[tex]FORMULE \:\ APLICATE \\\\\boxed{\left(a-b\right)^2=a^2-2ab+b^2}\\ \boxed{\cos ^2\left(x\right)+\sin ^2\left(x\right)=1}\\ \boxed{\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}\\\boxed{2\cos \left(x\right)\sin \left(x\right)=\sin \left(2x\right)}[/tex]
[tex]REZOLVARE\\\\\int \left(\sin \left(\frac{x}{2}\right)-\cos \left(\frac{x}{2}\right)\right)^2dx =\int \sin ^2\left(\frac{x}{2}\right)-2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)+\cos ^2\left(\frac{x}{2}\right)dx =>\\ \\\int \:-2\cos \left(\frac{x}{2}\right)\sin \left(\frac{x}{2}\right)+1dx => \\ -\int \:2\cos \left(\frac{x}{2}\right)\sin \left(\frac{x}{2}\right)dx+\int \:1dx => \\ \\\int \:2\cos \left(\frac{x}{2}\right)\sin \left(\frac{x}{2}\right)dx=-\cos \left(x\right)\\\int \:1dx=x[/tex]
[tex]De \:\ unde \:\ rezulta: \:\ -\left(-\cos \left(x\right)\right)+x => \cos \left(x\right)+C\\\\Rezultat \:\ final: \boxed{\int \left(\sin \left(\frac{x}{2}\right)-\cos \left(\frac{x}{2}\right)\right)^2dx=\cos \left(x\right)+x+C}[/tex]
