`(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)`

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))

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) 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 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 f(x)=sqrt(1+cos^(2)(x^(2))), then f'((sqrt(pi))/(2)) is (sqrt(pi))/(6)(b)-sqrt(pi/6)1/sqrt(6)(d)pi/sqrt(6)

tan^(-1)((x)/(y))-tan^(-1)((x-y)/(x+y)) is (A) (pi)/(2)(B)(pi)/(3)(C)(pi)/(4)(D)(pi)/(4) or 3(pi)/(4) is (A) (pi)/(2)(B)

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

int_(0)^((pi)/(2))(dx)/(1+sqrt(tan x))=int_(0)^((pi)/(2))(dx)/(1+sqrt(cot x))=(pi)/(4)

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