" (1) "cos^(2)(x)/(2)

int[(sin^(2).(x)/(2)-cos^(2).(x)/(2))^(2)+4sin^(2).(x)/(2).cos^(2).(x)/(2)]dx=

If cos2x+2cos x=1 then (2-cos^(2)x)sin^(2)x is equal to

sin x = 2 ^ (n) * cos (x) / (2) * cos (x) / (2 ^ (2)) * cos (x) / (2 ^ (3)) * ... * cos ( x) / (2 ^ (n)) * sin (x) / (2 ^ (n))

Solve: 5cos2x+2cos^(2)((x)/(2))+1=0

Evaluate lim_(xto0) (8)/(x^(8)){1-"cos"(x^(2))/(2)-"cos"(x^(2))/(4)+"cos"(x^(2))/(2)"cos"(x^(2))/(4)}.

Evaluate lim_(xto0) (8)/(x^(8)){1-"cos"(x^(2))/(2)-"cos"(x^(2))/(4)+"cos"(x^(2))/(2)"cos"(x^(2))/(4)}.

int sin^(2)(x/2)-cos^(2)(x/2)dx

Simplify [{:(cos^(2)x,sin^(2)x),(sin^(2)x,cos^(2)x):}]+[{:(sin^(2)x,cos^(2)x),(cos^(2)x,sin^(2)x):}]