Question

Find the integrals (primitives):
$\rm \displaystyle\int (\sin x-\cos x)^{2} \ dx$

Answer 1

here is ur answer ☺️☺️☺️☺️

Answer 2

We know that

${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ \\ ( {sin \: x - cos \: x)}^{2} = {cos}^{2}x + {sin}^{2}x - 2sin \: x \: cos \: x \\ \\ ( {sin \: x - cos \: x)}^{2} = 1 - 2sin \: x \: cos \: x \\ \\ ( {sin \: x - cos \: x)}^{2} = 1 - sin \: 2x$
$\int (\sin x-\cos x)^{2} \ dx = \int 1 - \: sin \: 2x\\ \\ = \int1.dx - \int \: sin \: 2x \: dx \\ \\ = x - \int \: sin \: 2x \: dx \\ \\ let \: 2x = t \\ \\ dx = \frac{1}{2} dt \\ \\ = x - \frac{1}{2} \int \: sin \: t \: dt\\ \\ = x - \frac{1}{2} ( - cos \: t) + c \\ \\ \int (\sin x-\cos x)^{2} \ dx = x + \frac{1}{2} cos \: 2x + c$
Hope it helps you.