If `(dy)/(dx)`+y sec x=tan x," then `(sqrt(2)`+1)`y((pi)/(4))` - y(0)=, (A) `sqrt(2)` - `(pi)/(4)` (B) `sqrt(2)` + `(pi)/(4)` (C) `sqrt(2)`-`(pi)/(2) `(D) `sqrt(2)`+`(pi)/(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

lim_(x rarr1^(-))(sqrt(pi)-sqrt(2sin^(-1)x))/(sqrt(1-x))=(A)sqrt((2)/(pi))(B)sqrt((pi)/(2))(C)(1)/(pi)(D)sqrt((1)/(pi))

If sin^(-1)x+sin^(-1)y=(pi)/(2) and sin2x=cos2y, then (a)x=(pi)/(8)+sqrt((1)/(2)-(pi^(2))/(64))(b)y=sqrt((1)/(2)-(pi^(2))/(64))-(pi)/(12)(c)x=(pi)/(12)+sqrt((1)/(2)-(pi^(2))/(64))(d)y=sqrt((1)/(2)-(pi^(2))/(64))-(pi)/(8)

If y=cos(sinx^2) ,then at x=sqrt((pi)/(2)) , (dy)/(dx)=

int_(0)^(pi//2)(dx)/((1+sqrt(tanx)))=(pi)/(4)

Prove that: int_(0)^((pi)/(4))(sqrt(tan x)+sqrt(cot x)dx=sqrt(2)*(pi)/(2)

If int_(0)^(oo)e^(-x^(2))dx=(sqrt(pi))/(2), then int_(0)^(oo)e^(-ax^(2))dx where a>0 is: (A)(sqrt(pi))/(2) (B) (sqrt(pi))/(2a)(C)2(sqrt(pi))/(a) (D) (1)/(2)(sqrt((pi)/(a)))

Let f(x) be a non-negative continuous function such that the area bounded by the curve y=f(x), the x -axis,and the ordinates x=(pi)/(4) and x=beta>(pi)/(4) is beta sin beta+(pi)/(4)cos beta+sqrt(2)beta Then f'((pi)/(2)) is ((pi)/(2)-sqrt(2)-1)( b) ((pi)/(4)+sqrt(2)-1)-(pi)/(2)( d) (1-(pi)/(4)-sqrt(2))

If y=cos^(-1)(cos x) then (dy)/(dx) at x=(5 pi)/(4) is (i) 1( ii) -1( iii) (1)/(sqrt(2))( iv) sqrt(2) at x=(5 pi)/(4) is (i)

The value of sin^(-1){cot(sin^(-1)sqrt((2-sqrt(3))/(4))+cos^(-1)(sqrt(12))/(4)+sec^(-1)sqrt(2))} is equal to (pi)/(4) (b) (pi)/(2)(c)0(d)-(pi)/(2)