`sin^(-1)(sin5)> x^2-4x` hold if `x=2-sqrt(9-2pi)` `x=2+sqrt(9-2pi)` `x >2+sqrt(9-2pi)` `x in (2-sqrt(9-2pi),2+sqrt(9-2pi))`

sin^(-1)(sin5)> x^2-4x hold if (a) x=2-sqrt(9-2pi) (b) x=2+sqrt(9-2pi) (c) x >2+sqrt(9-2pi) (d) x in (2-sqrt(9-2pi),2+sqrt(9-2pi))

lim_(xrarr1^(-)) (sqrtpi-sqrt(2sin^-1x))/sqrt(1-x)= (A) sqrt(2/pi) (B) sqrt(pi/2) (C) 1/pi (D) sqrt(1/pi)

A solution of sin^-1 (1) -sin^-1 (sqrt(3)/x^2)- pi/6 =0 is (A) x=-sqrt(2) (B) x=sqrt(2) (C) x=2 (D) x= 1/sqrt(2)

int_0^(4/pi) (3x^2sin(1/x)-xcos(1/x))dx= (A) (8sqrt(2))/pi^3 (B) (32sqrt(2))/pi^3 (C) (24sqrt(2))/pi^3 (D) sqrt(2048)/pi^3

If sin^(-1) x + sin^(-1) y = pi/2 , prove that x sqrt(1-y^2) + y sqrt(1-x^2) =1 .

Prove that : int_0^(pi/2)sqrt(1-sin2x)dx=2(sqrt(2)-1)

int_0^3(3x+1)/(x^2+9)dx = (pi^)/(12)+log(2sqrt(2)) (b) (pi^)/2+log(2sqrt(2)) (c) (pi^)/6+log(2sqrt(2)) (d) (pi^)/3+log(2sqrt(2))

The integral int_(0)^(pi)sqrt(1+4"sin"^(2)x/2-4"sin"x/2)dx is equals to (a) pi-4 (b) (2pi)/3-4-sqrt(3) (c) (2pi)/3-4-sqrt(3) (d) 4sqrt(3)-4-(pi)/3

The area bounded by f(x)=sin^(-1)(sinx) and g(x)=pi/2-sqrt(pi^(2)/2-(x-pi/2)^(2)) is

sin^(-1)(x^2/4+y^2/9)+cos^(-1)(x/(2sqrt2)+y/(3sqrt2)-2)