /2}0dx$
$=-{\left[x\right]}_0^{\pi /4}$
$=-\frac{\pi }{4}$