If `f(x)=2sin^(-1)sqrt(1-x)+sin^(-1)(2sqrt(x(1-x)))` , where `x in (0,1/2),t h e nf^(prime)(x)` is (a)`2/(sqrt(x(1-x)))` (b) zero (c)`-2/(sqrt(x(1-x)))` (d) `pi`

If f(x) = 2 sin^(-1) sqrt(1-x) + sin^(-1)(2 sqrt(x (1-x))) where x in (0, 1/2) , then f'(x) has the value equal to (i) 2/(xsqrt(1-x)) (ii) 0 (iii) -2/(xsqrt(1-x)) (iv) pi

int(e^x[1+sqrt(1-x^2)sin^-1x])/sqrt(1-x^2)dx

(d)/(dx)[sin^(-1)(x sqrt(1-x)-sqrt(x)sqrt(1-x^(2)))] is

int_(0)^(1)sin^(-1)(x sqrt(1-x)-sqrt(x)sqrt(1-x^(2)))dx

tan^(-1)sqrt((1-x)/(1+x))+sin^(-1)2x sqrt(1-x^(2))=(5 pi)/(12) if x=

If f(x)=(sin^(-1)x+tan^(-1)x)/(pi)+2sqrt(x), then the range of f(x) is

If x in[-1/2,1] then sin^(-1)(sqrt(3)/(2)x-1/2sqrt(1-x^(2)))

If f(x)=(sin^(-1)x)/(sqrt(1-x^(2))),then(1-x^(2))f'(x)-xf(x)=

(sin^(-1)sqrt(x)-cos^(-1)sqrt(x))/(sin^(-1)sqrt(x)+cos^(-1)sqrt(x)),x in[0,1]