" Dif ferentiate "cos x^(3)*sin^(2)x^(5)

The value of int(cos^(3)x+cos^(5))/(sin^(2)x+sin^(5)x)dx

int (sin ^ (2) x cos ^ (2) x) / ((sin ^ (5) x + cos ^ (3) x sin ^ (2) x + sin ^ (3) x cos ^ (2) x + cos ^ (5) x) ^ (2)) dx

int(sin^(2)x cos^(2)x)/((sin^(5)x+cos^(3)x sin^(2)x+sin^(3)x cos^(2)x+cos^(5)x)^(2))backslash dx

int(sin^(2)x cos^(2)x)/((sin^(5)x+cos^(3)x sin^(2)x+sin^(3)x cos^(2)x+cos^(5)x)^(2))backslash dx

2cosx-cos3x-cos5x= ............... A) 16 cos ^(3) x sin ^(2) x B) 16 sin^(2) x cos ^(2) x C) 4 cos ^(2) x sin ^(2) x D) 4 sin ^(2) x cos ^(2)x

" If "sin^(2)x+sin x cos x-6cos^(2)x=0" and "-(pi)/(2)ltxlt0," then value of "cos2x" is " -3/5 3/5 -4/5 4/5

(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 ((cos x) ^ (4)) / ((sin x) ^ (3) ((sin x) ^ (5) + (cos x) ^ (5)) ^ ((3) / (5)) ) dx

int(3sin x-2)cos x(dx)/(5-cos^(2)x-4sin x)