`int(sqrt(sin^4x+cos^4x))/(sin^3xcosx)dx , x in (0,pi/2)`

int(sqrt(sin^(4)x+cos^(4)x))/(sin^(3)x cos x)dx,x in(0,(pi)/(2))

int_0^pi sin^4x cos^4x dx

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

int_ (0) ^ ((pi) / (2)) (sin x) / (sin x + cos x) dx = int_ (0) ^ ((pi) / (2)) (cos x) / (sin x + cos x) dx = int_ (0) ^ ((pi) / (2)) (dx) / (1 + cot x) = int_ (0) ^ ((pi) / (2)) (dx) / ( 1 + time x) = (pi) / (4)

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

Prove that : int_(0)^(pi//2) (x sin x cos x)/(sin^(4) x+ cos^(4)x)dx =(pi)/(16)

int_(0)^( pi/2)(sin^(4)x)/(sin^(4)x+cos^(4)x)dx