`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`

int_(0)^(Proving)3x^(2)sin((1)/(x))-x cos((1)/(x))dx=(32sqrt(2))/(pi^(3))

The tangent to the graph of the function y=f(x) at the point with abscissae x=1, x=2, x=3 make angles pi/6,pi/3 and pi/4 respectively. The value of int_1^3f\'(x)f\'\'(x)dx+int_2^3f\'\'(x)dx is (A) (4-3sqrt(3))/3 (B) (4sqrt(3)-1)/(3sqrt(3)) (C) (4-3sqrt(3))/2 (D) (3sqrt(3)-1)/2

int_(0)^(3)(3x+1)/(x^(2)+9)dx=(pi)/(12)+log(2sqrt(2))(b)(pi)/(2)+log(2sqrt(2))quad (c)(pi)/(6)+log(2sqrt(2))(d)(pi)/(3)+log(2sqrt(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))

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)))

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)^((pi)/(4))sqrt(1+sin2x)dx

int_(-pi/2)^( pi/2)sqrt(cos x-cos^(3)x)dx= (A) 3/4 (B) 4/3 (C) 3/2 (D) 2/3

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)