" b6."(sin^(2)x)/(1+cos^(2)x)

Simplify: (sin^(6)x+cos^(6)x)/(sin^(2)x cos^(2)x)

Evaluate int (sin^(6) x + cos^(6)x)/(sin^(2)x cos^(2)x) dx

Evaluate int (sin^(6) x + cos^(6)x )/(sin^(2) x cos^(2) x) dx

Evaluate int (sin^(6) x + cos^(6)x)/(sin^(2)x cos^(2)x) dx

Integrate: (sin^(6)x+cos^(6)x)/(sin^(2)x cos^(2)x)

" if "A=[[cos^(2)x,sin^(2)x],[-sin^(2)x,-cos^(2)x]]" and "B=[[sin^(2)x,cos^(2)x],[-cos^(2)x,-sin^(2)x]]" then find "A+B"

Ltquad x rarr0 (1-cos ^ (2) (sin x) -cos (sin ^ (2) x) + cos ^ (2) (sin x) cos (sin ^ (2) x)) / (x ^ ( 6)) =

int(sin^(6)x+cos^(6)x+3sin^(2)x cos^(2)x)dx is equal to

Evaluate :int(sin^(6)x+cos^(6)x)/(sin^(2)x*cos^(2)x)dx

Evaluate : int(sin^(6)x+cos^(6)x)/(sin^(2)x.cos^(2)x)dx.