`intdx/(sin^4x+cos^4x)`

intdx/(sin^4xcos^4x)

intdx/(sin^2x+cosx)

intdx/(4sin^2x+9cos^2x)

intdx/(4sin^2x+9cos^2x)

If f(x) =(cos^2x +sin^4x)/(sin^2x +cos^4x) for x in R then f(2020) =

int_0^(pi/2) sin^4x/(sin^4x+cos^4x) dx=

(i) int_0^(pi//2) (sin^3x)/(sin^3x+cos^3x) dx (ii) int_0^(pi//2) (cos^3x)/(sin^3x+cos^3x) dx (iii) int_0^(pi//2) (sin^4x)/(sin^4x+cos^4x) dx (iv) int_0^(pi//2) (cos^5x)/(sin^5x+cos^5x) dx (v) int_0^(pi//2) (sin^5x)/(sin^5x+cos^5x) dx

Evaluate: int_0^(pi/2) (sin2x)/(sin^4x+cos^4x)dx